The volume of a spherical balloon is 20⅚πm³. Find the radius of the balloon.
Who can help me to answer this question? Please and thank you very much .​

Angle <QAB is =15° because the opposite angles of an isosceles triangle are equal.

The length of the straight line AB = 80cm

The angle at a point = 360°

Angle AQB= 360 - 210° = 150

But the angle that makes up a triangle= 180°

180-150= 30°

But <QAB = <QBA because triangle AQB is an isosceles triangle.

30/2 = 15°

To calculate the length of the straight line the following is carried out using the sine laws.

a/ sina, = b sinb

a= 8cm, sin a { sin 15)

b= ? , sin B = 150

make b the subject formula;

8/sin15= b/sin 150

b= 8 × sin 150/sin 15

b= 80cm

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The rational expression that results into x-4/2(x+4) is x²-16/2x+8 * x+4/x²+8x+16

Rational expressions are written in the form a/b where a and b are integers.

According to the question, we need to determine the expression that is equal to x-4/2(x+4)

From the fourth option;

x²-16/2x+8 * x+4/x²+8x+16

Factorize the quadratic part to have;

x²-16/2x+8 * x²+8x+16/x+4

(x+4)(x-4)/2(x+4) * x+4/(x+4)^2

Cancel out the common expression to have;

x-4 * 1/2(x+4)

x-4/2(x+4)

Hence the rational expression that results into x-4/2(x+4) is x²-16/2x+8 * x+4/x²+8x+16

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Answer:

$6455.

Step-by-step explanation:

The formula for this investment is:

A = P(1 + r/4)^4t      where P = amount deposited ,  A amount after t years,

                                r = rate (as a decimal fraction).

So we have:

7160.06 = P(1 + 0.026/4)^16

7160.06 = P * 1.109227

P = 7160.06 / 1.109227

  =  $6455.

The answer is C. 3√5.

Answer:

Step-by-step explanation:

The most basic quadratic function is f(x) = x2, whose graph appears below. ... us to graph an entire family of quadratic functions using transformations.

[tex] {\qquad\qquad\huge\underline{{\sf Answer}}} [/tex]

let's solve ~

[tex]\qquad \sf  \dashrightarrow \: \cfrac{4}{y - 6} + \cfrac{5}{y + 3} = \cfrac{7y - 4}{ {y}^{2} - 3y - 18} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \cfrac{4(y + 3) + 5(y - 6)}{(y - 6)(y + 3)} = \cfrac{7y - 4}{ {y}^{2} - 3y - 18} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \cfrac{4y + 12 + 5y - 30}{ {y}^{2} + 3y - 6y - 18 } = \cfrac{7y - 4}{ {y}^{2} - 3y - 18} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \cfrac{9y - 18}{ {y}^{2} - 3y - 18 } = \cfrac{7y - 4}{ {y}^{2} - 3y - 18} [/tex]

[tex]\qquad \sf  \dashrightarrow \: 9y - 18 = 7y - 4[/tex]

[ denominator is same, so numerator must have same value to be equal ]

[tex]\qquad \sf  \dashrightarrow \: 9y - 7y = - 4 + 18[/tex]

[tex]\qquad \sf  \dashrightarrow \: 2y = 14[/tex]

[tex]\qquad \sf  \dashrightarrow \: y = 7[/tex]

Ellipses that are represented by equations, whose eccentricities are less than 0.5 are:

To identify the ellipses:

For the standard-form equation of an ellipse: Ax^2+Bx+Cy^2+Dy+E=0

We can define:

Then the eccentricity can be shown to be:

For Eccentricity<0.5, we want:

Checking the values of p/q for the given equations, we have p/q= equations (A), (B), and (F) are of ellipses with eccentricity < 0.5.

Therefore, ellipses that are represented by equations, whose eccentricities are less than 0.5 are:

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The complete question is given below:

Identify the ellipses, represented by equations, whose eccentricities are less than 0.5.

(A) 49x2 -98x + 64yz +256y -2,831=0            

(B) 81x2 -648x +100yz +200y -6,704 =0                                      

(C) 6x2 -12x +54y2 +108y -426 =0                  

(D) 49x2 + 196x +36y2 +216y -1,244 =0                                        

(E) 4x +32x +25y2 - 250y +589 =0                

(F) 64x2 +512x +81y2 -324y -3,836 =0

[tex] {\qquad\qquad\huge\underline{{\sf Answer}}} [/tex]

Let's solve ~

By using distance formula :

[tex]\qquad \sf  \dashrightarrow \: \sqrt{(x_2 - x_1) {}^{2} + (y_2 - y_1) {}^{2} } [/tex]

[tex]\qquad \sf  \dashrightarrow \: \sqrt{(4 - ( - 1)) {}^{2} + (4 - ( - 1)) {}^{2} } [/tex]

[tex]\qquad \sf  \dashrightarrow \: \sqrt{(4 + 1) {}^{2} + (4 + 1) {}^{2} } [/tex]

[tex]\qquad \sf  \dashrightarrow \: \sqrt{2(5) {}^{2} } [/tex]

[tex]\qquad \sf  \dashrightarrow \: 5 \sqrt{2 }\: \: units[/tex]

[tex] \rm{AB= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} }[/tex]

Proof:

[tex] \rm{Let \: X'OX \: and \: YOY' \: be \: the \: z-axis \: and \: y-axis \: respectively. Then, O \: is \: the \: origin.}[/tex]

[tex] \rm{Let \: A(x_1,y_1) \: and \: B(x_2,y_2) \: be \: the \: given \: points.}[/tex]

[tex] \rm{Draw \: AL \perp \: OX, BM \perp \: OX \: and \: AN \perp BM }[/tex]

Now,

[tex] \rm{OL=x_1,OM=x_2,AL=y_1 \: and \: BM=y_2}[/tex]

[tex]\rm\therefore{AN=LM=(OM-OL)=(x_2-x_1)}[/tex]

[tex] \: \: \: \: \rm{BN=(BM-NM)=(BM-AL)=(y_2-y_1)}[/tex]

[tex] \rm{In \: right \: angled \: \triangle ANB, by \: Pythagorean \: theorem,}[/tex]

We have,

[tex] \: \: \: \: \rm{AB^2=AN^2+BN^2}[/tex]

[tex] \rm{or,AB^2=(x_2-x_1)^2+(y_2-y_1)^2}[/tex]

[tex] \rm\therefore AB= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} [/tex]

The Given question,

Find the distance of AG if A (4,4) and G (-1,-1).

Solution,

The given points are A(4,4) and G(-1,-1).

Then,

[tex] \rm{(x_1=4,y_1=4) and (x_2=-1,y_2=-1)}[/tex]

We know that,

The distance formula,

[tex] \rm{AG= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} }[/tex]

[tex] \: \: \: \: \: \: \: \: = \sqrt{ {( - 1 - 4)}^{2} + {( - 1 - 4)}^{2} } [/tex]

[tex] \: \: \: \: \: \: \: \: = \sqrt{ {( - 5)}^{2} + {( - 5)}^{2} } [/tex]

[tex] \: \: \: \: \: \: \: \: = \sqrt{25 + 25} [/tex]

[tex] \: \: \: \: \: \: \: \: = \sqrt{50} [/tex]

[tex] \: \: \: \: \: \: \: \: = \sqrt{(2)(25)} [/tex]

[tex] \: \: \: \: \: \: \: \: = \sqrt{(2)(5)(5)} [/tex]

[tex] \: \: \: \: \: \: \: \: = \sqrt{(2)( {5}^{2}) } [/tex]

[tex] \rm \: \: \: \: \: \: \: \: = 5 \sqrt{2} \: units[/tex]

Answered by:

[tex]\frak{\red{moonlight123429}}[/tex]

For the given data set

Minimum = 48

Lower quartile = 55

Median = 63.5

Upper quartile = 74

Maximum = 80

Interquartile range = 19

From the question, we are to determine the minimum, lower quartile, median, upper quartile, maximum, and interquartile range of the given data set

The given data set is

48, 50, 54, 56, 63, 63, 64, 68, 74, 74, 79, 80

Minimum = 48

Lower quartile = (54+56)/2

Lower quartile = 110/2

Lower quartile = 55

Median = (63+64)/2

Median = 127/2

Median = 63.5

Upper quartile = (74+74)/2

Upper quartile = 148/2

Upper quartile = 74

Maximum = 80

Interquartile range = Upper quartile - Lower quartile

Interquartile range = 74 - 55

Interquartile range = 19

Hence, for the given data set

Minimum = 48

Lower quartile = 55

Median = 63.5

Upper quartile = 74

Maximum = 80

Interquartile range = 19

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Answer:

Slope (m) = infinity

Step-by-step explanation:

Given points:

⇒ (9, 2) and (9, 27)

In order to find the slope of these points, we must use the slope formula.

So the formula is ⇒ y₂ - y₁ / x₂ - x₁

Plug in the given points:

⇒ 27 - 2 / 9 - 9

Solve both the top and bottom (numerator and denominator):

⇒ 25 / 0

Divide:

⇒ Undefined.

(Anything divided by zero is undefined).

Answer:

Step-by-step explanation:

The graph of the straight line [tex]y = -\frac{1}{3}x -2[/tex] is plotted. The graph is shown below.

A straight line is of the form y = mx + c, where m is the slope and c is the y-intercept.

To plot the given straight line [tex]y = -\frac{1}{3}x -2[/tex], follow the following steps:

Step 1: Substitute x = 0 in the given equation to obtain the point where the line intersects the y-axis.

y = 0 - 2

y = -2

The point at the y-axis is (0, -2).

Step 2: Substitute y = 0 in the given equation to obtain the point where the line intersects the x-axis.

[tex]0 = -\frac{1}{3}x -2\\x = -6[/tex]

The point at the x-axis is (-6, 0).

Step 3: Draw a straight line passing through both points.

Thus, the straight line [tex]y = -\frac{1}{3}x -2[/tex] is plotted.

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Answer:

x < -19/7

Step-by-step explanation:

Solve for x

2x+5  < (x+1)/4

Multiply each side by 4

4( 2x+5) <  (x+1)/4 *4

8x + 20  < x+1

Subtract x from each side

8x+20-x < x+1-x

7x + 20 < 1

Subtract 20 from each side

7x+20-20 < 1-20

7x< -19

Divide by 7

7x/7 < -19/7

x < -19/7

Answer:

[tex] \red{x < \frac{ - 19}{7} }[/tex]

Step-by-step explanation:

[tex]2x + 5 < \frac{x + 1}{4} \\ \text{use \: cross \: muliplication} \\ 4(2x + 5) < x + 1 \\ 8x + 20 < x + 1 \\ 8x - x < - 20 + 1 \\ 7x < - 19 \\ \frac{7x}{7} < \frac{ - 19}{7} \\ x < \frac{ - 19}{7} [/tex]

Step-by-step explanation:

there are 4 suit with 13 cards each

Multiply 9 with 9, and we get:

9 × 9 = 81

Multiply 10 with 10, and we get:

10 × 10 = 100

We get the final result as:

Hope this helps :)

The area of the polygon is 25.5 square units

From the figure, we have the following coordinates

A = (-2, 1)

B = (3, 2)

C = (5, -3)

D = (-1, -3)

The area of the polygon is then calculated as:

A = 0.5 * |(x1y2 - x2y1) + (x2y3 - x3y2) + (x3y4 - x4y3) + (x4y1 - x1y4)|

Substitute the known values in the above equation

Area = 0.5 * |(-2 * 2 - 3 * 1) + (3 * -3  - 5 * 2) + (5 * -3  + 1 * -3) + (-1 * 1 + 2 * -3)|

Evaluate the sum of products

Area = 0.5 * |-51|

Remove the absolute bracket

Area = 0.5 * 51

Evaluate the product

Area = 25.5

Hence, the area of the polygon is 25.5 square units

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The equivalent angle in the interval [0, 2pi] is 5pi/6.

Remember that for any given angle θ in radians, we define a family coterminal or equivalent angles as:

A = θ + n*(2pi)

Where n is a whole number.

Now we want to find an angle equivalent to 17pi/6 that is on the range between 0 and 2pi.

Then we can write:

A = 17pi/6 + n*(2pi)

If we use n = -1, then we get:

A = 17pi/6 - 2pi = 17pi/6 - 12pi/6 = 5pi/6

This is the equivalent angle in the desired range.

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Answer:

True

Step-by-step explanation:

You are taking away sand from the ground and adding sand to the bucket. There is less sand on the ground and more in the bucket making it inverse.

[tex]\textit{exponential form of a logarithm} \\\\ \log_a(b)=y \qquad \implies \qquad a^y= b \\\\[-0.35em] ~\dotfill\\\\ \begin{array}{lllllllll} \log_{\underline{10}}(10)=1&\implies &\underline{10}^1=10 ~~ \checkmark\\\\ \log_{10}(100)=2&\implies &10^2=100~~ \checkmark\\\\ \log_{10}(55)=1.5&\implies &10^{1.5}=55\implies 10^{\frac{3}{2}}=55\implies \sqrt[2]{10^3}=55\\\\ &&\sqrt{1000}\ne 55 ~~ \bigotimes \end{array}[/tex]

increments over the exponent, are not directly proportional to increments on the resulting values, namely the half-way from 10² and 10⁴ is not 10³, because 10⁴ is 100 times 10², and 10³ is simply 10 times 10².

The approximate solution to the given system of equation is (-2.71, 2.14)

From the question, we are to determine the approximate solution to the given system of linear equations.

The given equations are

y = 0.5x + 3.5

y = -2/3 x+ 1/3

From the given information, we are to show the solutions on a graph

The graph that shows the solution to the given system of equation is shown below.

The solution to the given system is the point of intersection of the lines. The coordinate of the point of intersection of the lines is (-2.71, 2.14). That is, x = -2.71, y = 2.14.

Hence, the approximate solution to the given system of equation is (-2.71, 2.14)

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There would not be enough time to reach the half marathon distance.

The distance run in a week can be represented with a linear equation. A linear equation is an equation that has a single variable raised to the power of 1.

The linear equation that would be used to represent the distance run would be in the form:

Miles run in a week = miles run the previous week - increase in miles

Increase in miles = 6 - 4.5 =  1.5 miles

Miles run in the fourth week = 1.5 + 6  = 7.5 miles

Miles run in the fifth week = 7.5 miles + 1.5 miles = 9 miles

Miles run in the sixth week = 9 + 1.5 = 10.5 miles

There would not be enough time to reach the half marathon distance because in the sixth week she would be able to run 10.50 miles

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Answer:

Solution Given:

let pitcher which is small be S medium be M and large be L and barrel be B.

By question

6S+3M+1L= 1B...... eq. 1

2S+1M+3L=1B.......eq. 2

The difference in 2L from the first to the second is 4S and 2M

Therefore, each L=2S+M

In equation 2

2S+1M+3L=1B

replacing 2S +1M by L,we get

L+3L=1B

4L=1B

Therefore, 4 large pitcher is required

Answer:

4 large pitchers

Step-by-step explanation:

Define the variables:

Create two equations with the given information and the defined variables.

Equation 1

If the barrel can be filled with 6 small pitchers, 3 medium pitchers and 1 large pitchers:

⇒ 6x + 3y + z = b

Equation 2

If the barrel can be filled with 2 small pitchers, 1 medium pitcher and 3 large pitchers:

⇒ 2x + y + 3z = b

Substitute Equation 2 into Equation 1

⇒ 6x + 3y + z = 2x + y + 3z

Subtract 6x from both sides:

⇒ 3y + z = -4x + y + 3z

Subtract y from both sides:

⇒ 2y + z = -4x + 3z

Subtract z from both sides:

⇒ 2y = -4x + 2z

Divide both sides by 2:

⇒ y = -2x + z

Substitute the found expression for y into Equation 1:

⇒ 6x + 3(-2x + z) + z = b

⇒ 6x - 6x + 3z + z = b

⇒ 4z = b

Therefore, 4 large pitchers are needed to fill the barrel.

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The answer is.........

25.99

The total time in seconds the ball will take to travel before returning the ground is 2.25 seconds.

The breaking or breakdown of an entity (such as an integer, a matrices, or a polynomials) into a products of another unit, or factors, whose multiplication results in the original number, matrix, etc., is known as factorisation or factoring in mathematics. 

Calculation for the time;

Let 't' be the time in second the ball will travel.

Let H(t) be the total height travelled by the ball.

The equation of the height is given as

[tex]H(t)=-16 t^{2}+36 t[/tex]

As soon as the ball will return to the ground the total height will become zero.

So, put equation of height equal to zero.

[tex]-16 t^{2}+36 t=0[/tex]

Factorise the above equation;

[tex]-16 t\left(t-\frac{9}{4}\right)=0[/tex]

Put each value equals to zero to get the value of time.

[tex]\begin{aligned}&t=0 \mathrm{sec} \\&t=9 / 4=2.25 \mathrm{sec}\end{aligned}[/tex]

As, time can not be zero

Therefore, the total time taken by the ball to return to the ground is 2.25 sec.

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Answer:

20.0

Step-by-step explanation:

We can use the distance formula, d = [tex]\sqrt{(x_{1}-x_{2}) ^{2} +(y_{1}-y_{2}) ^{2}}[/tex]  to find the length of three sides separately.

AB =  [tex]\sqrt{(-4+2) ^{2} +(1-3}) ^{2}}[/tex] = [tex]\sqrt{8}[/tex]

BC =  [tex]\sqrt{(-2-3) ^{2} +(3+4) ^{2}}[/tex] = [tex]\sqrt{74}[/tex]

CA =  [tex]\sqrt{(3+4) ^{2} +(-4-1) ^{2}}[/tex] = [tex]\sqrt{74}[/tex]

Perimeter = [tex]\sqrt{74}[/tex] + [tex]\sqrt{8}[/tex] + [tex]\sqrt{74}[/tex] = 20.0330776588 units

Answer:

Step-by-step explanation:

Answer:

You can count how many boxes that are not shaded to find how many eggs were not used, but that's to find out the fraction

Step-by-step explanation:

Count the boxes total in the grid. then count the number of boxes were not shaded. the number of total boxes is your denominator, and the number of boxes thar aren't shaded is your numerator.

I think...

The perimeter of the figure to the nearest tenth is 44.6 units

The perimeter of a figure is the sum of the whole sides.

Therefore,

A parallelogram has opposite sides equal to each other.

Therefore, the perimeter of the figure is as follows;

The figure combines a parallelogram and a semi circle.

Therefore,

perimeter of the figure = 4 + 14 + 14 + πr

perimeter of the figure = 32 + 3.14 × 4

perimeter of the figure = 32 + 12.56

perimeter of the figure = 44.56

perimeter of the figure ≈ 44.6 units

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The ounces of cheese Ellie uses to make tacos is 2

What is a unitary method?

The unitary method is a technique for solving a problem by first finding the value of a single unit, and then finding the necessary value by multiplying the single unit value.

Calculation :

1 pound = 16 ounces

Total cheese Ellie has = [tex]\frac{2}{8}[/tex] pound

Cheese required to make tacos = [tex]\frac{2}{8}[/tex]×[tex]\frac{1}{2}[/tex] = [tex]\frac{1}{8}[/tex] pounds

Ounces of cheese required to make cheese are [tex]\frac{1}{8}[/tex] × 16 = 2 ounces

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Coplanar circles that have the same center, but not necessarily the congruent radii are called concentric circles.

From the question, we have the following statements:

As a general rule, circles that have the above features are referred to as concentric circles.

This is so because concentric circles have the same center, and they do not intersect

Hence, coplanar circles that have the same center, but not necessarily the congruent radii are called concentric circles.

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Answer:

A: (4,7)

B: (2,1)

C: (6,1)

Step-by-step explanation:

To plot the points, go along the x-axis (the horizontal line with the x on the end) and stop where the dot is. Whatever number the dot is above is the   x-coordinate. Then, however many numbers you go up by to reach the dot is your y-coordinate.

hope this helps :)

Subtracting the volume of the cylinder from the volume of the prism, the volume of metal in the hex nut to the nearest tenth exists [tex]$$23.6 cm^3[/tex]

Diameter of the cylinder be d = 1.6 cm

Apothem of the hexagon be a = 2 cm

Thickness of the steel hex nut be t = 2 cm

Volume of the prism be [tex]V_p[/tex]

Volume of the cylinder be [tex]V_c[/tex]

Volume of metal in the hex nut,

[tex]$$V = V_p - V_c[/tex]

To estimate the volume of a prism,

[tex]$$V_p = A_b h[/tex]

Ab = n L a / 2

Number of the sides, n = 6

The side of the hexagon be L

Height of the prism, h = t = 2 cm

Central angle in the hexagon, A = 360°/n

A = 360°/6 = 60°

[tex]$tan (\frac{A}{2} )=(\frac{L/2}{a})[/tex]

simplifying the value of L, we get

[tex]$tan (\frac{60}{2} )=(\frac{L/2}{2})[/tex]

[tex]$tan 30}=(\frac{L/2}{2})[/tex]

[tex]$tan (\frac{\sqrt{3}}{3} )=(\frac{L/2}{2})[/tex]

Solving for L/2:

[tex]$\frac{2 \sqrt{3}}{3} =\frac{L}{2}[/tex]

Solving the value of L, we get

[tex]$2\frac{2 \sqrt{3}}{3} =L[/tex]

[tex]$\frac{4 \sqrt{3}}{3} =L[/tex]

[tex]$L=4 \sqrt{3}/3 cm[/tex]

Ab = n L a / 2

Substitute the values in the above equation, we get

[tex]$A_b=\frac{6 (4 \sqrt{3}/3)(2)}{2}[/tex]

[tex]$$A_b=24 \sqrt{3}/3 $$cm^2[/tex]

[tex]$A_b=8 \sqrt{3} cm^2[/tex]

[tex]V_p = A_b h[/tex]

substitute the values in the above equation, we get

[tex]$V_p=(8 \sqrt{3})(2)[/tex]

[tex]$V_p=16 \sqrt{3} cm^3[/tex]

[tex]$$V_p=16 (1.732) cm^3[/tex]

[tex]$$V_p=27.712 cm^3[/tex]

To estimate the volume of cylinder,

[tex]$V_c[/tex] = (π[tex]d^2[/tex]/4) h

Here, π = 3.14 and d = 1.6 cm

Height of the cylinder, h = t = 2 cm

substitute the values in the above equation, we get

[tex]$V_c=[3.14 (1.6) / 4] (2)[/tex]

[tex]$V_c=[3.14 (2.56) / 4] (2)[/tex]

[tex]$V_c=(2.0096) (2)[/tex]

[tex]$$V_c=4.019 cm^3[/tex]

Substitute the values in the equation, we get

[tex]$$V=V_p-V_c[/tex]

[tex]$$V=27.712 - 4.019[/tex]

[tex]$$V=23.693 cm^3[/tex]

[tex]$$V=23.6 cm^3[/tex]

Therefore, the volume of metal in the hex nut, to the nearest tenth exists [tex]$$23.6 cm^3[/tex].

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