Alex travels 46
miles per hour for
3.2 hours. How far
has he gone?

Answer:

4

Step-by-step explanation:

the time it takes for a plane to travel a distance varies directly as the speed of the plane because ts=16,200

A logarithmic equation exists as an equation that uses the logarithm of an expression containing a variable. The value of the logarithmic equation x = log 2.97/log 1.13.

A logarithmic equation exists as an equation that applies the logarithm of an expression having a variable. To estimate exponential equations, first, see whether you can note both sides of the equation as powers of the same number.

Given: [tex]$1.13^x = 2.97[/tex]

Taking log on both sides, we get

log [tex]$1.13^x[/tex] = log 2.97

x log 1.13 = log 2.97

x = log 2.97/log 1.13

Therefore, the value of logarithmic equation x = log 2.97/log 1.13.

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Answer:

[tex]y=-\frac{1}{3}x+7[/tex]

Step-by-step explanation:

Parallel lines have the same slope, so the slope of the line we need to find is -1/3.

Substituting into point-slope form and converting to slope-intercept form,

[tex]y-5=-\frac{1}{3}(x-6) \\ \\ y-5=-\frac{1}{3}x+2 \\ \\ y=-\frac{1}{3}x+7[/tex]

Answer:

C. slope

Greetings !

The symbol for slope is "m". Slope can be calculated by taking any two locations along a line and calculating the ratio of the RISE to the RUN (the ratio of the difference between vertical positions of the locations to the difference between horizontal positions of the locations).

Answer: slope.

Step-by-step explanation:

The statement that describes the quotient of 3 over 10 division sign2 over 5 is A. The maximum amount of oil the bowl can hold is Fraction 3 over 4 cup.

From the information given, we are told that a bowl holds fraction 3 over 10 cups of oil when it is Fraction 2 over 5 full.

Therefore, the statement that best describes the quotient of 3 over 10 division sign2 over 5 will be that the maximum amount of oil the bowl can hold is fraction 3 over 4 cup.

In conclusion, the correct option is A.

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Answer:

p(t) = 2950^3t

Step-by-step explanation:

I’m not sure if this is exactly what you wanted or not. Please let me know more info and I’ll write any more answers for this question in the comments. Have a great day!!

The geometric sequence when the first term is 1/5 and the common ratio is 5 is; f(n) = f(n-1) . 5.

The geometric sequence described in the task content is one whose first term is; f(1) = 1/5.

Additionally, it follows from convention that the recursive function for a geometric sequence is; a product of a previous term and the common ratio.

Hence, we have; f(n) = f(n-1) . 5.

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The value of x in the system of equation is 6

The system of equations is given as:

y = 1/2 * 2^x

y = 5x + 2

Substitute y = 1/2 * 2^x in y = 5x + 2

1/2 * 2^x = 5x + 2

Multiply through by 2

2^x = 10x + 4

Next, we use the trial by error.

Set x = 6

This gives

2^6 = 10 * 6 + 4

Evaluate the product and the exponent

64 = 60 + 4

Evaluate the sum

64 = 64

Hence, the value of x in the system of equation is 6

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Answer:

Option 3

Step-by-step explanation:

Since the denominators are the same, you can just subtract the numerators.

The true statement is that no number in base 10 can be written in multiple ways in base 4

The base of the numbers are given as:

Base 10 and base 4

Base 10 numbers are also referred to as decimal numbers, while base 4 numbers are quaternary numbers

There is only one equivalent of each number in each base.

This means that (for instance)

357 in base 10 is 11211 in base 4

The above number does not have any other representation in base 4 and it can not be written in another way.

This is the same for other numbers in base 10

Hence, the true statement is that no number in base 10 can be written in multiple ways in base 4

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The intersection of the sets has the following properties: Commutative law – A ∩ B = B∩ A. Associative law – (A ∩ B)∩ C = A ∩ (B∩ C) φ ∩ A = φ

do mark brainliest

The length of the altitude AD is 12 units.

The height of the triangle can be found as follows:

We have to find an angle using cosine law before  we can find the height.

Therefore,

20² = 13² + 21² - 2 × 21 × 13 cos B

400 = 169 + 441  - 546 cos B

400 - 610 = - 546 cos B

-210 = - 546 cos B

cos B = -210 / -546

cos B = 0.38461538461

B = cos⁻¹ 0.38461538461

B = 67.3810899783

B = 67.38°

Hence,

sin 67.38° = opposite / hypotenuse

sin 67.38° = AD / 13

cross multiply

AD = 13 sin 67.38°

AD = 11.9999882145

AD = 12 units

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The diagonals of the qudrilateral will not be perpendicular to each other.

A quadrilateral is a four-sided polygon with four edges and four corners that is used in geometry.

Slope or the gradient is the number or the ratio which determines the direction or the steepness of the line. If two segments are perpendicular, the multiplication of their slopes is -1.

The slope of a coordinate pair is given by the change in y divided by the change in x, that is:

For the diagonal AD the slope will be:-

Slope = ( -5 + 3 ) / ( -2 - 4) = ( 1 / 3 )

For the diagonal BC, the slope will be:-

Slope= ( 1 - 3) / ( 4 - 2 ) = ( -2 / 2) = -1

The multiplication of the two slopes will be,

M  =  ( 1 / 3 ) x -1 = ( - 1 / 3 )

Therefore,  the diagonals of the quadrilateral will not be perpendicular to each other. Because the product of the slope is not equal to -1.

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Answer:

The product of the slopes of the diagonals is -1.

Step-by-step explanation:

I had the question and got it right.

The term that is not possible in the domain of a sequence is:

-5.

The domain of a function is the set that contains all possible input values for the function. For a sequence, the domain is the set that contains all the indexes of the terms, starting at 0 and going until the nth term.

For example, suppose we have the following sequence: 3, 5, 7, ...

From what was explained above, which also can be visualized with the example, an index term of a sequence cannot be negative, hence the term that is not possible in the domain of a sequence is:

-5.

Which is the only negative number of the options.

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The two ratios in the tables shown which have a common denominator you could use to compare is; 3/12 and 4/12

Machael's ratio:

lemons to water = 1:4

Equivalent ratio

= 3 : 12

= 4 : 16

Sondra's ratio:

lemons to water = 2 : 6

Equivalent ratio

= 4 : 12

= 6 : 18

Therefore, the two ratios in the tables shown which have a common denominator you could use to compare is; StartFraction 3 Over 12 EndFraction and StartFraction 4 Over 12 EndFraction

Complete question

Michael and Sondra are mixing lemonade. In Michael’s lemonade, the ratio of lemons to water is 1:4. In Sondra’s lemonade, the ratio of lemons to water is 2:6. Several equivalent ratios for each mixture are shown in the ratio tables. Imagine that you want to compare Michael’s ratio to Sondra’s ratio. Which two ratios in the tables shown have a common denominator you could use to compare?

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Answer:

  67 square units

Step-by-step explanation:

The area using the left-hand sum is the sum of products of the function value at the left side of the interval and the width of the interval.

The attachment shows a table of the x-value at the left side of each interval, and the corresponding function value there. The interval width is 1 unit in every case, so the desired area is simply the sum of the function values.

The approximate area is 67 square units.

Split up the interval [0, 6] into 6 equally spaced subintervals of length [tex]\Delta x = \frac{6-0}6 = 1[/tex]. So we have the partition

[0, 1] U [1, 2] U [2, 3] U [3, 4] U [4, 5] U [5, 6]

where the left endpoint of the [tex]i[/tex]-th interval is

[tex]\ell_i = i - 1[/tex]

with [tex]i\in\{1,2,3,4,5,6\}[/tex].

The area under [tex]f(x)=x^2+2[/tex] on the interval [0, 6] is then given by the definite integral and approximated by the Riemann sum,

[tex]\displaystyle \int_0^6 f(x) \, dx \approx \sum_{i=1}^6 f(\ell_i) \Delta x \\\\ ~~~~~~~~ = \sum_{i=1}^6 \bigg((i-1)^2 + 2\bigg) \\\\ ~~~~~~~~ = \sum_{i=1}^6 \bigg(i^2 - 2i + 3\bigg) \\\\ ~~~~~~~~ = \frac{6\cdot7\cdot13}6 - 6\cdot7 + 3\cdot6 = \boxed{67}[/tex]

where we use the well-known sums,

[tex]\displaystyle \sum_{i=1}^n 1 = \underbrace{1 + 1 + \cdots + 1}_{n\,\rm times} = n[/tex]

[tex]\displaystyle \sum_{i=1}^n i = 1 + 2 + \cdots + n = \frac{n(n+1)}2[/tex]

[tex]\displaystyle \sum_{i=1}^n i^2 = 1 + 4 + \cdots + n^2 = \frac{n(n+1)(2n+1)}6[/tex]

Answer:

100 miles

Cost = $63

Step-by-step explanation:

Let us assume that the distance driven for both plans when they both equal in cost is X miles

Plan 1

Cost = 48  + 0.15X

Plan2

Cost = 53 + 0.1X

If they are both equal then

48 + 0.15X = 53 + 0.10X

Collecting like terms

0.15X - 0.10X = 53 - 48

0.05X = 5

X 5/0.05 = 100 miles

Plan 1 Cost = 48 + 0.15(100) = 48 + 15 = $63

Plan 2 Cost = 53 + 0.1(100) = 53 + 20 = $63

See below for the combination of the arithmetic operations and exactly five 3's

The conditions are given as:

There are no direct rules to this, except by trial and error.

After several trials, we have the following operations:

(3 * 3 - 3 * 3)/3 = 0

3 - 3/3 - 3/3 = 1

(3 + 3 - 3 / 3)-3 = 2

(3 * 3 - 3 + 3)/3 = 3

3 *3/3 + 3/3 = 4

3 +3/3 + 3/3= 5

3 + 3 + (3 - 3)/3 = 6

(3^3 - 3 - 3)/3 = 7

3 + 3 + (3 + 3)/3 = 8

3 + 3 + 3+ 3 -3 =9

3 + 3 + 3 + 3/3 = 10

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There is a maximum value of 7/6 located at (x, y) = (5/6, 7).

The function given to us is f(x, y) = xy.

The constraint given to us is 6x + y = 10.

Rearranging the constraint, we get:

6x + y = 10,

or, y = 10 - 6x.

Substituting this in the function, we get:

f(x, y) = xy,

or, f(x) = x(10 - 6x) = 10x - 6x².

To find the extremum, we differentiate this, with respect to x, and equate that to 0.

f'(x) = 10 - 12x ... (i)

Equating to 0, we get:

10 - 12x = 0,

or, 12x = 10,

or, x = 5/6.

Differentiating (i), with respect to x again, we get:

f''(x) = -12, which is less than 0, showing f(x) is maximum at x = 5/6.

The value of y, when x = 5/6 is,

y = 12 - 6x,

or, y = 12 - 6*(5/6) = 7.

The value of f(x, y) when (x, y) = (5/6, 7) is,

f(x, y) = xy,

or, f(x, y) = (5/6)*7 = 7/6.

Thus, there is a maximum value of 7/6 located at (x, y) = (5/6, 7).

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The number of contemporary titles and classic titles in Jarred DVDs collection is 2,921 and 579 respectively.

Simultaneous equation is an equation which involves the solving for two unknown values at the same time.

x + y = 3,500

x – y = 2,342

Add both be equation

x + x = 3,500 + 2,342

2x = 5,842

x = 5,842 ÷ 2

x = 2,921

Substitute x = 2,921 into

x – y = 2,342

2,921 - y = 2, 342

-y = 2,342 - 2,921

-y = -579

y = 579

Therefore, the number of contemporary titles and classic titles in Jarred DVDs collection is 2,921 and 579 respectively.

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

Here we go ~

Let's calculate its discriminant ~

[tex]\qquad \sf  \dashrightarrow \: {t}^{2} + \cfrac{17}{2} t - 5 = 0[/tex]

[ Multiply both sides by 2 ]

[tex]\qquad \sf  \dashrightarrow \: 2 {t}^{2} + 17t - 10[/tex]

[tex]\qquad \sf  \dashrightarrow \: discriminant = {b}^{2} - 4ac[/tex]

[tex]\qquad \sf  \dashrightarrow \: d = (17) {}^{2} - (4 \times 2 \times - 10)[/tex]

[tex]\qquad \sf  \dashrightarrow \: d = 289 - ( - 80)[/tex]

[tex]\qquad \sf  \dashrightarrow \: d = 369[/tex]

[tex]\qquad \sf  \dashrightarrow \: \sqrt {d }= 3 \sqrt{41} \approx19.209 [/tex]

So, by quadratic formula :

[tex]\qquad \sf  \dashrightarrow \: t = \dfrac{ - {b}^{} \pm \sqrt{d} }{2a} [/tex]

[tex]\qquad \sf  \dashrightarrow \: t = \dfrac{ - {17}^{} \pm \sqrt{369} }{2 \times 2} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \:t = \cfrac{ - 17 - 19.209}{4} \: \: and \: \: t = \dfrac{-17+19.209}{4} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \:t = \cfrac{ - 36.209}{4} \: \: and \: \: t = \dfrac{2.209}{4} [/tex]

[tex]\qquad \sf  \therefore \: t = - 9.052 \: \: \: or \: \: \: t = 0.552[/tex]

Answer:

23

Step-by-step explanation:

The question gives the first equation:

0.05N + 0.10D = 2.20

It can be rewritten as:

0.05$ times the number of nickels plus 0.10$ times the number of dimes is equivalent to $2.20.

Notice that 0.05$ or 5 cents is equivalent to one nickel and 0.10$ or 10 cents is equivalent to one dime.

The question asks for N+D. We know that N stands for the number of nickels Daniel has and D stands for the number of dimes Daniel has. We also know that Daniel only has dimes and nickels in his pocket. Therefore N+D most be equivalent to the total number of coins, which is given in the question to be equal to 23 coins.

Therefore...

N+D = 23

Answer:

m<C = 12.0 °

Step-by-step explanation:

Since we know that this is a non-right triangle, we must use either the law of sines or law of cosines to find m<C.  

The information we now have only allows us to use the law of sines, which focuses on proportions:

[tex]\frac{SinA}{a}=\frac{SinB}{b}=\frac{SinC}{c}[/tex]

Thus, we can use:

[tex]\frac{sin(98)}{19}=\frac{sinC}{4}\\ 4*sin(98)=19*sinC\\ \frac{4*sin(98)}{19}=sinC\\ sin^-1\frac{4*sin98}{19}=12.033=12.0[/tex]

The sin^-1 represents the sin inverse, which you must use to find angle measures.

Answer:

C) 98

Step-by-step explanation:

Hope this helps! sry  if I'm wrong

The value of x in the given equation 15x + 120 = 5x + 1100 is 98. Option C is correct.

An equation is a combination of numbers, variables, mathematical operations, and functions. It is basically a statement emphasising that the two or more expressions are equal to each other.

The given equation is

15x + 120 = 5x + 1100

Take the like terms together,

15x - 5x = 1100 - 120

10x = 980

x = [tex]\frac{980}{10}[/tex]

x = 98.

Thus, the option C is correct stating that the value of x in the given equation is 98.

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The complete question is as follows:

Find the value of x in the equation:

15x + 120 = 5x + 1100.

A. 66

B. 122

C. 98

D. 76

SUBMIT.

Answer:

[tex]a = 1, \ - 3[/tex]

Explanation:

[tex](a-6)(a+8) = -45[/tex]

distribute

[tex]a^2 + 8a - 6a - 48 = -45[/tex]

collect terms

[tex]a^2 + 8a - 6a - 48 + 45=0[/tex]

simplify

[tex]a^2 + 2a - 3=0[/tex]

factor

[tex](a - 1)(a + 3)= 0[/tex]

set to zero

[tex]a = 1, \ - 3[/tex]

Answer:

a = 1,  -3

Step-by-step explanation:

Given equation:

[tex](a-6)(a+8)=-45[/tex]

Expand the brackets:

[tex]\implies a^2+8a-6a-48=-45[/tex]

[tex]\implies a^2+2a-48=-45[/tex]

Add 45 to both sides:

[tex]\implies a^2+2a-48+45=-45+45[/tex]

[tex]\implies a^2+2a-3=0[/tex]

To factor a quadratic in the form [tex]ax^2+bx+c[/tex], find two numbers that multiply to [tex]ac[/tex] and sum to [tex]b[/tex].  

Two numbers that multiply to -3 and sum to 2 are: 3 and -1.  

Rewrite the middle term as the sum of these two numbers:

[tex]\implies a^2+3a-a-3=0[/tex]

Factorize the first two terms and the last two terms separately:

[tex]\implies a(a+3)-1(a+3)=0[/tex]

Factor out the common term (a + 3):

[tex]\implies (a-1)(a+3)=0[/tex]

Apply the zero product property:

[tex]\implies (a-1)=0 \implies a=1[/tex]

[tex]\implies (a+3)=0 \implies a=-3[/tex]

Verify the solutions by inputting the found values of a into the original equation:

[tex]a=1 \implies (1-6)(1+8) & =-5 \cdot 9 = -45[/tex]

[tex]a=-3 \implies (-3-6)(-3+8) & =-9 \cdot 5 = -45[/tex]

Hence both found values of a are solutions of the given equation.

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Answer:

Let Ji and Bi represent the initial amounts that Joe and Bill have at the start.  Number after J and B will be used to indicate subsequent steps in the problem.

We are told that "Joe has five times as much money as Bill," which we can write as:

   1)  Ji = 5Bi

We learn that "Joe pays Bill $5," which we can represent as:

   2)  J1 = Ji - 5

This would mean that Bill has added $5:

   3)  B1 = Bi + 5

We are then told that "Joe has just twice the amount Bill now has," which we can write as:

   4)  J1 = 2B1

===

We can rearrnage and substitute the above relationships to eliminate one of the two variables (B1 or J1)

J1 = Ji - 5                [from 2]

2B1 = Ji - 5              [Substitute 4 to eliminate J1]

Ji = 5Bi                    [from 1]

2B1 = 5Bi - 5           [Substitute 1 to eliminate Ji]

B1 = Bi + 5               [Rearrange]

2(Bi + 5) = 5Bi - 5    [Use the above expression in the previous equation to eliminate B1]

2Bi + 10 = 5Bi - 5    [Simplify]

-3Bi = -15                  [Simplify]

Bi = $5                       [Solve]

Ji = 5Bi                    [from 1]

Ji = 5*(5)                  [Since Bi = $5]

Ji = $25                    [Solve]  

===

CHECK:

Does Joe has five times as much money as Bill?

Ji = $25 and Bi = $5  YES

When Joe pays Bill $5 he owes him, does Joe has just twice the amount Bill now has?

J1 = $25 - $5 = $20

B1 = $5 + $5  = $10    YES

we verified the intermidiate value theorem applies to the function f(x) = x^2 + 7x  + 1 . And the value of c is 2.

According to the given question.

We have a function.

f(x) = x^2 + 7x  + 1

As, we know that "the Intermediate Value Theorem (IVT) states that if f is a continuous function on [a,b] and f(a)<M<f(b), there exists some c∈[a,b] such that f(c)=M".

Now, we will apply the theorem for the given function f(x).

So,

f(0) = 0^2 +7(0) + 1 = 1

And,

f(9)=9² + 7(9) + 1 = 81 + 63 + 1 = 145

Here,  f(0) = 1< 19< 145 = f(9).

So, f is continous since it is a polynomial. Then the IVT applies, and such c exists.

To find, c,

We have to solve the quadratic equation f(c) =19.

This equation is

c² + 7c + 1 = 19.

Rearranging, c²+ 7c - 18=0.

Factor the expression to get

c² + 9c - 2c -18 = 0

⇒ c(c + 9) - 2( c + 9) = 0

⇒ (c - 2)(c + 9) = 0

⇒ c = 2 or -9

c = -9 is not possible beacuse it is not in the interval [0, 9].

So, the value of c is 2.

⇒ f(2) = 2^2 + 7(2) + 1 = 4 + 14 + 1 = 19

Hence, we verified the intermidiate value theorem applies to the function f(x) = x^2 + 7x  + 1 . And the value of c is 2.

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The greatest common factor of the two expressions given as in the task content is; 3v².

It follows from the task content that the terms whose greatest common factor are to be determined are: 15v³ and 12 v².

15v³ and 12v²

= 3v²(5v) and 3v²(4)

= 3v²(5v) and (4)

Consequently, in a bid to factorise the two expressions by means of their greatest Common factor, the greatest common factor can be determined as; 3v².

The correct answer choice which therefore represents the greatest common factor as required in the task content is; 3v².

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[tex]\huge\text{Hey there!}[/tex]

[tex]\mathsf{Formula: a^2 + b^2 = c^2}[/tex]

[tex]\textsf{Solving:}[/tex]

[tex]\mathsf{2^2 + 6^2 = c^2}[/tex]

[tex]\mathsf{2 \times 2 + 6 \times6 = c^2}[/tex]

[tex]\mathsf{4 + 36 = c^2}[/tex]

[tex]\mathsf{40 = c^2}[/tex]

[tex]\large\textsf{Therefore, your answer should be:}[/tex]

[tex]\huge\boxed{\frak{\sqrt{40}}}\huge\checkmark[/tex]

[tex]\huge\text{Good luck on your assignment \& enjoy your day!}[/tex]

~[tex]\frak{Amphitrite1040:)}[/tex]