The graph of a linear relation passes through the point (-3,-5) and has a slope of 3.
Determine an equation for this linear relation in slope intercept form.

The number set(s) that - 10 belong to are rational numbers, integers and real numbers. Option C, D and E

A rational number can be defined as a number expressed as the ratio of two integers, where the denominator is not be equal to zer0

- 10 can be written as = 1/ 10

Integers are whole number that could be positive, negative and even zero

- 10 is a negative whole number

Real numbers are numbers with continuous quantity that can represent  distance along a number line

-10 can represent distance along a number line.

Thus, the number set(s) that - 10 belong to are rational numbers, integers and real numbers. Option C, D and E

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[tex]\sum^{\infty}_{n=1} (a/b)^n=5 \\ \\ =\frac{a/b}{1-\frac{a}{b}}=5 \\ \\ \frac{a}{b-a} =5 \\ \\ \frac{a}{b}=\frac{5}{6}[/tex]

So, we need to find

[tex]\sum^{\infty}_{n=1} n(5/6)^n

[/tex]

Let this sum be S.

Then,

[tex]S=(5/6)+2(5/6)^2 +3(5/6)^3+\cdots \\ \\ \frac{5}{6}S=(5/6)^2 + 2(5/6)^3+\cdots \\ \\ \implies \frac{1}{6}S=(5/6)+(5/6)^2+(5/6)^3+\cdots=5 \\ \\ \implies S=\boxed{30}[/tex]

If you know the slope of the lines, you can also use these methods:

- see if the slopes multiplied by each other are -1 (should be if they're perpendicular)

- test if the slopes are opposite reciprocals (if they are, they are perpendicular)

The true statement relating to a property of the function y =sin x is that the maximum and minimum values of the function are 1 and -1 respectively. Option B

The following are the properties of the sin trigonometric ratio of the function;

From the above listed deductions, we can see that the true statement about the function y = sin x is that the range which is always known as the maximum and minimum values of the function are 1 and - 1 respectively.

Thus, the true statement relating to a property of the function y =sin x is that the maximum and minimum values of the function are 1 and -1 respectively. Option B

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Answer:

2/9 - 2/15

Solution

LCM = 45

45 ÷9 ×2 = 10

45÷ 15 ×2 = 6

10 - 6 = 4

ANSWER = 4/45

Answer:

l

Step-by-step explanation:

take the lcm of 9 and 15. it will be 45 . than continue

Answer:7 21/100

Step-by-step explanation: We will covert 7.21 into a fraction. .21 = 21/100. Since 21 can only be divided by 3 and 7, the simplest form of 21/100 is 21/100. So, the mixed number is 7 and 21/100

Answer:

7 and 21/100

Step-by-step explanation:

Since .01 is 1/100, 7.21 will be 7 and 21/100. There is no way to further simplify this, therefore, it is our final answer. Hope this helps! :)

Answer:

x=2

Step-by-step explanation:

2^x *2^4=2^3x

using law of indices

2^x+4=2^3x

x+4=3x

4=3x-x

4=2x

2=x

          Both of these problems will be solved in a similar way, but with different numbers. First, we set up an equation with the values given. Then, we solve. Lastly, we plug into the original expressions to solve for the angles.

[23] ABD = 42°, DBC = 35°

(4x - 2) + (3x + 2) = 77°

4x+ 3x + 2 - 2  = 77°

4x+ 3x= 77°

7x= 77°

x= 11°

-

ABD = (4x - 2) = (4(11°) - 2) = 44° - 2 = 42°

DBC = (3x + 2) = (3(11°) + 2) = 33° + 2 = 35°

[24] ABD = 62°, DBC = 78°

(4x - 8) + (4x + 8) = 140°

4x + 4x + 8 - 8 = 140°

4x + 4x = 140°

8x = 140°

8x = 140°

x = 17.5°

-

ABD = (4x - 8) = (4(17.5°) - 8) = 70° - 8° = 62°

DBC =(4x + 8) = (4(17.5°) + 8) = 70° + 8° = 78°

The most general antiderivative of the given function g(t) is (8t + t³/3 + t²/2 + c).

The antiderivative of a function is the inverse function of a derivative.

This inverse function of the derivative is called integration.

Here the given function is: g(t) = 8 + t² + t

Therefore, the antiderivative of the given function is

∫g(t) dt

= ∫(8 + t² + t) dt

= ∫8 dt + ∫t² dt + ∫t dt

= [8t⁽⁰⁺¹⁾/(0+1) + t⁽²⁺¹⁾/(2+1) + t⁽¹⁺¹⁾/(1+1) + c]

= (8t + t³/3 + t²/2 + c)

Here 'c' is the constant.

Again, differentiating the result, we get:

d/dt(8t + t³/3 + t²/2 + c)

= [8 ˣ 1 ˣ t⁽¹⁻¹⁾ + 3 ˣ t⁽³⁻¹⁾/3 + 2 ˣ t⁽²⁻¹⁾/2 + 0]

= 8 + t² + t

= g(t)

The antiderivative of the given function g(t)is (8t + t³/3 + t²/2 + c).

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Answer: 243

Step-by-step explanation:

If you divide the 18 inch by two to get the r, you will get 9. So you should multiple the 3 and the 9 square. So it should be 3x81=243

Answer: 45

Step-by-step explanation: Given the way the formula is formatted, the first term is 1. The common difference can be found by subtracting a number from the number that follows (ex. 3-2 or 4-3), therefore it's 1. The desired term is what you're trying to find so 44-1=43. When you put it all together, the formula should be 2+1(44-1) which equals 45 when you follow the rules of PEMDAS.

The coordinates for P and Q are as follows,

P = (a, a)

Q = (0, a)

Finding the Missing Coordinates:

Triangle OPQ is an isosceles triangle, hence two of its legs are equal.

Since the coordinates of the end point of the leg OQ lies on the y-axis, and OP is parallel to x-axis, OQ ⊥ QP ............ (1)

Also, it indicates that the x- coordinate of point Q is 0.

⇒ The coordinates of point Q are (0, a)

From (1), we can infer that,

OP = OQ [∵ OP is the hypotenuse]

⇒ The distance of point P from x-axis = a

The distance of point P from y-axis =a

Hence, the coordinates of point P are given as (a, a).

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The average value of the given function is [tex]f_{avg} = \frac{-2a^{2} + 6a + 6 }{6}[/tex].

According to the statement

we have given that the function f(x) and we have to find the average value of that function.

So, For this purpose, we know that the

The given function f(x) is

[tex]f(x) = -x^{2} + 2x +1[/tex]

And now integrate this function with the limit 0 to a then

[tex]f_{avg} = \frac{1}{b - a} \int\limits^a_0 {f(x)} \, dx = -x^{2} + 2x +1[/tex]

Now integrate this then

[tex]f_{avg} = \frac{1}{a} \int\limits^a_0 {-x^{2} + 2x +1} \, dx[/tex]

Then the value becomes according to the integration rules is:

[tex]f_{avg} = \frac{1}{a} \int\limits^a_0 {-\frac{x^{3} }{3} + \frac{2x^{2} }{2} +x} \,[/tex]

Now put the limits then answer will become as output is:

[tex]f_{avg} = \frac{1}{a} [ {-\frac{a^{3} }{3} + \frac{2a^{2} }{2} +a} \,][/tex]

Now solve this equation then

[tex]f_{avg} = [ {-\frac{a^{2} }{3} + \frac{2a }{2} +1} \,][/tex]

Now

[tex]f_{avg} = \frac{-2a^{2} + 6a + 6 }{6}[/tex]

This is the value which represent the average of the given function in the statement.

So, The average value of the given function is [tex]f_{avg} = \frac{-2a^{2} + 6a + 6 }{6}[/tex].

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Answer:

3x + 16 + 30/x - 13

Step-by-step explanation:

In order to solve this IVP using Laplace transforms, we must first write f(t) in terms of the Heaviside function.

f(t)=0*(u(t)-u(t-Pi))+1*(u(t-Pi)-u(t-2Pi))+0*(u(t-2Pi))

f(t)=u(t-π)-u(t-2π)

So, the rewritten IVP is

y'' +y = u(t-π)-u(t-2π)y(0)=0, y'(0)=1

Taking the Laplace transform of both sides of the equation, we get:

s2L{y}-sy(0)-y'(0)+L{y}=(1/s)*e-πs-(1/s)*e-2πs

s2L{y}-1+L{y}=(1/s)*e-πs-(1/s)*e-2πs

(s2+1)L{y}=1+(1/s)*e-πs-(1/s)*e-2πs

L{y}=1/(s2+1)+(1/s(s2+1))e-πs-(1/s(s2+1))*e-2πs

Now, we must take the inverse transform of both sides to solve for y.

The first inverse transform is easy enough. By definition, it is sin(t).

The second two inverse transforms will be a little tougher, we will have to use partial fraction decomposition to break them down into terms that are easier to compute.

A/s+(Bs+C)/(s^2+1)=1/(s(s^2+1))

A(s^2+1)+(Bs+C)(s)=1

As^2+A+Bs^2+Cs=1

Rewriting this system in matrix form, we get:

1  1  0    A     0

0  0  1  * B  = 0

1  0  0    C     1

Using row-reduction we find that A=1, B=-1, and C=0. So, our reduced inverse transforms are:

L-1{(e-πs)(1/s-s/(s2+1))}

and

L-1{(e-2πs)(1/s-s/(s2+1))}

Using the first and second shifting properties, these inverse transforms can be computed as.

L-1{(e-πs)(1/s-s/(s2+1))}=u(t-π)-cos(t-π)u(t-π)

L-1{(e-2πs)(1/s-s/(s2+1))}=u(t-2π)-cos(t-2π)u(t-2π)

Combining all of our inverses transforms, we get the solution the IVP as:

y=sin(t)+u(t-π)-cos(t-π)u(t-π)+u(t-2π)-cos(t-2π)u(t-2π)

In mathematics, the Laplace transform, named after its discoverer Pierre Simon Laplace (/ləˈplɑːs/), transforms a function of real variables (usually in the time domain) into a function of complex variables (in the time domain). is the integral transform that Complex frequency domain, also called S-area or S-plane).

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Answer:

  (c)  f(g(3)) > g(f(3))

Step-by-step explanation:

The relationship between the function values can be found by evaluating the functions.

The order of operations tells us we must first evaluate g(3).

  g(x) = -3x +15

  g(3) = -3(3) +15 = 6

Then we can evaluate f(x) for x=6:

  f(x) = 7x

  f(6) = 7(6) = 42

So, the composition is ...

  f(g(3)) = 42

As above, we must first evaluate f(3):

  f(3) = 7(3) = 21

Then we can evaluate g(x) for x=21:

  g(21) = -3(21) +15 = -63 +15 = -48

That means the composition is ...

  g(f(3)) = -48

The first is greater than the second.

  42 > -48

  f(g(3)) > g(f(3))

Answer:

20 cookies

Step-by-step explanation:

8 in 1.5 minutes

so we want to find how many in 3.75 minutes since 3 + 45/60 = 3.75

so then its 1.5*2 = 3 so 8*2 = 16 to get that 16 cookies in 3 minutes

then we still have .75 left so then divide 8/2 to get 4 cookies in 0.75 minutes

16+4 = 20

you can also just find how many in 0.25 minutes (15 seconds) you get 6/8

multiply that by 3.75/0.25 = 15 you get 15*(8/6) = 20

The measure of the angle between the vectors

[tex]$\arccos[ (-\sqrt{13 } i )/ (\sqrt{5 }) ]\\\sqrt{5 }[/tex].

What is the measure of the angle between the vectors?

Given:

[tex]$\mathrm{u}=\langle -3,2\rangle$[/tex] and [tex]$v=\langle 2,1\rangle$[/tex]

Computing the angle between the vectors, we get

[tex]$\quad \cos (\theta)=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}| \cdot|\vec{b}|}$[/tex]

To estimate the lengths of the vectors, we get

Computing the Euclidean Length of a vector,

[tex]$\left|\left(x_{1}, \ldots, x_{n}\right)\right|=\sqrt{\sum_{i=1}^{n}\left|x_{i}\right|^{2}}$[/tex]

Let, [tex]$\mathrm{u} &=\langle -3,2\rangle \\[/tex] and [tex]$\mathrm{v} &=\langle 2,1\rangle \\[/tex]

If [tex]$\mathrm{u} &=\langle -3,2\rangle \\[/tex]

[tex]$|u| &=\sqrt{-3^{2}+(2)^{2}} \\[/tex]

[tex]$&=\sqrt{5}i \\[/tex] and

[tex]$\mathrm{v} &=\langle 2,1\rangle \\[/tex]

[tex]$|v| &=\sqrt{2^{2}+(1)^{2}} \\[/tex]

[tex]$&=\sqrt{5}[/tex]

Finally, the angle is given by:

Computing the angle between the vectors, we get

[tex]$ $\cos (\theta)=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}| \cdot|\vec{b}|}$[/tex]

[tex]$&\cos (\Phi)=-\sqrt{13 } i/ \sqrt{5 } \\[/tex]

simplifying the above equation, we get

[tex]$&\Phi=\arccos (\cos (\Phi))[/tex]

[tex]$=\arccos[ (-\sqrt{13 } i )/ (\sqrt{5 }) ]\\\sqrt{5 }[/tex]

Therefore, the measure of the angle between the vectors

[tex]$\arccos[ (-\sqrt{13 } i )/ (\sqrt{5 }) ]\\\sqrt{5 }[/tex].

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Answer: line ZX is congruent to line VX (option 4)

Step-by-step explanation:

We already know <X is congruent to <X, we also know that line YX is congruent to line XW. Now all we need is one more line adjacent to X which is going to be ZX ad VX

Step-by-step explanation:

The values of b, h and k are (c) b = 2, h = -2 and k = -9

The logarithmic function is given as:

f(x) = log₂(x + 2) - 9

A logarithmic function is represented as:

f(x) = logb(x - h) + k

By comparing both equations, we have

b = 2

h = -2

k = -9

Hence, the values of b, h and k are (c) b = 2, h = -2 and k = -9

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Answer:

1.0954.

Step-by-step explanation:

Standard error  =  std dev / √n

=  6 / √30

= 1.0954.

Answer:

  p(x) = x² +30x -375

Step-by-step explanation:

When a quadratic has real rational coefficients, any irrational or complex roots come in conjugate pairs.

A root of p means (x -p) is a factor of the polynomial. Here, we have roots of -15+10√6 and -15-10√6, so the factored form can be written ...

  p(x) = (x -(-15 +10√6))(x -(-15 -10√6))

Using the factoring of the difference of squares, we can write this as ...

  p(x) = (x +15)² -(10√6)²

Expanding the factored form, we can write the polynomial as ...

  p(x) = x² +30x +225 -600

  p(x) = x² +30x -375

[tex]x = \frac{\sqrt{5} + 1}{2} \approx 1.618033989[/tex]

[tex]\cos(x)[/tex] is an even function, while [tex]\sin(x)[/tex] is odd. This means

[tex]\cos(-x) = \cos(x) \text{ and } \sin(-x) = -\sin(x)[/tex]

[tex]\cot(x)[/tex] is defined by

[tex]\cot(x) = \dfrac{\cos(x)}{\sin(x)}[/tex]

so it is an odd function, since

[tex]\cot(-x) = \dfrac{\cos(-x)}{\sin(-x)} = \dfrac{\cos(x)}{-\sin(x)} = -\cot(x)[/tex]

Putting everything together, it follows that

[tex]\cot(-x) \cos(-x) \sin(-x) = (-\cot(x)) \cos(x) (-\sin(x)) \\\\~~~~~~~~= \cot(x) \cos(x) \sin(x) \\\\ ~~~~~~~~= \cos^2(x)[/tex]

The distance the player is from the right post is = 5.1 meters

The Pythagorean theorem states that the square of the length of the hypotenuse of a right triangle equals the sum of the squares of the lengths of the other two sides.

Formula for the Pythagorean theorem =

a²= b²+c²

From the diagram given,

The hypotenuse (a) = x²

b= 1.82²

c = 4.8²

x² = 1.82² + 4.8²

x²= 3.3124 + 23.04

x²= 26.3524

a= √26.3524

a= 5.1 meters.

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the area of the shaded part is 30. 89 cm²

We have the shape to be a rectangle

The area of the shaded part should be;

Area of rectangle - 2 ( area of a semi circle)

The formula for area of a rectangle

Area = length × width

Area = 12 × 12

Area = 144 cm²

Area of a semicircle = 1/2 πr²

Area = 1/ 2 × 3. 142 × 6²

Area = 56. 56 cm²

Area of shaded part = area of rectangle - 2( area of semicircle)

Area of shaded part = 144 - 2(56. 56)

Area of shaded part = 144 - 113. 11

Area of shaded part = 30. 89 cm²

Thus, the area of the shaded part is 30. 89 cm²

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Taylor series is [tex]f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2} }[/tex]

To find the Taylor series for f(x) = ln(x) centering at 9, we need to observe the pattern for the first four derivatives of f(x). From there, we can create a general equation for f(n). Starting with f(x), we have

f(x) = ln(x)

[tex]f^{1}(x)= \frac{1}{x} \\f^{2}(x)= -\frac{1}{x^{2} }\\f^{3}(x)= -\frac{2}{x^{3} }\\f^{4}(x)= \frac{-6}{x^{4} }[/tex]

.

.

.

Since we need to have it centered at 9, we must take the value of f(9), and so on.

f(9) = ln(9)

[tex]f^{1}(9)= \frac{1}{9} \\f^{2}(9)= -\frac{1}{9^{2} }\\f^{3}(x)= -\frac{1(2)}{9^{3} }\\f^{4}(x)= \frac{-1(2)(3)}{9^{4} }[/tex]

.

.

.

Following the pattern, we can see that for [tex]f^{n}(x)[/tex],

[tex]f^{n}(x)=(-1)^{n-1}\frac{1.2.3.4.5...........(n-1)}{9^{n} } \\f^{n}(x)=(-1)^{n-1}\frac{(n-1)!}{9^{n}}[/tex]

This applies for n ≥ 1, Expressing f(x) in summation, we have

[tex]\sum_{n=0}^{\infinite} \frac{f^{n}(9) }{n!} (x-9)^{2}[/tex]

Combining ln2 with the rest of series, we have

[tex]f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2} }[/tex]

Taylor series is [tex]f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2} }[/tex]

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