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2 years ago

If f(x) = -x + 5 and g(x) = x^2, what is (g° (-1)?

Mathematics

2 answers:

ipn [44]2 years ago

7 0

Answer:

$(g\circ f)(-1)=36$

Step-by-step explanation:

We are given the two functions:

$f(x)=-x+5\text{ and } g(x)=x^2$

And we want to find:

$(g\circ f)(-1)$

This is equivalent to:

$=g(f(-1))$

So, evaluating f(-1) first, we acquire:

$f(-1)=-(-1)+5=6$

Then by substitution:

$=g(6)$

So:

$=(6)^2=36$

Therefore:

$(g\circ f)(-1)=36$

Your answer is indeed correct.

KATRIN_1 [288]2 years ago

3 0

Answer:

please refer to the above attachment.

<h3>36 is the right answer.</h3>

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Suppose that you are offered the following "deal.

MissTica

Answer:

expected profit is - $0.50

Step-by-step explanation:

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2 years ago

Suppose the mean SAT verbal score is 525 with a standard deviation of 100, while the mean SAT math score is 575 with a standard

Ipatiy [6.2K]

Answer:

The mean of the combined math and verbal scores is 1100, while the standard deviation is 141.

Step-by-step explanation:

Normal variables

Normal variables have mean $\mu$ and standard deviation $\sigma$

When we add normal variables, the combined mean is the sum of both means, and the standard deviation is the square root of the sum of both variances. The distribution is still normal.

In this question:

Verbal: $\mu_{V} = 525, \sigma_{V} = 100$ .

Math: $\mu_{M} = 575, \sigma_{M} = 100$

Combined:

$\mu = \mu_{V} + \mu_{M} = 525 + 575 = 1100$

$\sigma = \sqrt{\sigma_{V}^{2}+\sigma_{M}^{2}} = \sqrt{100^2 + 100^2} = 141$

The mean of the combined math and verbal scores is 1100, while the standard deviation is 141.

6 0

2 years ago

5. An airplane takes 4 hr to travel a distance of 4100 km. Another airplane travels at a speed

Mazyrski [523]

Answer:

4.969696 hours

Step-by-step explanation:

The distance traveled is shown through the following equations

Distance = velocity * time

This means that

4100=velocity*4

This means that ariplane 1 travles at 1025 km/hr

We can then subtract 200 from this to find that airplane two travels at a speed of 825 km/hr

Now we need to find how long it takes for that plane to travel 4100

So using the same equation

4100=825*time

Divison will tell us that the answer is around 4.9696 hours which is pretty much 5

7 0

3 years ago

(10 points) Consider the initial value problem y′+3y=9t,y(0)=7. Take the Laplace transform of both sides of the given differenti

Rashid [163]

Answer:

The solution

$Y (s) = 9( -1 +3 t + e^{-3 t} ) + 7 e ^{-3 t}$

Step-by-step explanation:

Explanation:-

Consider the initial value problem y′+3 y=9 t,y(0)=7

Step(i):-

Given differential problem

y′+3 y=9 t

Take the Laplace transform of both sides of the differential equation

L( y′+3 y) = L(9 t)

Using Formula Transform of derivatives

L(y¹(t)) = s y⁻(s)-y(0)

By using Laplace transform formula

$L(t) = \frac{1}{S^{2} }$

Step(ii):-

Given

L( y′(t)) + 3 L (y(t)) = 9 L( t)

$s y^{-} (s) - y(0) + 3y^{-}(s) = \frac{9}{s^{2} }$

$s y^{-} (s) - 7 + 3y^{-}(s) = \frac{9}{s^{2} }$

Taking common y⁻(s) and simplification, we get

$( s + 3)y^{-}(s) = \frac{9}{s^{2} }+7$

$y^{-}(s) = \frac{9}{s^{2} (s+3}+\frac{7}{s+3}$

Step(iii):-

By using partial fractions , we get

$\frac{9}{s^{2} (s+3} = \frac{A}{s} + \frac{B}{s^{2} } + \frac{C}{s+3}$

$\frac{9}{s^{2} (s+3} = \frac{As(s+3)+B(s+3)+Cs^{2} }{s^{2} (s+3)}$

On simplification we get

9 = A s(s+3) +B(s+3) +C(s²) ...(i)

Put s =0 in equation(i)

9 = B(0+3)

B = 9/3 = 3

Put s = -3 in equation(i)

9 = C(-3)²

C = 1

Given Equation 9 = A s(s+3) +B(s+3) +C(s²) ...(i)

Comparing 'S²' coefficient on both sides, we get

9 = A s²+3 A s +B(s)+3 B +C(s²)

0 = A + C

put C=1 , becomes A = -1

$\frac{9}{s^{2} (s+3} = \frac{-1}{s} + \frac{3}{s^{2} } + \frac{1}{s+3}$

Step(iv):-

$y^{-}(s) = \frac{9}{s^{2} (s+3}+\frac{7}{s+3}$

$y^{-}(s) =9( \frac{-1}{s} + \frac{3}{s^{2} } + \frac{1}{s+3}) + \frac{7}{s+3}$

Applying inverse Laplace transform on both sides

$L^{-1} (y^{-}(s) ) =L^{-1} (9( \frac{-1}{s}) + L^{-1} (\frac{3}{s^{2} }) + L^{-1} (\frac{1}{s+3}) )+ L^{-1} (\frac{7}{s+3})$

By using inverse Laplace transform

$L^{-1} (\frac{1}{s} ) =1$

$L^{-1} (\frac{1}{s^{2} } ) = \frac{t}{1!}$

$L^{-1} (\frac{1}{s+a} ) =e^{-at}$

Final answer:-

Now the solution , we get

$Y (s) = 9( -1 +3 t + e^{-3 t} ) + 7 e ^{-3t}$

5 0

2 years ago

Solve for x: −2x + 5 < 7 x > −1 x < −1 x > −6 x < −6

mars1129 [50]

hopefully this helps :)

5 0

2 years ago

Read 2 more answers