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3 years ago

11

Andre went to a sporting goods store that was having a different sale. He bought a baseball glove and 2 packages of socks. The b

aseball glove normally cost 34$.

1 answer:

Anna35 [415]3 years ago

6 0

Answer:

idk.........................

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A teacher would like to study how students' seating locations in the classroom affects their test scores. If he were

LUCKY_DIMON [66]

Answer:

B, C, D, maybe A

Step-by-step explanation:

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4 years ago

75% of n is 45. n is what percent of 90?

Law Incorporation [45]

Answer:

66.67%

Step-by-step explanation:

75 /100 = .75

.75n = 45

45/.75 = 60

60 is n

60/90 = .666667

.666667 * 100 = 66.67%

3 0

3 years ago

Read 2 more answers

Consider the function on the interval (0, 2π). f(x) = 7 sin2(x) + 7 sin(x) (a) Find the open interval(s) on which the function i

-BARSIC- [3]

Answer with Step-by-step explanation:

Given

$f(x)=7sin(2x)+7sin(x)$

Differentiating both sides by 'x' we get

$14cos(2x)+7cos(x)=f'(x)$

Now we know that for an increasing function we have

$f'(x)>0\\\\14cos(2x)+7cos(x)>0\\\\2cos(2x)+cos(x)>0\\\\2(2cos^{2}(x)-1)+cos(x)>0\\\\4cos^{2}(x)+cos(x)-2>0\\\\(2cos(x)+\frac{1}{2})^2-2-\frac{1}{4}>0\\\\(2cos(x)+\frac{1}{2})^2>\frac{9}{4}\\\\2cos(x)>\frac{3}{2}-\frac{1}{2}\\\\\therefore cos(x)>\frac{1}{4}\\\\\therefore x=[0,cos^{-1}(1/4)]\cup [2\pi-cos^{-1}(1/4),2\pi ]$

Similarly for decreasing function we have

$[tex]f'(x)$

Part b)

To find the extreme points we equate the derivative with 0

$f'(x)=0\\\\cos(x)=\frac{1}{4}\\\\x=cos^{-1}(\frac{1}{4})$

Thus point of extrema is only 1.

4 0

4 years ago

Use the laplace transform to solve the given initial-value problem. y' + 5y = e3t, y(0) = 2

Iteru [2.4K]

In mathematics, the Laplace transform, named after its discoverer Pierre Simon Laplace, 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).

Learn more about Laplace transform here brainly.com/question/17062586

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4 0

2 years ago

In engineering, the modulus of elasticity is a way to measure how much an object deforms along an axis when opposing forces are

Allushta [10]

Step-by-step explanation:

Consider an engineering material of initial length Lo, Area (A), Modulus of elasticity (E) and applied a force P due to which change in the length of the material is δ2 from it’s original length (Lo)

Initial length of the material is Lo. Hence, at time t = 0 when no force applied on the material the length of the material will not change (i.e., at time t=0, δ1 = 0)

Modulus of elasticity of the material:

$E=\frac{P \cdot L_{o}}{A\left[\delta_{2}-\delta_{1}\right]}$

Area of the material:

$E=\frac{P \cdot L_{o}}{A\left[\delta_{2}-\delta_{1}\right]}$

$A=\frac{P \cdot L_{o}}{E\left[\delta_{2}-\delta_{1}\right]}$

Length of the material:

$E=\frac{P \cdot L_{0}}{A\left[\delta_{2}-\delta_{1}\right]}$

$L_{0}=\frac{E \cdot A\left[\delta_{2}-\delta_{1}\right]}{P}$

5 0

3 years ago