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3 years ago

Help with number 5 (c,d) pleaseee!!!

Mathematics

1 answer:

Advocard [28]3 years ago

4 0

D because of the input to the square radius of the units three squares over t the right

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What is the value of y?

V125BC [204]

Answer:

2nd answer : 6 $\sqrt{3}$

Step-by-step explanation:

144-36=108

$\sqrt{108)=$ 6 $\sqrt{3}$

5 0

3 years ago

If f(x)=x^2+3x+5 what is f(3+h) ?

andriy [413]

F(x)=x^2+3x+5
f(3+h)=(3+h)^2+3(3+h)+5
f(3+h)=9+6h+h^2+9+3h+5
f(3+h)=23+9h+h^2

7 0

3 years ago

Read 2 more answers

Trent is drawing a diagram of his house for a school project. He decides that every 5 feet in real life should be 2 inches in th

salantis [7]

Answer:

5 feet (60 inches) proportional to 2 inches is a 30:1 relationship.

15 feet (180 inches) in real life should equal 6 inches on the diagram.

180 divided by 30 = 6

Step-by-step explanation:

5 0

3 years ago

a grocery clerk has 16 oranges, 20 apples, and 24 pears. The clerk needs to put an equal number of apples, oranges, and pears in

wolverine [178]

Answer:

Step-by-step explanation:

Here you find the greatest common factor of 16,20 and 24

The greatest common factor of a set of whole numbers is the largest positive integer that divides evenly into all numbers with no remainder.

Here, the G.C.F will be

16 20 24

2 8 10 12

2 4 5 6

2×2=4

4 0

3 years ago

Find the general solution to 1/x dy/dx - 2y/x^2 = x cos x, y(pi) = pi^2

Finger [1]

Answer:

$\frac{y}{x^2}=\sin x+\pi$

Step-by-step explanation:

Consider linear differential equation $\frac{\mathrm{d} y}{\mathrm{d} x}+yp(x)=q(x)$

It's solution is of form $y\,I.F=\int I.F\,q(x)\,dx$ where I.F is integrating factor given by $I.F=e^{\int p(x)\,dx}$ .

Given: $\frac{1}{x}\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{2y}{x^2}=x\cos x$

We can write this equation as $\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{2y}{x}=x^2\cos x$

On comparing this equation with $\frac{\mathrm{d} y}{\mathrm{d} x}+yp(x)=q(x)$ , we get $p(x)=\frac{-2}{x}\,\,,\,\,q(x)=x^2\cos x$

I.F = $e^{\int p(x)\,dx}=e^{\int \frac{-2}{x}\,dx}=e^{-2\ln x}=e^{\ln x^{-2}}=\frac{1}{x^2}$ { formula used: $\ln a^b=b\ln a$ }

we get solution as follows:

$\frac{y}{x^2}=\int \frac{1}{x^2}x^2\cos x\,dx\\\frac{y}{x^2}=\int \cos x\,dx\\\\\frac{y}{x^2}=\sin x+C$

{ formula used: $\int \cos x\,dx=\sin x$ }

Applying condition: $y(\pi)=\pi^2$

$\frac{y}{x^2}=\sin x+C\\\frac{\pi^2}{\pi}=\sin\pi+C\\\pi=C$

So, we get solution as :

$\frac{y}{x^2}=\sin x+\pi$

4 0

3 years ago