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3 years ago

Julien threw x number of pitches at baseball practice. Jacob threw 1/5 fewer than that

Mathematics

1 answer:

Naily [24]3 years ago

5 0

Answer:

ok so jacob threw 1/5 of x

Step-by-step explanation:

if you say julien threw x and jacob threw 1/5 of that than jacob threw 1/5 of x

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You have a fishing line spool with an end that has an area of 20.0 cm2. How much fishing line do you need to wind around the spo

lana66690 [7]

<h3>Problem Solution</h3>

Assuming the spool is a cylinder and the circumference we're winding around is that of a circle with the given area, we can write the relation between circumference and area as

... C = 2√(πA)

10 times the circumference is then

... 10C = 20√(π·20 cm²) = 40√(5π) cm ≈ 159 cm

<h3>Formula Derivation</h3>

The usual formulas for circumference and area are

... C = 2πr

... A = πr²

If we multiply the area formula by π and take the square root, we get

... πA = (πr)²

... √(πA) = πr

Multiplying this by 2 gives circumference.

... C = 2√(πA) = 2πr

3 0

3 years ago

Match the following pair of angles with their respective definitions.HELP

Andreyy89

Answer:

Step-by-step explanation:

3 and 7 - corresponding

2 and 7 - alt exterior

4 and 5 - alt interior

3 and 6 - alt interior

8 0

3 years ago

Read 2 more answers

The answer is 585 R4 and could also be written as 585 4/8. Simplify the answer.

Digiron [165]

Answer:

585 1/2 ?

Step-by-step explanation:

i dont know if you put the whole question but i simplified 4/8 and got 1/2.

so i guess the answer your looking for is 585 1/2 ...

4 0

2 years ago

D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x

MA_775_DIABLO [31]

The solution depends on the value of $k$ . To make things simple, assume $k>0$ . The homogeneous part of the equation is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0$

and has characteristic equation

$r^2-16k=0\implies r=\pm4\sqrt k$

which admits the characteristic solution $y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}$ .

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be $y_p=ae^{4x}+be^x$ . Then

$\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x$

So you have

$16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x$
$(16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x$

This means

$16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}$
$b(1-16k)=30\implies b=\dfrac{30}{1-16k}$

and so the general solution would be

$y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x$

8 0

2 years ago

What is y=x-4;(-2,3)

Yuki888 [10]

Answer:

Step-by-step explanation:

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2 years ago