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4 years ago

15x-37=38 what is the answer to that i really really need it

2 answers:

6 0

Add 37 to each side of the equation to get 75. Now get x alone by dividing 15 on both sides. X gets isolated or by itself and 75 divided by 15 is 2,
so your final answer is: x=5.
Hope this helps:)

stealth61 [152]4 years ago

3 0

15x - 37 = 38
 + 37 + 37
15x = 75
÷15 ÷15
x = 5

To check:
15x - 37 = 38
15(5) - 37 = 38
75 - 37 = 38
38 = 38

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HELP PLEASE ASAP!!!

Bas_tet [7]

A. It's a composite function, so basically, wherever you see a p, replace it with 5t, because we are given that information. So, your answer is:
$A[p(t)] = 5t \pi 2=10t \pi$

B. Let's use the function we created, and just plug in 2 for t:
$A[p(2)] = 10(2) \pi$
$A[p(2)] = 62.83$

So, your answer is (approximately) 62.83 units².

3 0

3 years ago

Read 2 more answers

Solve the equation: <img src="https://tex.z-dn.net/?f=5%5Ex%5E%28%5Ex%5E-%5E1%5E%29%3D1" id="TexFormula1" title="5^x^(^x^-^1^)=1

Tomtit [17]

The concept:

We are given the equation:

$5^{x(x-1)} = 1$

Which can be simplified as:

$5^{x^{2} - x} = 1^{1}$

Since any number to the power '0' is 1

x² - x must be equal to 0 for the given equation to be true

Solving for x:

x² - x = 0

x(x-1) = 0

now, we can divide both sides by either x OR x-1

So we will see what we get for either choice:

x = 0/(x-1) x-1 = 0/x

x = 0 x = 1

Hence, the value of x is either 0 or 1

7 0

3 years ago

What are 1 of the 4 tips to make a fraction simpler?

bonufazy [111]

Hello there.

Question: What are 1 of the 4 tips to make a fraction simpler?

Answer: The easiest way is to factor the numerator and denominator.

Hope This Helps You!
Good Luck Studying ^-^

3 0

3 years ago

A giant tank in a shape of an inverted cone is filled with oil. the height of the tank is 1.5 metre and its radius is 1 metre. t

skad [1K]

The given height of the cylinder of 1.5 m, and radius of 1 m, and the rate

of dripping of 110 cm³/s gives the following values.

1) The rate of change of the oil's radius when the radius is 0.5 m is r' ≈ 9.34 × 10⁻⁵ m/s

2) The rate of change of the oil's height when the height is 20 cm is h' ≈ 1.97 × 10⁻³ m/s

3) The rate the oil radius is changing when the radius is 10 cm is approximately 0.175 m/s

<h3>How can the rate of change of the radius & height be found?</h3>

The given parameters are;

Height of the tank, h = 1.5 m

Radius of the tank, r = 1 m

Rate at which the oil is dripping from the tank = 110 cm³/s = 0.00011 m³/s

$1) \hspace{0.15 cm}V = \frac{1}{3} \cdot \pi \cdot r^2 \cdot h$

From the shape of the tank, we have;

$\dfrac{h}{r} = \dfrac{1.5}{1}$

Which gives;

h = 1.5·r

$V = \mathbf{\frac{1}{3} \cdot \pi \cdot r^2 \cdot (1.5 \cdot r)}$

$\dfrac{d}{dr} V =\dfrac{d}{dr} \left( \dfrac{1}{3} \cdot \pi \cdot r^2 \cdot (1.5 \cdot r)\right) = \dfrac{3}{2} \cdot \pi \cdot r^2$

$\dfrac{dV}{dt} = \dfrac{dV}{dr} \times \dfrac{dr}{dt}$

$\dfrac{dr}{dt} = \mathbf{\dfrac{\dfrac{dV}{dt} }{\dfrac{dV}{dr} }}$

$\dfrac{dV}{dt} = 0.00011$

Which gives;

$\dfrac{dr}{dt} = \mathbf{ \dfrac{0.00011 }{\dfrac{3}{2} \cdot \pi \cdot r^2}}$

When r = 0.5 m, we have;

$\dfrac{dr}{dt} = \dfrac{0.00011 }{\dfrac{3}{2} \times\pi \times 0.5^2} \approx 9.34 \times 10^{-5}$

The rate of change of the oil's radius when the radius is 0.5 m is r' ≈ 9.34 × 10⁻⁵ m/s

2) When the height is 20 cm, we have;

h = 1.5·r

$r = \dfrac{h}{1.5}$

$V = \mathbf{\frac{1}{3} \cdot \pi \cdot \left(\dfrac{h}{1.5} \right) ^2 \cdot h}$

r = 20 cm ÷ 1.5 = $13.\overline3$ cm = $0.1\overline3$ m

Which gives;

$\dfrac{dr}{dt} = \dfrac{0.00011 }{\dfrac{3}{2} \times\pi \times 0.1 \overline{3}^2} \approx \mathbf{1.313 \times 10^{-3}}$

$\dfrac{d}{dh} V = \dfrac{d}{dh} \left(\dfrac{4}{27} \cdot \pi \cdot h^3 \right) = \dfrac{4 \cdot \pi \cdot h^2}{9}$

$\dfrac{dV}{dt} = \dfrac{dV}{dh} \times \dfrac{dh}{dt}$

$\dfrac{dh}{dt} = \dfrac{\dfrac{dV}{dt} }{\dfrac{dV}{dh} }$

$\dfrac{dh}{dt} = \mathbf{\dfrac{0.00011}{\dfrac{4 \cdot \pi \cdot h^2}{9}}}$

When the height is 20 cm = 0.2 m, we have;

$\dfrac{dh}{dt} = \dfrac{0.00011}{\dfrac{4 \times \pi \times 0.2^2}{9}} \approx \mathbf{1.97 \times 10^{-3}}$

The rate of change of the oil's height when the height is 20 cm is h' ≈ 1.97 × 10⁻³ m/s

3) The volume of the slick, V = π·r²·h

Where;

h = The height of the slick = 0.1 cm = 0.001 m

Therefore;

V = 0.001·π·r²

$\dfrac{dV}{dr} = \mathbf{ 0.002 \cdot \pi \cdot r}$

$\dfrac{dr}{dt} = \mathbf{\dfrac{0.00011 }{0.002 \cdot \pi \cdot r}}$

When the radius is 10 cm = 0.1 m, we have;

$\dfrac{dr}{dt} = \dfrac{0.00011 }{0.002 \times \pi \times 0.1} \approx \mathbf{0.175}$

The rate the oil radius is changing when the radius is 10 cm is approximately 0.175 m

Learn more about the rules of differentiation here:

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3 0

3 years ago

What is Point-Slope Form? What is Slope Intercept Form? What is the Slope Formula?

kipiarov [429]

Answer:

For the first part

1.) Negative

2.)zero

3.) undefined

4.) positive

The point slope formula is y-y1=m(x-x1). It is best to use if you already know the slope and one other point on the line. It is used to find another point on the line.

The slope-intercept form is used to show a line. To use this form, you need to know the slope and the y-intercept.

The slope formula is y=mx+b. M is the slope while b is the y-intercept.

8 0

3 years ago