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Delvig [45]
3 years ago
13

For a line with the equation y=2/3x−4, the slope is:

Mathematics
1 answer:
statuscvo [17]3 years ago
5 0

Answer:

The answer is B. 2/3 because y=mx+b mx us always the slope and b is the y intercept

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TUU SILICU.
Shalnov [3]

Answer:

4.013 units

Step-by-step explanation:

We know that the volume of a cone is given by the following equation:

v = pi / 3 * r ^ 2 * h

we need to know the height, therefore we solve for h:

h = 3 * v / (pi * r ^ 2)

replacing v = 105 and r = 5

h = 3 * 105 / (3.14 * 5 ^ 2)

h = 4.013

Which means that the height measures 4.013 units

4 0
3 years ago
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I NEED HELP PLEASE I have been on this question for a hour
Alexus [3.1K]

Answer:

A-8

Step-by-step explanation:

Any other higher number plugged in makes the equation false

7 0
3 years ago
FInd the value of C.
My name is Ann [436]

Answer:

The side C equals to 10. hence the answer is letter A

Step-by-step explanation:

To solve this, we need to use trigonometric functions.

we know that sin (α) = Lo / H for a triangle. Being Lo : Length of the opposite side and H: Length of the hypotenuse.

Given α= 45º and Lo= 5√2 and replacing in the equation:

sin (45º) = 5√2 / C (1)

Using trigonometric identities we know that sin(45º) =(√2)/2. Replacing in equation (1):

sin (45º) = (√2)/2 = 5√2 / C ⇒ C = 2 *5 *√2 / (√2) ⇒ C=10

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3 years ago
Solve 5x^2=10x + 4<br><br> Help please I literally forgot what topic this is! Thanksssss !!!!
faltersainse [42]
X = 4/15. The topic is solving for unknowns in an equation.
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3 years ago
7-18 use part 1 of the fundamental theorem of calculus to find the derivative of the function.
blondinia [14]

\displaystyle h(x)=\int\limits_{1}^{\sqrt{x}}~\cfrac{z^2}{z^4+1}dz~\hspace{10em}\cfrac{dh}{du}\cdot \stackrel{chain~rule}{\cfrac{du}{dx}\implies \cfrac{dh}{dx}} \\\\[-0.35em] ~\dotfill\\\\ u=\sqrt{x}\implies \cfrac{du}{dx}=\cfrac{1}{2\sqrt{x}} \\\\[-0.35em] ~\dotfill

\cfrac{dh}{dx}\implies \displaystyle \cfrac{d}{du}\left[ \int\limits_{1}^{u}~\cfrac{z^2}{z^4+1}dz \right]\cdot \cfrac{1}{2\sqrt{x}}\implies \left[ \cfrac{u^2}{u^4+1} \right]\cdot \cfrac{1}{2\sqrt{x}} \\\\\\ \stackrel{substituting~back}{\left[ \cfrac{(\sqrt{x})^2}{(\sqrt{x})^4+1} \right]\cdot \cfrac{1}{2\sqrt{x}}}\implies \cfrac{x}{x^2+1}\cdot \cfrac{1}{2\sqrt{x}}\implies \cfrac{\sqrt{x}}{2x^2+2}

7 0
2 years ago
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