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2 years ago

I don't know how to do this. If you can help that would be awesome.

Mathematics

1 answer:

AfilCa [17]2 years ago

4 0

Use the volume of a cylinder equation, V = pi*r^2*h to find the volume of the cylinder. To find the answer to the second one, find the volume of the square and subtract the volume of the cylinder from it.

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Write the sentence as an equation, using x as the variable. Then find the solution of the equation from the set of integers bet

Inessa05 [86]

Answer:

The solution of given equation is x = -4

Step-by-step explanation:

We are given the following in the question:

Let x be the variable such that

$x \in (-12,12)$

Relation:

The quotient of a number and 4 is negative 1

We can write the equation as:

$\dfrac{x}{4} = -1$

Solution:

$\dfrac{x}{4} = -1\\\\\Rightarrow x = 4\times -1\\\Rightarrow x = -4$

Thus, the solution of given equation is x = -4

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3 years ago

How to find instantaneous rate of change algebraically?

frosja888 [35]

You use (y2-y1)/(x2-x1) so if you have two points (ex: (1,2) and (2,4)), plug in those number into the formula
(4-2)/(2-1) = 2/1 = 2 so the slope (or instantaneous rate of change) is 2 for those two points<span />

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3 years ago

Review Lines and Parabolas

Virty [35]

Answer:

Step-by-step explanation:

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2 years ago

Help I don’t understand at all

zloy xaker [14]

Answer:

$1)\ \ 4h^2-13h+6\\2)\ \ 7x^3y^2-x^2y+1\\3)\ \ -7n+2\\4)\ \ -8m+4$

Step-by-step explanation:

Simplify the expression by combining like terms. Remember, like terms have the same variable part, to simplify these terms, one performs operations between the coefficients. Please note that a variable with an exponent is not the same as a variable without the exponent. A term with no variable part is referred to as a constant, constants are like terms.

$2h^2-7h+2h^2-h+6+4h-9h$

$(2h^2+2h^2)+(-7h-h+4h-9h)+(6)$

$4h^2-13h+6$

Use a very similar method to solve this problem as used in the first. Please note that all of the rules mentioned in the first problem also apply to this problem; for that matter, the rules mentioned in the first problem generally apply to any pre-algebra problem.

$8x^3y^2-7x^2y+8x-4-x^3y^2+2x^2y+4x^2y-8x+5$

$(8x^3y^2-x^3y^2)+(-7x^2y+2x^2y+4x^2y)+(8x-8x)+(-4+5)$

$7x^3y^2-x^2y+1$

Use the same rules as applied in the first problem. Also, keep the distributive property in mind. In simple terms, the distributive property states the following ( $a(b+c)=(a)(b)+(a)(c)=ab+ac$ ). Also note, a term raised to an exponent is equal to the term times itself the number of times the exponent indicates. In the event that the term raised to an exponent is a constant, one can simplify it. Apply these properties here,