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2 years ago

7

The volume of a sphere is 81 cubic centimeters. Which equation can be used to solve for the radius of the sphere?

1 answer:

sashaice [31]2 years ago

5 0

Answer:

81π=4/3πr^3. Hope this helps

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What is the volume of this square pyramid?

tatuchka [14]

Answer: 1296 cm³

Step-by-step explanation:

Since, the volume of a square pyramid is,

$V = \frac{a^2h}{3}$

Where a = side of the square base,

h = height of the pyramid,

By the given diagram,

a = 18 cm,

h = 12 cm,

Hence, the volume of the given pyramid,

$V = \frac{18^2\times 12}{3}$

$= \frac{324\times 12}{3}$

$=\frac{3888}{3}$

$=1296$ cube cm.

⇒ First option is correct.

5 0

3 years ago

The high temperature was recorded as 25.6°F. The low temperature that same day was recorded as −2.7°F. What was the difference b

VMariaS [17]

28.3

Step-by-step explanation:

the difference is the answer to a subtraction problem so we can plug in our numbers like this

25.6-(-2.7) which equals 28.3

a good thing to remember when subtracting negative numbers is that a positive and negative won't always make a negative like in this example

4 0

2 years ago

What is the value of A when we rewrite... (PLZ HELP QUICK)

Marina86 [1]

Answer:

<h2> $\frac{133}{8}$ </h2>

Step-by-step explanation:

Given,

${( \frac{5}{2} )}^{x} + {( \frac{5}{2} )}^{x + 3}$

$= {( \frac{5}{2}) }^{x} + {( \frac{5}{2}) }^{x} \times {( \frac{5}{2} )}^{3}$

$= ( \frac{5}{2} ) ^{x} (1 + {( \frac{5}{2} )}^{3}$

$= {( \frac{5}{2} )}^{x} (1 + \frac{125}{8} )$

$= {( \frac{5}{2} )}^{x} ( \frac{1 \times 8 + 125}{8} )$

$= {( \frac{5}{2}) }^{x} ( \frac{8 + 125}{8} )$

${( \frac{5}{2} )}^{x} ( \frac{133}{8} )$

Comparing with A • ${( \frac{5}{2}) }^{x}$

A = $\frac{133}{8}$

Hope this helps...

Good luck on your assignment...

7 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7Bx%2By%3D1%7D%20%5Catop%20%7Bx-2y%3D4%7D%7D%20%5Cright.%20%5C%5C%5Clef

brilliants [131]

Answer:

<em>(a) x=2, y=-1</em>

<em>(b) x=2, y=2</em>

<em>(c)</em> $\displaystyle x=\frac{5}{2}, y=\frac{5}{4}$

<em>(d) x=-2, y=-7</em>

Step-by-step explanation:

<u>Cramer's Rule</u>

It's a predetermined sequence of steps to solve a system of equations. It's a preferred technique to be implemented in automatic digital solutions because it's easy to structure and generalize.

It uses the concept of determinants, as explained below. Suppose we have a 2x2 system of equations like:

$\displaystyle \left \{ {{ax+by=p} \atop {cx+dy=q}} \right.$

We call the determinant of the system

$\Delta=\begin{vmatrix}a &b \\c &d \end{vmatrix}$

We also define:

$\Delta_x=\begin{vmatrix}p &b \\q &d \end{vmatrix}$

And

$\Delta_y=\begin{vmatrix}a &p \\c &q \end{vmatrix}$

The solution for x and y is

$\displaystyle x=\frac{\Delta_x}{\Delta}$

$\displaystyle y=\frac{\Delta_y}{\Delta}$

(a) The system to solve is

$\displaystyle \left \{ {{x+y=1} \atop {x-2y=4}} \right.$

Calculating:

$\Delta=\begin{vmatrix}1 &1 \\1 &-2 \end{vmatrix}=-2-1=-3$

$\Delta_x=\begin{vmatrix}1 &1 \\4 &-2 \end{vmatrix}=-2-4=-6$

$\Delta_y=\begin{vmatrix}1 &1 \\1 &4 \end{vmatrix}=4-3=3$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{3}{-3}=-1$

The solution is x=2, y=-1

(b) The system to solve is

$\displaystyle \left \{ {{4x-y=6} \atop {x-y=0}} \right.$

Calculating:

$\Delta=\begin{vmatrix}4 &-1 \\1 &-1 \end{vmatrix}=-4+1=-3$

$\Delta_x=\begin{vmatrix}6 &-1 \\0 &-1 \end{vmatrix}=-6-0=-6$

$\Delta_y=\begin{vmatrix}4 &6 \\1 &0 \end{vmatrix}=0-6=-6$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{-6}{-3}=2$

The solution is x=2, y=2

(c) The system to solve is

$\displaystyle \left \{ {{-x+2y=0} \atop {x+2y=5}} \right.$

Calculating:

$\Delta=\begin{vmatrix}-1 &2 \\1 &2 \end{vmatrix}=-2-2=-4$

$\Delta_x=\begin{vmatrix}0 &2 \\5 &2 \end{vmatrix}=0-10=-10$

$\Delta_y=\begin{vmatrix}-1 &0 \\1 &5 \end{vmatrix}=-5-0=-5$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-10}{-4}=\frac{5}{2}$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{-5}{-4}=\frac{5}{4}$

The solution is

$\displaystyle x=\frac{5}{2}, y=\frac{5}{4}$

(d) The system to solve is

$\displaystyle \left \{ {{6x-y=-5} \atop {4x-2y=6}} \right.$

Calculating:

$\Delta=\begin{vmatrix}6 &-1 \\4 &-2 \end{vmatrix}=-12+4=-8$

$\Delta_x=\begin{vmatrix}-5 &-1 \\6 &-2 \end{vmatrix}=10+6=16$

$\Delta_y=\begin{vmatrix}6 &-5 \\4 &6 \end{vmatrix}=36+20=56$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{16}{-8}=-2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{56}{-8}=-7$

The solution is x=-2, y=-7

4 0

3 years ago

a rectangular prism and its dimensions are shown in the drawing .What us the total surface area of the rectangular prism in Squa

Xelga [282]

Answer:

Superficie de un prisma rectangular fórmula

Área de un prisma rectangular = 2la + 2ah + 2lh.

Step-by-step explanation:

4 0

3 years ago