Ask question

Ask question

All categories

3 years ago

5

Jeffrey has an aquarium in the shape of a rectangular prism. He is going to place a layer of sand over the base for his iguana.

The area of the base of the aquarium can be represented by the expression (42x3−14x2+21x) feet3. If the width of the aquarium is represented by the expression (7x) feet, what is the length of the aquarium?

1 answer:

nikitadnepr [17]3 years ago

8 0

Answer:

$Length = 6x^2 - 2x + 3$

Step-by-step explanation:

Given

$Area = 42x^3 - 14x^2 + 21x$

$Width = 7x$

Required

Determine the length of the aquarium

The area of a rectangular base is calculated as:

$Area = Length * Width$

Substitute values for Area and Width

$42x^3 - 14x^2 + 21x = Length * 7x$

Make Length the subject

$Length = \frac{42x^3 - 14x^2 + 21x}{7x}$

Factorize the numerator

$Length = \frac{7x(6x^2 - 2x + 3)}{7x}$

$Length = 6x^2 - 2x + 3$

You might be interested in

On the map of a town, the library is located at (–8, 8), and the seafood market is located at (–8, 2). Mary walks from the libra

sertanlavr [38]

6 units. have a great day!!!!

6 0

3 years ago

Read 2 more answers

Find the value of x and y <br> (5y-23) <br> (2x+13)<br> (3x)<br> 47

liq [111]

Answer:

x = 24

y = 19

Step-by-step explanation:

7 0

3 years ago

Circle A is shown. Secant W Y intersects tangent Z Y at point Y outside of the circle. Secant W Y intersects circle A at point X

timama [110]

Circle A is missing, so i have attached it.

Answer:

∠XYZ = 35°

Step-by-step explanation:

We want to find the angle ∠XYZ in the image attached.

To solve that, we will use the formula in the theorem for angle formed by secants or tangents. Thus;

∠XYZ = ½(arc WZ - arc XZ)

From the image, arc WZ = 175° and arc XZ = 105°

Thus;

∠XYZ = ½(175 - 105)

∠XYZ = ½(70)

∠XYZ = 35°

5 0

3 years ago

Read 2 more answers

Simplify using order of operations 9 divided by 3 + 4 times 3 divided by 6

Alchen [17]

Answer:

$5$

Step-by-step explanation:

$1. \: 3 + 4 \times 3 \div 6 \\ 2. \: 3 + 12 \div 6 \\ 3. \: 3 + 2 \\ 4. \: 5$

7 0

3 years ago

Match each series with the equivalent series written in sigma notation

PIT_PIT [208]

Answer:

$3 + 12 + 48 + 192 + 768 = \sum\limits^4_{n=0} 3 * 4^n$

$4 + 32 + 256 + 2048 + 16384 = \sum\limits^4_{n=0} 4 * 8^n$

$2 + 6 + 18 + 54 + 162 = \sum\limits^4_{n=0} 2* 3^n$

$3 + 15 + 75 + 375 + 1875 = \sum\limits^4_{n=0} 3* 5^n$

Step-by-step explanation:

Given

See attachment for complete question

Required

Match equivalent expressions

Solving (a):

$3 + 12 + 48 + 192 + 768$

The expression can be written as:

$3 \to 3*4^{0$ --- 0

$12 \to 3 * 4^{1$ ---- 1

$48 \to 3 * 4^{2$ --- 2

$192 \to 3 * 4^{3$ ---- 3

$768 \to 3 * 4^{4$ ---- 4

For the nth term, the expression is:

$Term = 3 * 4^{n$ ---- n

So, the summation is:

$3 + 12 + 48 + 192 + 768 = \sum\limits^4_{n=0} 3 * 4^n$

Solving (b):

$4 + 32 + 256 + 2048 + 16384$

The expression can be written as:

$4 \to 4 * 8^0$ --- 0

$32 \to 4 * 8^1$ ---- 1

$256 \to 4 * 8^2$ --- 2

$2048 \to 4 * 8^3$ ---- 3

$16384 \to 4 * 8^4$ ---- 4

For the nth term, the expression is:

$Term \to 4 * 8^n$ ---- n

So, the summation is:

$4 + 32 + 256 + 2048 + 16384 = \sum\limits^4_{n=0} 4 * 8^n$

Solving (c):

$2 + 6 + 18 + 54 + 162$

The expression can be written as:

$2 \to 2 * 3^0$ --- 0

$6 \to 2 * 3^1$ ---- 1

$18 \to 2 * 3^2$ --- 2

$54 \to 2 * 3^3$ ---- 3

$162 \to 2 * 3^4$ ---- 4

For the nth term, the expression is:

$Term \to 2 * 3^n$ ---- n

So, the summation is:

$2 + 6 + 18 + 54 + 162 = \sum\limits^4_{n=0} 2* 3^n$

Solving (d):

$3 + 15 + 75 + 375 + 1875$

The expression can be written as:

$3 \to 3 * 5^0$ --- 0

$15 \to 3 * 5^1$ ---- 1

$75 \to 3 * 5^2$ --- 2

$375 \to 3 * 5^3$ ---- 3

$1875 \to 3 * 5^4$ ---- 4

For the nth term, the expression is:

$Term \to 3 * 5^n$ ---- n

So, the summation is:

$3 + 15 + 75 + 375 + 1875 = \sum\limits^4_{n=0} 3* 5^n$

5 0

3 years ago