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adelina 88 [10]

3 years ago

15

Solve for x I don't know how to solve it

1 answer:

mixer [17]3 years ago

8 0

Answer:

x=-12

Step-by-step explanation:

32+2x=6+x+14

2x-x=6+14-32

x=-12

ST=6

TU=-12+14=2

SU=6+2=8

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the net of the rectangular prism is shown below. use a ruler to measure the dimensions of the net to the nearest half centimeter

Arte-miy333 [17]

Answer:

the answer is D. 40.5 cm

8 0

3 years ago

Correctly order the proceed for solving for x using the algebraic properties of equality.

Snezhnost [94]

Answer:

3

1

2

Step-by-step explanation:

Step 1: Subtract 9

50.24 = 3.14x²

Step 2: Divide by 3.14

x² = 16

Step 3: Square root both sides

x = ±4

6 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

2 years ago

Can some help me please<br>

Darina [25.2K]

Answer:

i cant see the first few

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Jenny's aquarium holds 25 quarts of water. She buys a new one that holds 32 quarts.

Virty [35]

Answer:

7 quarts

Step-by-step explanation:

you subtract the amount that the first aquarium can hold from the second amount

5 0

2 years ago