All categories

3 years ago

Which three expressions are equivalent to 10x +11

Mathematics

1 answer:

BARSIC [14]3 years ago

8 0

Answer:

B, C, and E

Step-by-step explanation:

<u>Simplify all answer choices:</u>

5(2x+10)+1=

10x+50+1=

10x+51

7(x+2)+3x-3=

7x+14+3x-3=

10x+11

3(3x+4)+x-1=

9x+12+x-1=

10x+11

2(6x+4)+2x+5=

12x+8+2x+5=

14x+13

2(6x+5)-2x+1=

12x+10-2x+1=

10x+11

Answer choices B, C, and E simplify to 10x+11

You might be interested in

what is an equivalent fraction to 3/8 ? Find a common denominator for the pair of fractions. Then, write equivalent fractions

Ann [662]

Answer:

6 1/6

Step-by-step explanation:

A fraction that is equivalent to 3/8 is 6 1/6

5 0

3 years ago

Beth’s music playlist contains songs she bought and songs her mother bought. The ratio of her songs to her mother’s songs is 3:2

denpristay [2]

Answer:

IDK

nvm its writ 5 in beths colum

sorry for my spelling

Step-by-step explanation:

i got it right on apex

3 0

3 years ago

So what is a dilation and how will it effect the image and the preimage(i have 8th grade math).

Gennadij [26K]

Answer:

In geometry, a transformation is an operation that moves, ﬂips, or changes a shape to create a new shape. A dilation is a type of transformation that enlarges or reduces a ﬁgure (called the preimage) to create a new ﬁgure (called the image).

Step-by-step explanation:

6 0

3 years ago

9(w - 2) = – 18<br> someone help me

drek231 [11]

Step-by-step explanation:

First multiply

9w - 18 = - 18

9w = - 18 + 18

9w = 0

w = 0 / 9 = 0

w = 0

6 0

3 years ago

Read 2 more answers

I need help writing the first 5 terms of the sequence!

Angelina_Jolie [31]

The first five terms of the sequence are 1, 4, 7, 10, 13.

Solution:

Given data:

$a_{1}=1$

$a_{n}=a_{n-1}+3$

General term of the arithmetic sequence.

$a_{n}=a_{n-1}+d$ , where d is the common difference.

d = 3

$a_{n}=a_{n-1}+3$

Put n = 2 in $a_{n}=a_{n-1}+3$ , we get

$a_{2}=a_1+3$

$a_{2}=1+3$

$a_2=4$

Put n = 3 in $a_{n}=a_{n-1}+3$ , we get

$a_{3}=a_2+3$

$a_{3}=4+3$

$a_3=7$

Put n = 4 in $a_{n}=a_{n-1}+3$ , we get

$a_{4}=a_3+3$

$a_{4}=7+3$

$a_4=10$

Put n = 5 in $a_{n}=a_{n-1}+3$ , we get

$a_{5}=a_4+3$

$a_{5}=10+3$

$a_5=13$

The first five terms of the sequence are 1, 4, 7, 10, 13.

3 0

3 years ago