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Nimfa-mama [501]
3 years ago
12

Please help me I have 5 minutes to do

Mathematics
2 answers:
vivado [14]3 years ago
4 0
A: 2 (once the red circles are removed)
B: 3+ (-5) = 3-5 = -2
C: (-1)+3 = 2
D: 4-(-1) = 4+1 = 5

Could be two answers, in my opinion:
B because it’s the only negative number or D because absolute value of all others is 2
DochEvi [55]3 years ago
3 0

Answer:

B

Step-by-step explanation:

IM 100% SURE

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Can the -1 in front of - (y - 1) be distributed?
olchik [2.2K]

-1(y-1)

-y+1 would be the simplified version so yes, it can be distributed

I hope this helps

-Ayden

6 0
3 years ago
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I WILL MARK BRAINLEST!!! FOR YOU CAN ANSWER THESE TWO
ludmilkaskok [199]

Answer:

First question: 1.4

Second question: Option 3

Step-by-step explanation:

First question:

You simplify the left equation first,

-0.4 x -3.5 = 1.4 , -0.4 x 2.9 = -1.16

<u>1.4k - 1.16</u>

For this to equal the right equation, the number missing must be 1.4.

Second question:

You simplify the equation given to you first,

1/4 x 8 = 2 , 1/4 x 12 = 3

<u>2x + 3</u>

The option that is equivalent to the answer is option 3; 2x + 3

8 0
3 years ago
What is the midpoint of the line segment with endpoints (3.2,2.5) and (1.6-4.5)?
NikAS [45]

Answer:

(2.4, -1)

Step-by-step explanation:

To find the midpoints of two points in the format (x,y), we find the mean for the values of x and y.

In this question:

(3.2, 2.5) and (1.6, -4.5)

Mean for the values of x:

(3.2 + 1.6)/2 = 2.4

Mean for the values of y:

(2.5 - 4.5)/2 = -1

Midpoint:

(2.4, -1)

4 0
4 years ago
Solve for x! Please help!
erica [24]

Answer:

<h2>x = 12</h2>

Step-by-step explanation:

ΔADC and ΔCDB are similar (AA). Therefore the sides are in proportion:

\dfrac{AD}{DC}=\dfrac{DC}{DB}

We have AD = 16, DC = x, DB = 9. Substitute:

\dfrac{16}{x}=\dfrac{x}{9}            <em>cross multiply</em>

x^2=(16)(9)\to x=\sqrt{(16)(9)}\\\\x=\sqrt{16}\cdot\sqrt9\\\\x=4\cdot3\\\\x=12

3 0
3 years ago
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Answer:

five thousand one hundred seventy nine

Step-by-step explanation:

6 0
3 years ago
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