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mart [117]
4 years ago
11

Fill in the blanks. Please answer as shown in the boxes. Please fill out all boxes. Will mark brainiest if correct

Mathematics
1 answer:
jarptica [38.1K]4 years ago
6 0

Answer:

(5x + 7y)² = 25x² + 70xy + 49y²

Step-by-step explanation:

(5x + 7y)² = 25x² + 70xy + 49y²

Because,

(5x + b)² = (5x)² + 2(5x)(b) + b²

2(5x)(b) = 70xy

b = 7y

b² = (7y)² = 49y²

You might be interested in
What is the greatest common factor of...
gladu [14]

Answer:

3 a²b

Step-by-step explanation:

This is what I did:

3 a²b ( 5b² - 7a²)

when you try to expand this, it will give you back the original expansion so 3a²b is the greatest common factor

5 0
3 years ago
Elvaluate 3^-2 jgigjdkcjcivkxjgfnv
BARSIC [14]

Answer:

1/9

Step-by-step explanation:

when you have a negative exponent, you move the term to the denominator: 1/3^2

then you just do 3^2: 1/9

4 0
4 years ago
Cinemark is having a summer special.If you get a membership one-time cost is $12.50 plus $2.00 per movie that you watch.Write a
lys-0071 [83]

Answer:

Y=12.50x+2.00

Step-by-step explanation:

8 0
3 years ago
You are looking for a new cell phone plan. The first company, Cellular-Tastic (f), charges a
Step2247 [10]

Answer:

200 minutes

Step-by-step explanation:

Given that:

Cellular - Tastic (f) :

Monthly fee = $35

Per minute fee = $0.11

Dirt cheap call (g) :

Monthly fee = $55

Per minute fee = $0.01

How many minutes would you need to use for the cell phones to cost the same

amount?

Let number of minutes for cost to be equal = x

35 + 0.11x = 55 + 0.01x

Collect like terms

0.11x - 0.01x = 55 - 35

0.1x = 20

x = 20/0.1

x = 200

Hence, for cost to be the Same, 200 minutes must be used

5 0
3 years ago
V =x × y × z solve for x
kogti [31]

Hello from MrBillDoesMath!

Answer:

x =  V/(y*z)


Discussion:

V = x * y*  z    =>          (divide both sides by y*z)

V/(y*z) = (x*y*z) / (y*z) =>

V/(y*z) = (y*z) * x /  (y*z) =>               ( as (y*z)/(y*z) = 1)

V/(y*z) = 1 * x  =>    

V/(y*z) = x


Thank you,

MrB



3 0
3 years ago
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