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Sunny_sXe [5.5K]
3 years ago
11

What is %90, percent of 20?

Mathematics
1 answer:
Nonamiya [84]3 years ago
4 0

Answer:

18

Step-by-step explanation:

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I need answers for 1 , 2, 4​
andreyandreev [35.5K]

Answer:

(3) x ≥ -3

(4) 2.5 gallons

(4) -12x + 36

Step-by-step explanation:

<em>Hey there!</em>

1)

Well its a solid dot meaning it will be equal to.

So we can cross out 1 and 2.

And it's going to the right meaning x is greater than or equal to -3.

(3) x ≥ -3

2)

Well if each milk container has 1 quart then there is 10 quarts.

And there is 4 quarts in a gallon, meaning there is 2.5 gallons of milk.

(4) 2.5 gallons

4)

16 - 4(3x - 5)

16 - 12x + 20

-12x + 36

(4) -12x + 36

<em>Hope this helps :)</em>

3 0
3 years ago
Find the product. (4m^2 – 5)(4m^2 + 5)
aliya0001 [1]

Answer: 16m^4−25


Step-by-step explanation:

(4m^2−5)(4m^2+5)

=(4m^2+−5)(4m^2+5)

=(4m^2)(4m^2)+(4m^2)(5)+(−5)(4m^2)+(−5)(5)

=16m^4+20m^2−20m^2−25

=16m^4−25

3 0
3 years ago
-3(7w-3)+8w=-5(w+1) what is the value of w
bogdanovich [222]

Answer:

2

Step-by-step explanation:

Step 1:

- 21w + 9 + 8w = - 5w - 5

Step 2:

- 13w + 9 = - 5w - 5

Step 3:

- 7w + 9 = - 5

Step 4:

- 7w = - 14

Step 5:

14 = 7w

Answer:

w = 2

Hope This Helps :)

6 0
3 years ago
Viết các lũy thừa sau dưới dạng một lũy thừa hoặc các tích các lũy thừa có số nguyên và số nguyên A.8 mũ 2 B. 654​
Semmy [17]

Step-by-step explanation:

câu trả lời là trong hình ảnh trên

5 0
3 years ago
Find the value of the discriminant for <img src="https://tex.z-dn.net/?f=7x%5E%7B2%7D%20%2B5x%2B1%3D0" id="TexFormula1" title="7
kondor19780726 [428]

Answer:

No real roots

Step-by-step explanation:

Given

7x² + 5x + 1 = 0 ← in standard form

with a = 7, b = 5, c = 1

To determine the nature of the roots use the discriminant

Δ = b² - 4ac

• If b² - 4ac > 0 then roots are real and distinct

• If b² - 4ac = 0 then roots are real and equal

• If b² - 4ac < 0 then the roots are not real

Here

b² - 4ac = 5² - (4 × 7 × 1) = 25 - 28 = - 3

Thus the 2 roots are not real

7 0
3 years ago
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