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Leya [2.2K]
3 years ago
9

Add 5 + (-8) The answer is -3 TRUE OR FALSE

Mathematics
2 answers:
dimaraw [331]3 years ago
5 0

Answer:

True

Step-by-step explanation:

5 + (-8) is the same thing as 5-8

Hope this helps!!!

Yuri [45]3 years ago
3 0

Answer:

True

Step-by-step explanation:

5+ -8

Rearranging the order

-8 +5

-3

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Which best describes the numbers satisfying the inequality x &lt;7?
likoan [24]

Answer:

(B) all numbers less than 7

Step-by-step explanation:

Given the inequality:

x<7

It means that:

  • x cannot be equal to 7
  • x cannot be greater than 7

Therefore, x<7 consists <u>of all numbers less than 7.</u>

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Match each value to what it represents.
Serhud [2]

Answer:

Step-by-step explanation:

(5^3/2)^-2/3 = 0.2 ⇒ E

(256^0.5)^1.25 = 32 ⇒ C

(81^1/6)^3/2 = 3 ⇒ F

(1024^0.03)^20 = 64 ⇒ D

(1000^4/7)^7/3 = 10,000 ⇒ B

(49^5/2)^0.2 = 7 ⇒ A

Step-by-step explanation:

* Lets explain how to solve the problem

- If we have base x to a power n and all to the power of m then we

multiply the two powers to be one power on the base

- [(x^n)^m] = x^(nm)

* Lets solve the problem

∵

∴ (5^3/2)^-2/3 = 0.2 ⇒ E

∵

∵ 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2^8

∴

∴ (256^0.5)^1.25 = 32 ⇒ C

∵

∵ 81 = 3 × 3 × 3 × 3 = 3^4

∴

∴ (81^1/6)^3/2 = 3 ⇒ F (answer F must be 3 not 0.333)

∵

∵ 1024 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2^10

∴

∴ (1024^0.03)^20 = 64 ⇒ D

∵

∵ 1000 = 10 × 10 × 10 = 10³

∴

∴ (1000^4/7)^7/3 = 10,000 ⇒ B

∵

∴ (49^5/2)^0.2 = 7 ⇒ A

5 0
3 years ago
What are the solutions to 7x^2+10x-3=x^2+3x
marysya [2.9K]

Answer:

x = 1 / 3

x = -1.5

Step-by-step explanation:

First move everything onto one side.

6x^2 + 7x - 3 = 0

Divide by 6:

x^2 + 7/6x - 0.5 = 0

Now, you can move the 0.5 to the other side because it is determined that this is not factorable.

x^2 + 7/6x = 0.5

Now you have to complete the square.

Take 1 / 2 of 7 / 6 and square it.

49 / 144

x^2 + 7/6x + 49/144 = 0.5 + 49/144

(x + 7 / 12)^2 = 121 / 144

Square root:

x + 7 / 12 = ±11 / 12

x = 1 / 3, x = -1.5

7 0
4 years ago
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