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Anika [276]
1 year ago
7

What is the solution set of this compound inequality? 1 ≤ |x + 3| ≤ 4​

Mathematics
1 answer:
Anika [276]1 year ago
3 0

The solution is:

-7\leq x\leq -4 or -2\leq x\leq 1

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A turtle was swimming 2 feet below the surface of a pond. While looking for food, he went
Snezhnost [94]

Answer:

9 feet down

Step-by-step explanation:

2+7=9

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WORTH 20 POINTS PLS HELP!
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4 0
3 years ago
Solve this system of linear equations. Separate
alex41 [277]

Answer:

y = 7 and x =3

Step-by-step explanation:

use elimination method

so we gonna eliminate y first

9x = 1 - 4y

-7x = -7 + 4y

7| 9x = 1 - 4y

9| -7x = -7 + 4y

63x = 7 - 28y

-63x = -63 + 36y

add eqtn 1 to eqtn2

you will get

0 = -56 + 8y

56 = -56 + 56 + 8y

56 = 8y

56÷8 = 8y÷ 8

7 = y

also eliminate x as we have done wth y

4| 9x = 1 - 4y

4| -7x = -7 + 4y

36x = 4 - 16y

-28x = -28 + 16y

add the two equations

u will get

8x = -24 + 0

8x = -24

8x÷8 = 24÷ 8

x = 3

5 0
2 years ago
Need help ASAP Will mark u brainlyest
balandron [24]
Step 2 is wrong: 4x - x = 5x
X = -11 is the actual answer, pretty sure.
Hope this helps and please mark as brainliest. If you have any other doubts, feel free to ask.
4 0
2 years ago
The point P(2k, k) is equidistant from A(-2, 4) and B (7,-5).<br>Find the value of k.<br>​
mario62 [17]

Answer:

k = 3

Step-by-step explanation:

Using the distance formula

d = √ (x₂ - x₁ )² + (y₂ - y₁ )²

with (x₁, y₁ ) = A (- 2, 4 ) and (x₂, y₂ ) = P (2k, k)

AP = \sqrt{(2k+2)^2+(k-4)^2}

Repeat

with (x₁, y₁ ) = B (7, - 5) and P = (2k, k)

BP = \sqrt{(2k-7)^2+(k+5)^2}

Given that AP = BP, then

\sqrt{(2k+2)^2+(k-4)^2} = \sqrt{(2k-7)^2+(k+5)^2}

Square both sides

(2k + 2)² + (k - 4)² = (2k - 7)² + (k + 5)² ← expand factors on both sides

4k² + 8k + 4 + k² - 8k + 16 = 4k² - 28k + 49 + k² + 10k + 25

Simplify both sides by collecting like terms

5k² + 20 = 5k² - 18k + 74 ( subtract 5k² from both sides )

20 = - 18k + 74 ( subtract 74 from both sides )

- 54 = - 18k ( divide both sides by - 18 )

k = 3

7 0
3 years ago
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