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gayaneshka [121]

2 years ago

12

On a number line, show all points whose coordinates satisfy the following inequalities: |x−20|<5

2 answers:

ivann1987 [24]2 years ago

5 0

Answer:

Step-by-step explanation:

VLD [36.1K]2 years ago

4 0

Answer:

On the number line integers from 16 to 24 is the answer.

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Any one smart and can help me catch up? ill give many points

Sergio [31]

Answer:

4/15

Step-by-step explanation:

1) find common denominator, 15 in this problem

$\frac{4}{5}= \frac{12}{15}$

2) solve

$\frac{12}{15}- \frac{8}{15} = \frac{4}{15}$

7 0

2 years ago

Please answer this question for me. I’ll mark you as the brainliest.

kirza4 [7]

Step-by-step explanation:

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5 0

2 years ago

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Solve . 3/27 = w/54

WINSTONCH [101]

$\frac{3}{27}=\frac{w}{54}$ Multiply 54 to both sides to get "w" by itself

$(54)\frac{3}{27}=\frac{w}{54}(54)$

$\frac{162}{27} =w$

6 = w

6 0

2 years ago

Read 2 more answers

Use the following image to assist in your answer.<br> a =<br> - 6<br> - 9<br> - 4

vfiekz [6]

Step-by-step explanation:

Pythagoras' theorem for the smallest one :

$c^2 = 4^2 + 6^2$

$c^2 = 16 + 36$

$c^{2}$ = 52

Pythagoras' theorem for the middle one :

$b^{2}$ = $6^{2}$ + $a^{2}$

Pythagoras' theorem for the biggest one :

$(4+a)^2 = c^2 + b^2$

$16 + 8a + a^2 = 52 + b^2$

Using the formula before (for $b^2$ ) it becomes :

$16 + 8a + a^2 = 52 + (6^2 + a^2)$

$16 + 8a + a^2 = 52 + 6^2 + a^2$

$16 + 8a = 52 + 6^2$

16 + 8a = 52 + 36

16+8a = 88

8a = 88-16

8a = 72

$a = \frac{72}{8}$

a = 9

Verifying :

$b^2 = 6^2 + a^2$

$b^2 = 36 + 81$

$b^2 = 117$

$b^{2}$ = 117

The biggest one :

$(4+a)^2 = c^2 + b^2$

$(4+9)^2 = 52 + 117$

$13^2 = 169$

True

8 0

2 years ago

Which of the following is NOT a rational number?

Tatiana [17]

Answer:

D

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers