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3 years ago

5

HHEEEEEEEELLLLPPPPPPPP 11 POINTS

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svp [43]3 years ago

7 0

The answer is C........

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Please help me with pythagoras.

lana [24]

Answer:

(q)

$cb \: = \sqrt{4 {}^{2} + {3}^{2} }$

$cb = 5$

$q = \sqrt{ {13}^{2} - {5}^{2} }$

$q = 12$

(r)

$cb = \sqrt{ {13}^{2} - {5}^{2} }$

$cb = 12$

$r = \sqrt{12 {}^{2} - 5 {}^{2} }$

$r = \sqrt{119}$

$r = 10.91$

(s)

$ab = \sqrt{20 {}^{2} - 16 {}^{2} }$

$ab = 12$

$s = \sqrt{12 {}^{2} + 8 {}^{2} }$

$s = 4 \sqrt{13}$

$s = 14.42$

Step-by-step explanation:

8 0

3 years ago

How do you rewrite one and 2 thirds using twelfths

TiliK225 [7]

Answer:

1 2/3 = 1 8/12

Step-by-step explanation:

2/3 * 4/4 because 3*4= 12 is 8/12 hope this helps :)

3 0

3 years ago

Read 2 more answers

1. Find the unit rate for lessons at Sit and Stay School. Show your work. Write your answer in the space below.

timofeeve [1]

Answer: $8

Step-by-step explanation:

96/12=8/1

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3 years ago

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PLEASE HELP!! GIVING BRAINIEST WHEN TWO PEOPLE ANSWER!!!

solniwko [45]

Answer:

Bright lemonade

Step-by-step explanation:

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3 years ago

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How to evaluate the limit

anzhelika [568]

$\displaystyle\lim_{x\to2}\frac{x^2-x+6}{x+2}$

Both the numerator and denominator are continuous at $x=2$ , which means the quotient rule for limits applies:

$\dfrac{\displaystyle\lim_{x\to2}(x^2-x+6)}{\displaystyle\lim_{x\to2}(x+2)}=\dfrac{2^2-2+6}{2+2}=\dfrac84=2$

Perhaps you meant to write that $x\to-2$ instead? In that case, you would have

$\displaystyle\lim_{x\to-2}\frac{x^2-x+6}{x+2}=\lim_{x\to-2}\frac{(x+2)(x-3)}{x+2}=\lim_{x\to-2}(x-3)=-2-3=-5$

4 0

3 years ago