3 years ago

Im deleting brainly so heres my points!

Mathematics

2 answers:

ASHA 777 [7]3 years ago

5 0

Answer:

Thanks!

Step-by-step explanation:

nadezda [96]3 years ago

3 0

Answer:

thx you so much and why are you deleting it

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The Pythagorean Theorem can be used to find the missing side length of a triangle when two other side lengths are known.

lubasha [3.4K]

Answer:

The value of b is 21

Step-by-step explanation:

29=√20^2+b^2

or, 29=√400+b^2

or, 841=400+b^2

or, 841-400=b^2

or, √441=b

b=21

5 0

2 years ago

Carl solved 20 math problems in 55 minutes how many math problems will carl solve in 75mins

-BARSIC- [3]

You need to find out how many problems Carl can do in a minute (unit rate), which is 20/55 or .36363636363. If you multiply it by 75 mins, you get 27.27, so the answer would be 27.

5 0

3 years ago

Read 2 more answers

Hello, precalc, need help on finding csc

Ainat [17]

Recall the double angle identity for cosine:

$\cos(2x) = \cos^2(x) - \sin^2(x) = 1 - 2 \sin^2(x)$

It follows that

$\sin^2(x) = \dfrac{1 - \cos(2x)}2 \implies \sin(x) = \pm \sqrt{\dfrac{1-\cos(2x)}2} \implies \csc(x) = \pm \sqrt{\dfrac2{1-\cos(2x)}}$

Since 0° < 22° < 90°, we know that sin(22°) must be positive, so csc(22°) is also positive. Let x = 22°; then the closest answer would be C,

$\csc(22^\circ) = \sqrt{\dfrac2{1-\cos(44^\circ)}} = \sqrt{\dfrac2{1-\frac5{13}}} = \dfrac{\sqrt{13}}2$

but the problem is that none of these claims are true; cot(32°) ≠ 4/3, cos(44°) ≠ 5/13, and csc(22°) ≠ √13/2...

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3 years ago

Hector wants to run 18 hours by his first track meet . suppose he runs the same amount of hours in week 2 and week 3 of his trai

Harman [31]

Is this division or mutilation?
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4 0

4 years ago

Read 2 more answers

Look at the picture please hell

ohaa [14]

Answer:

2 one that's the right one

7 0

3 years ago

Im deleting brainly so heres my points!​

Im deleting brainly so heres my points!