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3 years ago

ASAP! can someone help solve this with work?

Mathematics

1 answer:

KengaRu [80]3 years ago

7 0

Answer:

I don't just want to tell you the answer but I can tell you how.

Step-by-step explanation:

Use the pythagorean theorem, since this is a right triangle. If you don't know what this is its a^2+b^2=c^2

I hope this helps, if you have questions comment or look online.

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Fill in the blank in the sentence below with the correct term. The _________ of the numbers -4 and 4 is 4

Ray Of Light [21]

Answer:

Absolute value

Step-by-step explanation:

Absolute value is how many numbers it takes to get back to zero therefore it can never be negative, I think the term absolute value fits the blank perfectly

5 0

3 years ago

Factor n(2n+3)+4(2n-3)

zavuch27 [327]

The answer would be 2n^2+3n+ 8n-1

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2 years ago

For the following set of 20 numbers:

patriot [66]

Answer:

I cannot answer all of this, so I will only answer what I can:

Q1: 13

Q3: 22.5

IQR: 13.5

IQR(1.5): 20.25

Q3 + IQR(1.5) = 43

Q1 - IQR(1.5) = -7.25

I don't know if my calculations are all right but I hope this helps! :)

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2 years ago

A composite figure is made up of

son4ous [18]

a composit figure is made up of two figures

6 0

3 years ago

Read 2 more answers

Prove :<br>sin²θ + cos²θ = 1<br><br><br>thankyou ~

Gnom [1K]

Answer:

See below

Step-by-step explanation:

Here we need to prove that ,

$\sf\longrightarrow sin^2\theta + cos^2\theta = 1$

Imagine a right angled triangle with one of its acute angle as $\theta$ .

The side opposite to this angle will be perpendicular .
Also we know that ,

$\sf\longrightarrow sin\theta =\dfrac{p}{h} \\$

$\sf\longrightarrow cos\theta =\dfrac{b}{h}$

And by Pythagoras theorem ,

$\sf\longrightarrow h^2 = p^2+b^2 \dots (i)$

Where the symbols have their usual meaning.

Now , taking LHS ,

$\sf\longrightarrow sin^2\theta +cos^2\theta$

Substituting the respective values,

$\sf\longrightarrow \bigg(\dfrac{p}{h}\bigg)^2+\bigg(\dfrac{b}{h}\bigg)^2\\$

$\sf\longrightarrow \dfrac{p^2}{h^2}+\dfrac{b^2}{h^2}\\$

$\sf\longrightarrow \dfrac{p^2+b^2}{h^2}$

From equation (i) ,

$\sf\longrightarrow\cancel{ \dfrac{h^2}{h^2}}\\$

$\sf\longrightarrow \bf 1 = RHS$

Since LHS = RHS ,

Hence Proved !

I hope this helps.

5 0

2 years ago