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2 years ago

11

Mke y the subject of the formula. w=x^2-2yz

1 answer:

charle [14.2K]2 years ago

7 0

Answer:

W = x^2 - 2yz / - w

0 = x^2 - 2yz - w / +2yz

2yz = x^2 - w // 2z

y = (x^2 -w ) / 2z

So correct answer is D

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SpyIntel [72]

$QUADRATIC \: \: RESOLUTIONS \\ \\ \\\\4 {x}^{2} \: + \: 40x \: + \: 84 \: = \: 0 \\ \\ 4 \: ( \: {x}^{2} + 10x + 21) \: = \: 0 \\ \\ 4 \: can \: never \: be \: equal \: to \: \: 0 \\ \\ {x}^{2} \: + \: 10x \: + \: 21 = \: 0 \\ \\ {x}^{2} + \: 7x \: + 3x \: + \: 21 = 0 \\ \\ x(x + 7) + 3(x + 7) = 0 \\ \\ ( \: x + 7 \: ) \: ( \: x + 3 \: ) \: = \: 0 \\ \\ On \: solving \: \: , \: \: \: We \: get \: \: \: \: \: - \\ \\ \\ x \: = \: - 7 \: \: \: or \: \: \: x = \: - 3 \: \: \: \: \: \: \: \: Ans.$

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2 years ago

Please answer this question! 32 points and brainliest!!

stepladder [879]

Answer:

3x^2 -7x +2

Step-by-step explanation:

(15x^2 -35x+10) ÷5

(15/5x^2 -35/5x+10/5)

3x^2 -7x +2

This will factor

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5 0

3 years ago

Please help! what is the polar equation r = -5 theta in parametric form?<br><br> thank you!

Rudiy27

Use these equations when converting polar equations to parametric equations:

$x=r\cos\theta$

$y=r\sin\theta$

We know that $r=5\theta$ , so substitute that into both equations for x and y.

$x=5\theta\cos\theta$

$y=5\theta\sin\theta$

Now, replace $\theta$ with any variable that you want to represent the parametric equations in. I'll use the standard variable, $t$

$x=5t\cos t$

$y=5t\sin t$

Thus, $r=5\theta$ represented in parametric form is:

$(x,y)=(5t\cos t, 5t \sin t)$

Let me know if you need any clarifications, thanks!

~ Padoru

4 0

3 years ago

Please Answer ASAP

Masteriza [31]

hence, the common ratio of the given G.P. is 3 or −3.

6 0

2 years ago

Read 2 more answers

Given that k is a positive integer. Find, in terms of k, an expression for Sn, which is the sum of the integers from 2k to 4k in

Elden [556K]

Answer:

$S_n=3k\cdot (2k+1)=6k^2 +3k$

Step-by-step explanation:

k is a positive integer.

Consider an arithmetic sequence:

$a_1=2k\\ \\a_n=4k\\ \\d=1$

First, find n:

$a_n=a_1+(n-1)\cdot d\\ \\4k=2k+(n-1)\cdot 1\\ \\2k=n-1\\ \\n=2k+1$

Now, find the sum of these 2k+1 terms:

$S_n=\dfrac{a_1+a_n}{2}\cdot n\\ \\S_n=\dfrac{2k+4k}{2}\cdot (2k+1)=\dfrac{6k}{2}\cdot (2k+1)=3k\cdot (2k+1)=6k^2 +3k$

7 0

3 years ago