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3 years ago

If square root of 5 - x = 4 , then x=

Mathematics

2 answers:

Arada [10]3 years ago

8 0

I'm guessing here, but I may be right. The answer is x=0
If I'm wrong please correct me!!

velikii [3]3 years ago

8 0

√5 - x = 4
let x = -11
√5 - -11 = √5 + 11 = √16 = 4 Ans

So x = -11

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PART TWO MATH TEST :)<br><br> Please Help! I will give brainliest and donuts <33

ra1l [238]

Hey there,

Question #1

The answer would be in the attachment below.
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Question #2

The answer would be in the attachment below.
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Question #3

The answer would be in the attachment below.
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Question 4#

The last one was kind of tricky. But, as I saw this attachment, I noticed on how the rectangle was actually 3/4 on the base and for the height, it was 1/2. So by doing this,we need to find the area, and we would multiply these both. 1/2 x 3/4 = 3/8 but by looking at your options, those are not simplified so . . .your answer would be 6/16 because 3x2=6 & 8x2=16.
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I really hope this can help you Amanda.

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6 0

3 years ago

I WILL GIVE U BRANLIEST What is the coefficient of x^2 in the simplified product of

Vikki [24]

Answer:

-14

Step-by-step explanation:

Okay first we must simplify the expression.

so this:(3x^2 + x-2)(x - 5)

becomes this:3x^3-14x^2-7x+10

then we look at the coefficient of x^2 and we see that it's -14

3 0

3 years ago

Read 2 more answers

If tan e = 1.3, what is the value of cot (90° - 9)?

ella [17]

Answer:

Step-by-step explanation:

We are given that:

$\tan(\theta)=1.3$

And we want to find:

$\cot(90^\circ -\theta)$

Remember that tangent and cotangent are co-functions. In other words, they follow the cofunction identities:

$\tan(\theta)=\cot(90^\circ-\theta)\text{ and } \cot(90^\circ-\theta)=\tan(\theta)$

Therefore, since tan(θ) = 1.3 and cot(90° - θ) = tan(θ), then cot(90° - θ) must also be 1.3.

Our answer is A.

7 0

3 years ago

Find the domain of the function. <br><br> y = <img src="https://tex.z-dn.net/?f=%5Csqrt%7Bx%7D%20%2B%201" id="TexFormula1" title

Step2247 [10]

Answer:

[0, infinity)

Step-by-step explanation:

A square root can only use positive numbers, otherwise it becomes imaginary. A function needs to be real to output real values. Therefore, the domain is everything from 0 onwards.

5 0

3 years ago

Find f · dr c for the given f and

ruslelena [56]

Parameterize the path $\mathcal C$ by

$\mathbf r(t)=x(t)\,\mathbf i+y(t)\,\mathbf j$

with

$x(t)=3\cos t$
$y(t)=3\sin t$

and $0\le t\le\pi$ . Then

$\displaystyle\int_{\mathcal C}\mathbf f(x,y)\cdot\mathrm d\mathbf r=\int_{t=0}^{t=\pi}\mathbf f(x(t),y(t))\cdot\dfrac{\mathrm d\mathbf r}{\mathrm dt}\,\mathrm dt$
$=\displaystyle\int_0^\pi(9\cos^2t\,\mathbf i+9\sin^2t\,\mathbf j)\cdot(-3\sin t\,\mathbf i+3\cos t\,\mathbf j)\,\mathrm dt$
$=27\displaystyle\int_0^\pi(\cos t\sin^2t-\sin t\cos^2t)\,\mathrm dt=-18$

5 0

3 years ago