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Natalka [10]
3 years ago
9

Create a question like

Mathematics
1 answer:
vovikov84 [41]3 years ago
3 0

Answer:

x = (+ - ) 9

Step-by-step explanation:

x^2 = 81

u know how u can sometimes cancel something out by doing the opposite.

This is one of those times as well. the opposite of ^2 = square root.

so if we take the square root of both sides, this will cancel out the^2 on the left side.

sqrt (x^2) = sqrt(81)

x = (+-) 9.....thats a positive and a negative sign...because the sqrt of 81 is both 9 and -9.

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A machine requires three hours to make a unit of Product A and five hours to
svlad2 [7]

Let A unit be a; B unit be b

a + b = 95

b = 95 - a

3a + 5b = 395

3a + 5(95 -a) = 395

3a + 475 - 5a = 395

-2a = -80

a = 40

a + b = 95

40 + b = 95

b = 55

Therefore, a = 40; b = 55.

Hope this helps

8 0
3 years ago
−4+25=−4+25 solve for x show working
pychu [463]

Answer:

Its already solved, -4+25 does equal -4+25

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
What is the answer to 0.7(1.5 + y) = 3.5y - 1.47
Brums [2.3K]

Answer:

0.9

Step-by-step explanation:

0.7*(1.5 + y) = 3.5y - 1.47

or, (0.7*1.5) + 0.7y = 3.5y - 1.47

or, 1.05 + 1.47 = 3.5y - 0.7y

or, 2.52 = 2.8y

or, y = 2.52/2.8

or, y = 0.9

Hope this will help. Please mark me brainliest.

4 0
2 years ago
Complete the statement, f( ) = -5.<br><br> Blank 1: <br> Plz answer asap!
tankabanditka [31]
It’s 12 that’s the simple answer
4 0
3 years ago
Integrate ​G(x,y,z)equalsz over the parabolic cylinder yequalszsquared​, 0less than or equalsxless than or equals2​, 0less than
yKpoI14uk [10]

I gather you're supposed to compute the integral of G(x,y,z)=z over a surface S that is the part of the parabolic cylinder y=z^2 with 0\le x\le2 and 0\le z\le\frac{\sqrt{15}}2.

We can parameterize S by

\vec s(x,z)=x\,\vec\imath+z^2\,\vec\jmath+z\,\vec k

with the given constraints on x and z. Take the normal vector to S to be

\vec s_x\times\vec s_z=-\vec\jmath+2z\,\vec k

so that the surface element is

\mathrm dS=\|\vec s_x\times\vec s_z\|\,\mathrm dx\,\mathrm dz=\sqrt{1+4z^2}\,\mathrm dx\,\mathrm dz

Then in the integral, we have

\displaystyle\iint_Sz\,\mathrm dS=\int_0^2\int_0^{\sqrt{15}/2}z\sqrt{1+4z^2}\,\mathrm dz\,\mathrm dx=\boxed{\frac{21}2}

5 0
3 years ago
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