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frez [133]
3 years ago
10

How do you solve 3(x-10)= -6 ?

Mathematics
2 answers:
Sergio039 [100]3 years ago
8 0

Answer:

X=12

Step-by-step explanation:

Step 1

3(x−10)=6

(3)(x)+(3)(−10)=6

3x+−30=6

3x−30=6

Step 2

3x−30+30=6+30

3x=36

Step 3

3x

3

=

36

3

x=12

Andrei [34K]3 years ago
6 0

Answer:

x=8

Step-by-step explanation:

3(x-10)=-6

3x-30=-6

3x=24

x=8

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2 years ago
Barry bought 18.50 meters of light rope at ​$0.480 per meter and three meters of heavy rope at​ $1.08 per meter. How much change
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Answer:$2.88

Step-by-step explanation:

7 0
3 years ago
The value of -7 + 24 ÷ (-3)(2) is _____.<br><br> Show work
densk [106]

Answer:

\boxed{=-23}

Explanation: You had to used pemdas stands for: p-parenthesis, e-exponents, m-multiply, d-divide, a-add, and s-subtracting that came from left to right.

First you had to multiply and divide from left to right.

24/(-3)=-8(2)

8*2=16

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Then you add and subtract from left to right.

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=-23

Final answer: \boxed{=-23}

Hope this helps!

And thank you for posting your question at here on brainly, and have a great day.

-Charlie

3 0
3 years ago
Consider the following region R and the vector field F. a. Compute the​ two-dimensional curl of the vector field. b. Evaluate bo
Shalnov [3]

Looks like we're given

\vec F(x,y)=\langle-x,-y\rangle

which in three dimensions could be expressed as

\vec F(x,y)=\langle-x,-y,0\rangle

and this has curl

\mathrm{curl}\vec F=\langle0_y-(-y)_z,-(0_x-(-x)_z),(-y)_x-(-x)_y\rangle=\langle0,0,0\rangle

which confirms the two-dimensional curl is 0.

It also looks like the region R is the disk x^2+y^2\le5. Green's theorem says the integral of \vec F along the boundary of R is equal to the integral of the two-dimensional curl of \vec F over the interior of R:

\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\iint_R\mathrm{curl}\vec F\,\mathrm dA

which we know to be 0, since the curl itself is 0. To verify this, we can parameterize the boundary of R by

\vec r(t)=\langle\sqrt5\cos t,\sqrt5\sin t\rangle\implies\vec r'(t)=\langle-\sqrt5\sin t,\sqrt5\cos t\rangle

\implies\mathrm d\vec r=\vec r'(t)\,\mathrm dt=\sqrt5\langle-\sin t,\cos t\rangle\,\mathrm dt

with 0\le t\le2\pi. Then

\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\int_0^{2\pi}\langle-\sqrt5\cos t,-\sqrt5\sin t\rangle\cdot\langle-\sqrt5\sin t,\sqrt5\cos t\rangle\,\mathrm dt

=\displaystyle5\int_0^{2\pi}(\sin t\cos t-\sin t\cos t)\,\mathrm dt=0

7 0
3 years ago
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Your answer will be 2
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