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Andrei [34K]
2 years ago
5

A(5x + 3) = b + 15x a = (?) b = 5ax + 3a - 15x (?)

Mathematics
2 answers:
prisoha [69]2 years ago
8 0

Answer:

nice name man

Step-by-step explanation:

Paul [167]2 years ago
4 0
The answer is in the image attached. Sorry if it’s wrong.

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I don't know how to solve this please help.
snow_lady [41]

Simplify the expression.

Your answer is: 3738

3 0
3 years ago
Read 2 more answers
Can someone please<br> Help me?
Mila [183]

9514 1404 393

Answer:

  see below

Step-by-step explanation:

It is easiest to compare the equations when they are written in the same form.

The first set can be written in slope-intercept form.

  y = 2x +7

  y = 2x +7 . . . . add 2x

These equations are <em>identical</em>, so have infinitely many solutions.

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The second set can be written in standard form.

  y +4x = -5

  y +4x = -10

These equations <em>differ only in their constant</em>, so have no solutions.

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The third set is already written in slope-intercept form. The equations have <em>different slopes</em>, so have exactly one solution.

5 0
3 years ago
Sebastian wanted to order pizzas for $7 each. The delivery charge is $3.50. He has $20.
docker41 [41]

Answer:

he can buy 2

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Find the missing side of the right triangle. Round to the nearest 10th if necessary:
denpristay [2]
M=12.2
(7^2)+(10^2)=149
the square root of 149 is 12.2
3 0
2 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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