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2 years ago

Solve the equation for b

Mathematics

2 answers:

Alona [7]2 years ago

6 0

Answer:

b=7a

Step-by-step explanation:

11a+b/3=6a

or, 11a+b=6a*3( when there is divide in one side it will be multiply when it is in another side)

or, 11a+b=18a(multiplying the number)

or, b=18a-11a( when there is + sign in one side it will be - sing in another side )

or, b=7a

Keith_Richards [23]2 years ago

5 0

Answer:

D. b=7a

Step-by-step explanation:

Add -11/3a to both sides.

$\frac{1}{3}b=\frac{7}{3}a$

Divide both sides by 1/3

1 /3 b ÷1/ 3 = 7/ 3 ÷a 1 /3

and the answer is

b=7a

ik this doesnt make any sense but hope this helps

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olga2289 [7]

Answer:

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Step-by-step explanation:

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2 years ago

Read 2 more answers

What number is 25% of 400 please show work thank you

ANEK [815]

Answer:

100

Step-by-step explanation:

well, you can just use a calculator, or your mind if that's possible, to solve this.

at least, what i think you're asking.

400 × 0.25 = 100.

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2 years ago

Graph the inverse circular function by hand, given f(x) = 3sin(2x) ;-pi/4 ≤ ≤pi/4

nasty-shy [4]

Part a)

a) The given function is

$f(x) = 3 \sin(2x)$

We let

$y = 3 \sin(2x)$

Interchange x and y.

$x= 3 \sin(2y)$

Solve for y;

$\frac{x}{3} = \sin(2y)$

$y = \frac{1}{2} { \sin}^{ - 1}( \frac{x}{3} )$

${f}^{ - 1}(x) = \frac{1}{2} { \sin}^{ - 1}( \frac{x}{3} )$

Part b) The range of f(x) refers to y-values for which f(x) exists.

The range of f(x) is

$- 3 \leqslant y \leqslant 3$

This is because the function is within y=-3 and y=3.

c) The range of

${f}^{ - 1} (x)$

$- \frac{ \pi}{4} \leqslant y \leqslant \frac{\pi}{4}$

The domain is -3≤x≤3

This is because the domain and range of a function and its inverse swaps.

Part d) The graph is shown in the attachment.

3 0

3 years ago

What’s the equation for this graph?

harkovskaia [24]

Answer:

y=-3x+2

Step-by-step explanation:

as x goes up on y goes down 3

slope is change in y over change in x

-3/1=-3

the y intercept is (0,2) so the y intercept is +2

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2 years ago

Question partpointssubmissions usedverify that the divergence theorem is true for the vector field f on the region

maxonik [38]

$\nabla\cdot\mathbf f(x,y,z)=\dfrac{\partial(3x)}{\partial x}+\dfrac{\partial(xy)}{\partial y}+\dfrac{\partial(5xz)}{\partial z}=3+x+5x=3+6x$

By the divergence theorem, the flux across the boundary of the given region $\mathcal E$ is

$\displaystyle\iint_{\partial\mathcal E}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\iiint_{\mathcal E}\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV$
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3 years ago

Solve the equation for b ​

Solve the equation for b