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3 years ago

X+21+3x+9 solve for x

Mathematics

1 answer:

frez [133]3 years ago

7 0

Answer:

This is not an equation. It is an expression, you can't solve for x in an expression

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Need help with part c

Artyom0805 [142]

First understand that this is a linear graph. Find 2 points on the graph. We can use (0,1) and (3,-3).

Look at how much the x increases, in this case the x value increases by 0+3, so 3.

Then see how much the y value increases (make sure to evaluate them in the same order) 1 + (-3) = -2.

So you know that the y value decreases by 2 units for every 3 unit increase in x. Therefore the slope is y=(-2/3)x

Then figure out what you add to the end. The y intercept is (0,1), so add 1 to the end of y=(-2/3)x to move it up.

Your resulting eq is y=(-2/3)x+1

8 0

3 years ago

Fernanda's expenditures for this month are: rent, $750.00; groceries, $346.00;

Ivahew [28]

The answer is 1719.09 because that’s the total

7 0

3 years ago

Read 2 more answers

Please Help). Use the figure below to answer the questions. A) Describe in words a sequenc of transformations that maps □ABC to

lesya [120]

Answer:

See below

Step-by-step explanation:

The initial ordered-pairs are $(x, y)$

We have a rotation of 90 degrees counterclockwise with respect to origin

Note. Previously the points were

$A(-3, 4), B(-3, 0), C(-1, 3)$

After the rotation, we have

$A(-4, -3), B(0, -3), C(-3, -1)$

Thus, $(x, y) \rightarrow (-y, x)$

Then shifting horizontally to the right 2 units, we get ΔA'B'C'

Thus,

$(x, y) \rightarrow (-y, x) \rightarrow (x+2, y)$

7 0

3 years ago

A trough of water is 20 meters in length and its ends are in the shape of an isosceles triangle whose width is 7 meters and heig

Vaselesa [24]

Answer:

a) Depth changing rate of change is $0.24m/min$ , When the water is 6 meters deep

b) The width of the top of the water is changing at a rate of $0.17m/min$ , When the water is 6 meters deep

Step-by-step explanation:

As we can see in the attachment part II, there are similar triangles, so we have the following relation between them $\frac{3.5}{10} =\frac{a}{h}$ , then $a=0.35h$ .

a) As we have that volume is $V=\frac{1}{2} 2ahL=ahL$ , then $V=(0.35h^{2})L$ , so we can derivate it $\frac{dV}{dt}=2(0.35h)L\frac{dh}{dt}$ due to the chain rule, then we clean this expression for $\frac{dh}{dt}=\frac{1}{0.7hL}\frac{dV}{dt}$ and compute with the knowns $\frac{dh}{dt}=\frac{1}{0.7(6m)(20m)}2m^{3}/min=0.24m/min$ , is the depth changing rate of change when the water is 6 meters deep.

b) As the width of the top is $2a=0.7h$ , we can derivate it and obtain $\frac{da}{dt}=0.7\frac{dh}{dt} =0.7*0.24m/min=0.17m/min$ The width of the top of the water is changing, When the water is 6 meters deep at this rate