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2 years ago

15

The graph that correctly represents the inequality 2x > 50 is graph . The graph that correctly represents the inequality x +

6 < 32 is graph .

2 answers:

Nikitich [7]2 years ago

4 0

Answer:

Step-by-step explanation:Simplify the inequalities and match them with the graphs that represent them.

2x > 12

7x < 42

x + 1 < 8

3x > 15

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Nata [24]2 years ago

3 0

1) If <span>2x > 50 , then x > 25. Draw a circle at x=25 and then extend a ray in the positive x direction away from x = 25.

2) If x + 6 < 32, subtract 6 from both sides, obtaining x < 26.
Draw a cirlce at x=26 and extend a ray from that x to the left.</span>

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HELP it needs good reasoning

Sidana [21]

Answer:

Company B.

Step-by-step explanation:

On the graph, when x is 1, y is 50. That means it is $50 for one window. On the chart, when there is 1 window, the cost is $55. So, company B is more expensive!

5 0

2 years ago

Factor this trinomial completely.

zmey [24]

First you want to write 7x as a sum or difference: 15x-8x
Now, the expression would look like this: 10x^2+15x-8x-12
Next, factor out the 5x and -4 from the expression: 5x(2x+3)-4(2x+3)
Now, factor out the 2x+3 from the expression and you have your answer:
(2x+3)(5x-4)

4 0

3 years ago

If a= lw and if l= 1.5 and w = 0.75 what does a equal

stira [4]

Oi :)

a= l*w
a=1.5 * 0.75
a= 1.125

3 0

3 years ago

Read 2 more answers

What's -8 + (-1 1/8)

gogolik [260]

I think the answer is 9.375 if it needs to be in decimal form. If it needs to be in fraction form it would be improper, 9 3/8 would be the fraction. Hope it helps!

6 0

2 years ago

Solve these linear equations in the form y=yn+yp with yn=y(0)e^at.

WINSTONCH [101]

Answer:

a) $y(t) = y_{0}e^{4t} + 2$ . It does not have a steady state

b) $y(t) = y_{0}e^{-4t} + 2$ . It has a steady state.

Step-by-step explanation:

a) $y' -4y = -8$

The first step is finding $y_{n}(t)$ . So:

$y' - 4y = 0$

We have to find the eigenvalues of this differential equation, which are the roots of this equation:

$r - 4 = 0$

$r = 4$

So:

$y_{n}(t) = y_{0}e^{4t}$

Since this differential equation has a positive eigenvalue, it does not have a steady state.

Now as for the particular solution.

Since the differential equation is equaled to a constant, the particular solution is going to have the following format:

$y_{p}(t) = C$

So

$(y_{p})' -4(y_{p}) = -8$

$(C)' - 4C = -8$

C is a constant, so (C)' = 0.

$-4C = -8$

$4C = 8$

$C = 2$

The solution in the form is

$y(t) = y_{n}(t) + y_{p}(t)$

$y(t) = y_{0}e^{4t} + 2$

b) $y' +4y = 8$

The first step is finding $y_{n}(t)$ . So:

$y' + 4y = 0$

We have to find the eigenvalues of this differential equation, which are the roots of this equation:

$r + 4 =$

$r = -4$

So:

$y_{n}(t) = y_{0}e^{-4t}$

Since this differential equation does not have a positive eigenvalue, it has a steady state.

Now as for the particular solution.

Since the differential equation is equaled to a constant, the particular solution is going to have the following format:

$y_{p}(t) = C$

So

$(y_{p})' +4(y_{p}) = 8$

$(C)' + 4C = 8$

C is a constant, so (C)' = 0.

$4C = 8$

$C = 2$

The solution in the form is

$y(t) = y_{n}(t) + y_{p}(t)$

$y(t) = y_{0}e^{-4t} + 2$

6 0

3 years ago