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3 years ago

Gina solved the inequality below. 16≥2y−4 What is the solution to Gina’s inequality? A 8≥y B y≤40 C y≤10 D 6≥y

Mathematics

1 answer:

Vinvika [58]3 years ago

4 0

Answer: y ≤ 10

Step-by-step explanation:

16 ≥ 2y -4

Add 4 to both sides

16 + 4 ≥ 2y - 4 + 4

20 ≥ 2y

divide through by 2

20/2 ≥ 2y / 2

10 ≥ y

Therefore :

y ≤ 10

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Use integration by parts to find the integrals in Exercise.<br> ∫^3_0 3-x/3e^x dx.

Viefleur [7K]

Answer:

8.733046.

Step-by-step explanation:

We have been given a definite integral $\int _0^3\:3-\frac{x}{3e^x}dx$ . We are asked to find the value of the given integral using integration by parts.

Using sum rule of integrals, we will get:

$\int _0^3\:3dx-\int _0^3\frac{x}{3e^x}dx$

We will use Integration by parts formula to solve our given problem.

$\int\ vdv=uv-\int\ vdu$

Let $u=x$ and $v'=\frac{1}{e^x}$ .

Now, we need to find du and v using these values as shown below:

$\frac{du}{dx}=\frac{d}{dx}(x)$

$\frac{du}{dx}=1$

$du=1dx$

$du=dx$

$v'=\frac{1}{e^x}$

$v=-\frac{1}{e^x}$

Substituting our given values in integration by parts formula, we will get:

$\frac{1}{3}\int _0^3\frac{x}{e^x}dx=\frac{1}{3}(x*(-\frac{1}{e^x})-\int _0^3(-\frac{1}{e^x})dx)$

$\frac{1}{3}\int _0^3\frac{x}{e^x}dx=\frac{1}{3}(-\frac{x}{e^x}- (\frac{1}{e^x}))$

$\int _0^3\:3dx-\int _0^3\frac{x}{3e^x}dx=3x-\frac{1}{3}(-\frac{x}{e^x}- (\frac{1}{e^x}))$

Compute the boundaries:

$3(3)-\frac{1}{3}(-\frac{3}{e^3}- (\frac{1}{e^3}))=9+\frac{4}{3e^3}=9.06638$

$3(0)-\frac{1}{3}(-\frac{0}{e^0}- (\frac{1}{e^0}))=0-(-\frac{1}{3})=\frac{1}{3}$

$9.06638-\frac{1}{3}=8.733046$

Therefore, the value of the given integral would be 8.733046.

6 0

3 years ago

So I know I ask a lot of questions but I really don't know this one

Tresset [83]

A & C are the Answer

7 0

3 years ago

Cost to store 155 mark up 30

pickupchik [31]

Answer:

201.50

Step-by-step explanation:

SP = CP + markup

155 + 30% =
155*1.3 =
201.50

5 0

2 years ago

Find the approximate perimeter of ABC plotted below.

maksim [4K]

Answer:

B. 21.2

Step-by-step explanation:

Perimeter of ∆ABC = AB + BC + AC

A(-4, 1)

B(-2, 3)

C(3, -4)

✔️Distance between A(-4, 1) and B(-2, 3):

$AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$AB = \sqrt{(-2 - (-4))^2 + (3 - 1)^2} = \sqrt{(2)^2 + (2)^2)}$

$AB = \sqrt{4 + 4}$

$AB = \sqrt{16}$

AB = 4 units

✔️Distance between B(-2, 3) and C(3, -4):

$BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$BC = \sqrt{(3 - (-2))^2 + (-4 - 3)^2} = \sqrt{(5)^2 + (-7)^2)}$

$BC = \sqrt{25 + 49}$

$BC = \sqrt{74}$

BC = 8.6 units (nearest tenth)

✔️Distance between A(-4, 1) and C(3, -4):

$AC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$AC = \sqrt{(3 - (-4))^2 + (-4 - 1)^2} = \sqrt{(7)^2 + (-5)^2)}$

$AC = \sqrt{47 + 25}$

$AC = \sqrt{74}$

AC = 8.6 units (nearest tenth)

Perimeter of ∆ABC = 4 + 8.6 + 8.6 = 21.2 units

8 0

3 years ago

Brainliest goes to whoever answers correctly try to show work ONLY if you can also if you want more points answer my other quest

dolphi86 [110]

Answer:

Hi, there the answer is C. y=33/2x+425

I tried uploading a picture on how I got, but it keep saying I'm using offensive word, which I didn't but trust me the answer is right

Step-by-step explanation:

5 0

2 years ago