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defon
3 years ago
8

What is the solution to this equation?

Mathematics
2 answers:
Marina86 [1]3 years ago
6 0

Answer:

D. s=32

Step-by-step explanation:

To solve the equation, we want to find out what s is. In order to do this, we have to get s by itself. Perform the opposite of what is being done to the equation. Keep in mind, everything done to one side, has to be done to the other.

–5(s – 30) = –10

s-30 is being multiplied by -5. The opposite of multiplication is division. Divide both sides by -5.

–5(s – 30)/-5= –10/-5

s-30=-10/-5

s-30=2

30 is being subtracted from s. The opposite of subtraction is addition. Add 30 to both sides.

s-30+30=2+30

s=2+30

s=32

Let's check our solution. Substitute 32 in for s in the original equation. If our solution is correct, the left side of the equation should match the right side.

–5(s – 30) = –10

-5(32-30)=-10

-5(2)= -10

-10=-10

The statement above is true, so we know that s is equal to 32, and D is the correct answer.

VikaD [51]3 years ago
4 0

Answer:

s = 32

Step-by-step explanation:

–5(s – 30) = –10

Divide each side by -5

–5/ -5(s – 30) = –10/-5

s - 30 = 2

Add 30 to each side

s-30+30 = 2+30

s = 32

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What is the answer to 8x10^6 + 5x10^9=
Lesechka [4]

Answer:

13x10^15

Step-by-step explanation:

8+5 = 13

6+9 = 15

3 0
3 years ago
Exercise 3.5. For each of the following functions determine the inverse image of T = {x ∈ R : 0 ≤ [x^2 − 25}.
masya89 [10]

a. The inverse image of f(x) is f⁻¹(x) = ∛(x/3)

b. The inverse image of g(x) is g^{-1}(x) = e^{x}

c. The inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9

<h3 /><h3>The domain of T</h3>

Since T = {x ∈ R : 0 ≤ [x^2 − 25} ⇒ x² - 25 ≥ 0

⇒ x² ≥ 25

⇒ x ≥ ±5

⇒ -5 ≤ x ≤ 5.

<h3>Inverse image of f(x)</h3>

The inverse image of f(x) is f⁻¹(x) = ∛(x/3)

f : R → R defined by f(x) = 3x³

Let f(x) = y.

So, y = 3x³

Dividing through by 3, we have

y/3 = x³

Taking cube root of both sides, we have

x = ∛(y/3)

Replacing y with x we have

y = ∛(x/3)

Replacing y with f⁻¹(x), we have

So,  the inverse image of f(x) is f⁻¹(x) = ∛(x/3)

<h3>Inverse image of g(x)</h3>

The inverse image of g(x) is g^{-1}(x) = e^{x}

g : R+ → R defined by g(x) = ln(x).

Let g(x) = y

y =  ln(x)

Taking exponents of both sides, we have

e^{y} = e^{lnx} \\e^{y} = x

Replacing x with y, we have

y = e^{x}

Replacing y with g⁻¹(x), we have

So, the inverse image of g(x) is g^{-1}(x) = e^{x}

<h3>Inverse image of h(x)</h3>

The inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9

h : R → R defined by h(x) = x − 9

Let y = h(x)

y = x - 9

Adding 9 to both sides, we have

y + 9 = x

Replacing x with y, we have

x + 9 = y

Replacing y with h⁻¹(x), we have

So, the inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9

Learn more about inverse image of a function here:

brainly.com/question/9028678

5 0
2 years ago
Can anyone help please
lesantik [10]
355 ~ 350
102~ 100

350
- 100
= 250

3 0
3 years ago
Read 2 more answers
Will give brainlist if correct.
11111nata11111 [884]

Answer:

502.65

Step-by-step explanation:

Voulume=pi×raduis²×height

V=pi×4²×10

V=502.65

4 0
2 years ago
Am I doing number 1 right???
enot [183]
Yes you are but Id suggest writing just a bit clearer, so your teacher doesnt misread your answers!
7 0
3 years ago
Read 2 more answers
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