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kotegsom [21]
3 years ago
12

Solve for the variable

Mathematics
1 answer:
krek1111 [17]3 years ago
5 0

Answer:

G = 2/3

Step-by-step explanation:

-13G + 10 = 2(-5G + 4)  Distribute the 2 to (-5G + 4)

-13G + 10 = -10G + 8  

+13G           +13G         Add 13G to both sides

10 = 3G + 8

-8           -8                  Subtract 8 from both sides

2 = 3G                          Divide both sides by 3

2/3 = G

If this answer is correct, please make me Brainliest!

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Vinil7 [7]
Dydyeufogofififudugjdid7fogo

7 0
3 years ago
Solve for j.<br>-13j - 20 = -8j + 20​
aleksley [76]

Answer:

j = -8

Step-by-step explanation:

-13j - 20 = -8j + 20​

Add 13 j to each side

-13j+13j - 20 = -8j+13j + 20

-20 = 5j+20

subtract 20 from each side

-20-20 = 5j +20-20

-40 = 5j

Divide by 5

-40/5 = =5j/5

-8 =j

8 0
3 years ago
Explain how an enlargement or a reduction in the dimensions of a building would cause a change in the Scale factor.​ I need asap
murzikaleks [220]

Answer:

With an enlargement the scale factor would be greater than zero. With a reduction the scale factor would be a fraction or maybe even a decimal.

6 0
3 years ago
The altitude of Death Valley California is 282 ft below sea level of hot air balloon is launched From The Bottom Of Death Valley
asambeis [7]

Answer:

177 ft below sea level.

Step-by-step explanation:

Initial altitude of the hot air balloon = 282 below sea level

Rate of rise of balloon = 10 ft / second

Time = 12 seconds

Total rise in 12 seconds = Rate of rise of balloon \times Time

\Rightarrow 10 \times 12 = 120 ft

Now, we will have to subtract to find the new altitude.

New altitude = 282 - 120 = 162 ft

Now, it is given that balloon Captain lowers the balloon 15 ft.

To find the final altitude, we will have to add it to the previous altitude.

Therefore, the final altitude = 162 + 15 = <em>177 ft below sea level </em>

6 0
3 years ago
A lake polluted by bacteria is treated with an antibacterial chemical. Aftertdays, thenumber N of bacteria per milliliter of wat
jeyben [28]

Answer:

N(1)=50 is a minimum

N(15)=4391.7 is a maximum

Step-by-step explanation:

<u>Extrema values of functions </u>

If the first and second derivative of a function f exists, then f'(a)=0 will produce values for a called critical points. If a is a critical point and f''(a) is negative, then x=a is a local maximum, if f''(a) is positive, then x=a is a local minimum.  

We are given a function (corrected)

N(t) = 20(t^2-lnt^2)+ 30

N(t) = 20(t^2-2lnt)+ 30

(a)

First, we take its derivative

N'(t) = 20(2t-\frac{2}{t})

Solve N'(t)=0

20(2t-\frac{2}{t})=0

Simplifying

2t^2-2=0

Solving for t

t=1\ ,t=-1

Only t=1 belongs to the valid interval 1\leqslant t\leqslant 15

Taking the second derivative

N''(t) = 20(2+\frac{2}{t^2})

Which is always positive, so t=1 is a minimum

(b)

N(1)=20(1^2-2ln1)+ 30

N(1)=50 is a minimum

(c) Since no local maximum can be found, we test for the endpoints. t=1 was already determined as a minimum, we take t=15

(d)

N(15)=20(15^2-2ln15)+ 30

N(15)=4391.7 is a maximum

7 0
3 years ago
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