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valentina_108 [34]
2 years ago
9

Help me with this problem

Mathematics
1 answer:
sveta [45]2 years ago
6 0

Answer:

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Step-by-step explanation:

11/9/2001

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Solve each equation for the ind<br>2. 2(x-7)=10​
SVETLANKA909090 [29]

Answer:

x= 12

Step-by-step explanation:

2(x-7) = 10

Distribute the 2. 2x-14 = 10

Add the 14 onto both sides to cancel it out on the original side and following the rules- the same must be done to the other side. 2x-14 (+14) = 10+14 -----> 2x = 24

Divide both sides by 2 to get the X alone. 2x/2 = 24/2 ---------> x= 12

3 0
3 years ago
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An athlete eats 75 g of protein per day while training. How much protein will she eat during 21 days of training? Write your ans
lisov135 [29]

Answer:

1.575 kg

Step-by-step explanation:

75g x 21

1575 g

convert to kg

1.575 kg

Brainliest Appreciated!

7 0
3 years ago
Steps to solving y=2x+3 x=y-8
tatuchka [14]
First, plug in x=y-8 to y=2x+3
Next, it would be y=2(y-8)+3
Then, you need to distribute: y= 2y-16+3
After that, move 2y to the other side: -y= -13 (notice I also did -16+3)
Finally, divide negatives into both sides: y=13

Now to find x just plug in 13 to either of the equations: 13=2x+3
Then solve: x=5
So the FINAL answer is: (5,13)
Hope this helps!!!!!!!

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3 years ago
Mrs. Torres is mailing a package that weighs 12.5 pounds. The post office charges by the ounce to mail a package. How much does
LiRa [457]

Answer:

the package weights 200 ounces

8 0
2 years ago
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Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter. X = t5 1, y = t6 t;
kifflom [539]

The equation of the tangent to the curve at the point corresponding to the given value of the parameter will be x + y = c.

<h3>How to calculate the tangent of the parameter curves at a point?</h3>

The curves are given below.

x = t⁵ + 1 and y = t⁶ + t

Then differentiate the functions with respect to t, then we have

\rm \dfrac{dx}{dt} = 5t^4\\   ...1

and

\rm \dfrac{dy}{dt} = 6t^5 + 1   ...2

Divide equation 2 by equation 1, then we have

\rm \dfrac{\dfrac{dy}{dt} }{\dfrac{dx}{dt} } = \dfrac{dy}{dx} = \dfrac{6t^5 + 1}{5t^4}

Then the slope of the equation at t = −1, then we have

dy/dx = [6(-1)⁵ + 1]/[5(-1)⁴]

dy/dx = (-6+1)/5

dy/dx = -5/5

dy/dx = -1

Then the equation of the tangent line will be

y = -x + c

x + y = c

Where c is a constant.

More about the tangent of the parameter curves at a point link is given below.

brainly.com/question/12648555

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