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Usimov [2.4K]
3 years ago
11

Sarah's age is four more than twice her brothers age. Write two equivalent algebraic equations that can be used to find their ag

es.
Mathematics
1 answer:
marissa [1.9K]3 years ago
3 0

Answer:

Step-by-step explanation:

Let her brother's age be x

Sarah is 2x + 4

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Solve x in 2|x+ 6| + 9 = 19
xenn [34]
Answer: x=-1

Step by step:

2|x+6|+9=19

Subtract 9 from both sides.

2|x+6|=10

Divide both sides by 2

|x+6|=5

Use the definition of absolute value.

x+6=5

Subtract 6 on both sides

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Hope this helped! :)
5 0
2 years ago
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A kite flying in the air has a 12 line attached to it. Its line is pulled taut and casts a 9 shadow. Find the height of the kite
alexira [117]

Answer:

8 (7.94)

Step-by-step explanation:

You can think of it as a geometry problem.

What is formed here is a triangle, which sides ate: the line, the line's shadow, and the height from the ground to the kite (here I attach a drawing).

What you need to find is the height. We will call it H.

As the triangle formed is a right one, we can use Pitágoras' theorem. The height H squared plus the squared of the shadow is equal to the squared of the line (the hypotenuse). This is:

H^2 + 9^2 = 12^2

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H^2 = 63

Applying squared root in both sides

H = √63

H = 7,94

So, the height is approximately 8.

4 0
3 years ago
-4(j+k+g) plz help using box method ​
disa [49]

Answer:

-4j-4k-4g

Step-by-step explanation:

8 0
3 years ago
The admission fee for a charity event is $7 for children and $10 for adults. The event was attended by 700 people, and the total
maria [59]
If all were children, revenue would be 700*$7 = $4900. Revenue is actually $6400 -4900 = $1500 more than that. Each adult admission that replaces a child's admission adds $10 -7 = $3 to the revenue, so there must have been
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There were 200 children at the show.
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6 0
3 years ago
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I need help with d and I need to show my work.
cupoosta [38]
Ok
So when you multiply by a fraction just multiply top by top and bottom by bottom
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So answer is 8/15
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7 0
3 years ago
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