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k0ka [10]
3 years ago
10

a square has a perimeter of 38 inches. We want to know the length of each side. choose the best model for the problem

Mathematics
2 answers:
lianna [129]3 years ago
8 0
Its 9.5inches on each side. im not sure if that was your question :)
galina1969 [7]3 years ago
3 0
What are the models? If I see them I can probably help! :)
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Solve the equation<br>algebra 2<br><br>-6=3x-6y<br>4x=4+5x
Talja [164]
3x = 6y - 6
x = 2y - 2

4 ( 2y - 2 ) = 4 + 5 ( 2y - 2 )
8y - 8 = 4 + 10y - 10
2y = - 2
y = - 1

x = 2 ( - 1 ) - 2
x = - 4

i am a mathematics teacher. if anything to ask please pm me
6 0
3 years ago
Pls answer these questions correctly for brainly
FinnZ [79.3K]

40 - 3k + 10k = -10 - 8k - 10

40 + 7k = -10 - 8k - 10

40 + 7k = -8k - 20

40 + 7k - 40 = -8k - 20 - 40

7k = -8k - 60

7k + 8k = -8k - 60 + 8k

15k = -60

15k/15 = -60/15

k = -4

8 0
3 years ago
joni made a down payment of 1,600 on a car. that was 20% of the total price. what was the total price?
Alja [10]
Let t represent "total price."  Then 0.20t = $1600, and t = $1600/0.20 =
$8000 (answer)
4 0
3 years ago
Read 2 more answers
John wants to measure the height of a tree.he walks exactly 100 ft from the base of the tree and looks up.the angle from the gro
kati45 [8]

Answer: The height of the tree is 64.94ft

Step-by-step explanation:

Using the trigonometry of angles

Tan theta = opposite/adjacent

Tan 33° = height of the tree/100

Height of the tree= tan 33° * 100

= 0.6494*100

= 64.94ft

The height of the tree is 64.94ft

6 0
3 years ago
In a right triangle ABC, CD is an altitude, such that AD=BC. Find AC, if AB=3 cm, and CD= 2 cm.
konstantin123 [22]

Consider right triangle ΔABC with legs AC and BC and hypotenuse AB. Draw the altitude CD.

1. Theorem: The length of each leg of a right triangle is the geometric mean of the length of the hypotenuse and the length of the segment of the hypotenuse adjacent to that leg.

According to this theorem,

BC^2=BD\cdot AB.

Let BC=x cm, then AD=BC=x cm and BD=AB-AD=3-x cm. Then

x^2=(3-x)\cdot 3,\\ \\x^2=9-3x,\\ \\x^2+3x-9=0,\\ \\D=3^2-4\cdot (-9)=9+36=45,\\ \\\sqrt{D}=\sqrt{45}=3\sqrt{5},\\ \\x_1=\dfrac{-3-3\sqrt{5} }{2}0.

Take positive value x. You get

AD=BC=\dfrac{-3+3\sqrt{5} }{2}\ cm.

2. According to the previous theorem,

AC^2=AD\cdot AB.

Then

AC^2=\dfrac{-3+3\sqrt{5} }{2}\cdot 3=\dfrac{-9+9\sqrt{5} }{2},\\ \\AC=\sqrt{\dfrac{-9+9\sqrt{5} }{2}}\ cm.

Answer: AC=\sqrt{\dfrac{-9+9\sqrt{5} }{2}}\ cm.

This solution doesn't need CD=2 cm. Note that if AB=3cm and CD=2cm, then

CD^2=AD\cdot DB,\\ \\2^2=AD\cdot (3-AD),\\ \\AD^2-3AD+4=0,\\ \\D

This means that you cannot find solutions of this equation. Then CD≠2 cm.

8 0
3 years ago
Read 2 more answers
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