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anygoal [31]
3 years ago
9

If $2887.51 is 60%, What is 100%?

Mathematics
1 answer:
alina1380 [7]3 years ago
8 0
4812.52. I got this answer by seeing what number do you multiply 60 to get 2887.51. Then I multiplied that number by 100.
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EMERGENCY QUESTION! No Decimal answers plz!
telo118 [61]

Answer:

1,145,375 cm³

Step-by-step explanation:

Like with an earlier question you had, there is a chunk missing. If that chunk was filled in, this would be a 205 * 70 * 85 rectangular prism.

205 * 70 * 85

17,425 * 70

1,219,750

The cut out chunk is a 25 * 70 * 85 triangular prism.

1/2 (25 * 70 * 85)

1/2(2,125 * 70)

1/2(148,750)

74,375

Now that we know what the cut-out chunk is, subtract that from the first value we got.

1,219,750 -74,375

1,145,375 cm³

The volume of the Canadian Post mailbox is 1,145,375 cm³.

8 0
3 years ago
Read 2 more answers
CAN SOMEONE EXPLAIN AND MAYBE SHOW ME STEP BY STEP WHY THE ANSWER IS 75degrees. HURRY 50 points and brainiest!!!
KonstantinChe [14]

Answer:

75 degrees, we know that already so look below lol.

Step-by-step explanation:

This triangle is an isosceles triangle, LP is equivalent to LQ, and angles LQP and LPQ are also equivalent. We understand that PLQ is equal to 30 degrees, since its complementary to triangle LQM.

180 - 30 = 150

150/2 = 75.

<em>Have a nice day, fam. Spread The Love. Thanks for the opportunity.</em>

3 0
3 years ago
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Someone good at math pls help me w this I’ll mark u as brainliest ❤️:)
zhenek [66]
Angle 1= 45
Angle 2= 76
Angle 3= 80
8 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
The area of the entire large rectangle ​
Luda [366]

There is no rectangle but the formula is

a = l \times w

5 0
3 years ago
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