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schepotkina [342]
3 years ago
13

What is 10000000000053304950353953059490*0.8

Mathematics
2 answers:
LenKa [72]3 years ago
8 0

Answer: 8e+30

Step-by-step explanation:

bonufazy [111]3 years ago
4 0

Answer:

8000000000042643960283162447592 or in Scientific Notation 8.000000000042643960283162447592×10^30

Step-by-step explanation:

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Look at the number line below. Write the letter for each fraction in the correct box. ​
sp2606 [1]

Answer:

DCAB

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Inverse sine is used to find a given angle measure when we know the _ and the _​
andrey2020 [161]

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opposite side of an angle

hypotenuse

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V-16=-32<br>I need help with the math worksheet how to solve the question
sergeinik [125]

You solve an equation like this by adding the opposite of the constant to both sides of the equation.

... V -16 +16 = -32 +16 . . . . . addition property of equality: if a=c, then a+b = c+b

... V + 0 = -16 . . . . . . . . . . . . additive inverse property of integers: -16+16 = 0

... V = -16 . . . . . . . . . . . . . . . identity element of addition: V+0 = V

_____

<em>You can always do the same thing to both sides of an equation.</em> Here, it is useful to add the opposite of -16 to both sides. That way the constant on the left becomes zero, so you only have the variable by itself—which is what you want.

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
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Mariana [72]

find the area A=b*h/2

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A=11*23/2

A=126.5 square unit

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