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ahrayia [7]
2 years ago
7

Convert the following to scientific notation A) 0.315 B) 5,820,000,000

Mathematics
1 answer:
Savatey [412]2 years ago
3 0

Answer:

Step-by-step explanation:

a. 0.315 = 3.15 x 10^-1

b. 5,820,000,000 = 5.82 x 10^9

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Find the missing angle measure A=51°, B=82°, C=5x+2
Shalnov [3]

I'm assuming you mean triangle angles?
If so, all angles add up to 180 so combine them and set them equal to 180.
51+82+5x+2=180
135+5x=180
5x=45
x=9

8 0
3 years ago
The perimeter of the base of a regular quadrilateral prism is 60 cm, the area of a lateral face is 105 cm2. Find: the volume of
Harman [31]
Since the base is a regular quadrilateral, each of its 4 sides must have length
  s = P/4
  s = (60 cm)/4 = 15 cm

The area of one lateral face is the product of side length and height.
  A = s×h
  105 cm² = (15 cm)×h
Then the height of the prism is
  h = (105 cm²)/(15 cm) = 7 cm

The area of the base is then
  B = s²
  B = (15 cm)² = 225 cm²

The volume of the prism is the product of its base area and height.
  V = Bh
  V = (225 cm²)×(7 cm) = 1575 cm³


The volume is 1575 cm³.
5 0
3 years ago
Solve.<br> 3 - 2 | 2-5 | + 3
expeople1 [14]

3-2\left|2-5\left|+3\right\\= -23

5 0
2 years ago
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Please answer !!!!!!!!!!!!!!!!!!!!!!
rewona [7]

Answer:

umm iI not sure but i think it is

Step-by-step explanation:

write out expression

8 0
2 years ago
Please someone help me to prove this. ​
dem82 [27]
<h3><u>Answer</u> :</h3>

We know that,

\dag\bf\:sin^2A=\dfrac{1-cos2A}{2}

\dag\bf\:sin2A=2sinA\:cosA

<u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u>

<u>Now, Let's solve</u> !

\leadsto\:\bf\dfrac{sin^2A-sin^2B}{sinA\:cosA-sinB\:cosB}

\leadsto\:\sf\dfrac{\frac{1-cos2A}{2}-\frac{1-cos2B}{2}}{\frac{2sinA\:cosA}{2}-\frac{2sinB\:cosB}{2}}

\leadsto\:\sf\dfrac{1-cos2A-1+cos2B}{sin2A-sin2B}

\leadsto\:\sf\dfrac{2sin\frac{2A+2B}{2}\:sin\frac{2A-2B}{2}}{2sin\frac{2A-2B}{2}\:cos\frac{2A+2B}{2}}

\leadsto\:\sf\dfrac{sin(A+B)}{cos(A+B)}

\leadsto\:\bf{tan(A+B)}

5 0
3 years ago
Read 2 more answers
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