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steposvetlana [31]
3 years ago
9

Write each height using scientific notation 1700000000 and 152000000

Mathematics
1 answer:
Gnesinka [82]3 years ago
5 0
1.7 times 10 to the 9 power 1.52 times10 to the 8 power
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Needs to be solved in 10 min. I don't under stand.
GarryVolchara [31]

4a. 150*6=900 g

4b. 450/150=3 eggs

5a. 11.96/4=$2.99/bag

5b. 2.99*9=$26.91 for 9 bags

Hope this helps!

8 0
4 years ago
8+ <br> 3<br> z<br> ​ <br> =12 its z over 3 btw how do i show my work
otez555 [7]

8+z/3=12

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3 0
3 years ago
Write (-7).(-7).5.5.5.5?using exponents
Reptile [31]

(-7)^2 and 5^3 are your answers

3 0
3 years ago
Read 2 more answers
Mr. Naxvip wrote the expression below to represent the number of
aliya0001 [1]

Answer:

19

Step-by-step explanation:

b×3+5-1

=5×3+5-1

= 19

4 0
4 years ago
Please find the perimeter and area of these shapes
ladessa [460]

Answer:

ABC shaded area = 36\pi - 72   cm²

ABC shaded area perimeter = 6\pi +12\sqrt{2}    cm

ABCD area = \dfrac52 \pi  cm²

ABCD perimeter = 3\pi +2   cm

Step-by-step explanation:

<u>Shape ABC</u>

Assuming you want the area and perimeter of the shaded part of the shape only...

<u>Area</u>

Area of a sector = \dfrac12r^2\theta (where r is the radius and \theta<em> </em>

⇒ area of a sector = \dfrac12 \times 12^2\times \dfrac{\pi}{2} =36\pi  \ \textsf{cm}^2

Area of triangle = 1/2 x base x height

⇒ area of triangle = 1/2 x 12 x 12 = 72 cm²

Therefore, area of shaded area = area of sector - area of triangle

⇒ area = 36\pi - 72 cm²

<u>Perimeter</u>

Arc length = r\theta (where r is the radius and \theta<em> </em>

⇒ arc length = 12\times\dfrac12\pi =6\pi  \ \textsf{cm}

Hypotenuse of triangle = \sqrt{a^2+b^2} (where a and b are the legs of the right triangle)

⇒ hypotenuse = \sqrt{12^2+12^2} =12\sqrt{2} cm

Therefore, perimeter = arc length + hypotenuse

⇒ perimeter = 6\pi +12\sqrt{2}  cm

<u>Shape ABCD</u>

<u>Area</u>

Area of a semicircle = \dfrac12 \pi r^2 (where r is the radius)

⇒ area of large semicircle ABC = \dfrac12 \times \pi \times 2^2=2\pi  \ \textsf{cm}^2

⇒ area of small semicircle AD = \dfrac12 \times \pi \times 1^2=\dfrac12\pi  \ \textsf{cm}^2

⇒ area of shape ABCD = \dfrac12 \pi + 2 \pi=\dfrac52 \pi \ \textsf{cm}^2

<u>Perimeter</u>

1/2 circumference = \pi r

⇒ perimeter = 2\pi +2+\pi=3 \pi+2 \ \textsf{cm}

7 0
2 years ago
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