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3 years ago

What is the quotient of 9.8 x 107 and 7.0 x 104 expressed in scientific notation?

Mathematics

1 answer:

frez [133]3 years ago

4 0

Answer: its not 13

Step-by-step explanation:

i honestly just needed points

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A. How many lines of reflection symmetry does this shape have? (2 points)

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Answer:

A Hexagon

Step-by-step explanation:

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3 years ago

Expand the expression 2/5 (2d+3)

Jlenok [28]

Hello from MrBillDoesMath!

Answer:

(4/5)d + 6/5

Discussion:

2/5(3d+3) =

2(2d+3)/5 =

(4d + 6)/ 5 =

(4/5)d + 6/5

Thank you,

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5 0

3 years ago

“An architect is planning to incorporate several stone spheres of different sizes into the landscaping of a public park, and wor

Artist 52 [7]

Answer:

Part 1) The value that is closest to the cost of finishing a sphere with a 5.50-meter circumference is $900

Part 2) The value that is closest to the cost of finishing a sphere with a 7.85-meter circumference is $1,800

Step-by-step explanation:

Step 1

Find the radius of each sphere

we know that

The circumference of a circle is equal to

$C=2\pi r$

Find the radius of the sphere with a 5.50-meter circumference

For $C=5.50\ m$

assume

$\pi =3.14$

substitute and solve for r

$5.50=2(3.14)r$

$r=5.50/[2(3.14)]=0.88\ m$

Find the radius of the sphere with a 7.85-meter circumference

For $C=7.85\ m$

assume

$\pi =3.14$

substitute and solve for r

$7.85=2(3.14)r$

$r=7.85/[2(3.14)]=1.25\ m$

step 2

Find the surface area of each sphere

The surface area of sphere is equal to

$SA=4\pi r^{2}$

Find the surface area of sphere with a 5.50-meter circumference

For $r=0.88\ m$

assume

$\pi =3.14$

substitute

$SA=4(3.14)(0.88)^{2}$

$SA=9.73\ m^{2}$

Find the surface area of sphere with a 7.85-meter circumference

For $r=1.25\ m$

assume

$\pi =3.14$

substitute

$SA=4(3.14)(1.25)^{2}$

$SA=19.63\ m^{2}$

step 3

Find the cost of finishing each sphere

we know that

To find out the cost , multiply the surface area by $92 per square meter

Find the cost of sphere with a 5.50-meter circumference

$9.73*(92)=\$895.16$

therefore

The value that is closest to the cost of finishing a sphere with a 5.50-meter circumference is $900

Find the cost of sphere with a 7.85-meter circumference

$19.63*(92)=\$1,805.96$

therefore

The value that is closest to the cost of finishing a sphere with a 7.85-meter circumference is $1,800

6 0

3 years ago

5(g+8)-7=103 what's the answer?

castortr0y [4]

Do you need the steps? g=14

3 0

3 years ago

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