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NeTakaya
3 years ago
13

The diameter of Mercury is approximately 4.9×10^3 kilometers. The diameter of Earth is approximately 1.3×10^4 kilometers. About

how many times greater is the diameter of Earth than the diameter of Mercury? 2.7 27 270 2700
Mathematics
1 answer:
JulijaS [17]3 years ago
5 0
 \text {Number of times = } \dfrac{1.3 \times 10^4}{4.9 \times 10^3 }

\text {Number of times = }  \dfrac{1.3}{4.9} \times 10^{4-3}

\text {Number of times = }  0.27 \times 10

\text {Number of times = }  2.7
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Monica is mixing some custom-colored paint for her house. The color she wants requires 3⁄4 ounce of blue dye and 2⁄3 of an ounce
Genrish500 [490]

Ounce of blue dye mixed with one gallon of white paint base = 3/4

Ounce of blue dye mixed with eight gallons of white paint base = \frac{3}{4} (8)

= 3 × 2

=6

Ounce of pink dye mixed with one gallon of white paint base = 2/3

Ounce of pink dye mixed with eight gallons of white paint base = \frac{2}{3} (8)

= 16/3

= 5.3

Hence, 6 ounce of blue dye and 5.3 ounce of pink dye are needed to mix with 8 gallons of paint.

=6


4 0
3 years ago
Read 2 more answers
Identify the terms and like terms in the expression - x - 9x*2 + 12x*2 + 7
alexgriva [62]
The expression - x - 9x^2 + 12x^2 + 7 has 4 terms:
-x, -9x^2, 12x^2, 7

Like terms are:
-9x^2, 12x^2
4 0
3 years ago
(2x + 3) - (-8 - 2x)
masha68 [24]

Answer: 4x+11

Step-by-step explanation: Distribute the negative to the -8 and -2x. This will make 2x+3+8+2x. Combine like terms and you will get 4x+11. Hope this helps!

7 0
3 years ago
2. A solar lease customer built up an excess of 6,500 kilowatt hours (kwh) during the summer using his solar
tia_tia [17]

9514 1404 393

Answer:

  a)  E = 6500 -50d

  b)  5000 kWh

  c)  the excess will last only 130 days, not enough for 5 months

Step-by-step explanation:

<u>Given</u>:

  starting excess (E): 6500 kWh

  usage: 50 kWh/day (d)

<u>Find</u>:

  a) E(d)

  b) E(30)

  c) E(150)

<u>Solution</u>:

a) The exces is linearly decreasing with the number of days, so we have ...

  E(d) = 6500 -50d

__

b) After 30 days, the excess remaining is ...

  E(30) = 6500 -50(30) = 5000 . . . . kWh after 30 days

__

c) After 150 days, the excess remaining would be ...

  E(150) = 6500 -50(150) = 6500 -7500 = -1000 . . . . 150 days is beyond the capacity of the system

The supply is not enough to last for 5 months.

3 0
2 years ago
-3(X - 8) - (x + 5) - 23
rjkz [21]

Answer: −

3

X

−

x

−

4

Step-by-step explanation:

7 0
3 years ago
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