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UNO [17]
2 years ago
14

During the day, the temperature in Nome, Alaska rose 35°. The low temperature for the day was −22°F. What was the high temperatu

re for the day? Show your work.
Mathematics
1 answer:
Marysya12 [62]2 years ago
5 0

For this one add 35 to the -22 degrees.  So the equation is 35+(-22) which would equal 13 degrees F for a high.

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it is going to take 18 miles in 1 hour

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2 years ago
Factor 6x4 – 5x2 + 12x2 – 10 by grouping. What is the resulting expression?
Vadim26 [7]

Answer:

(6x² - 5)(x² + 2)

Step-by-step explanation:

Given

6x^{4} - 5x² + 12x² - 10 ( factor the first/second and third/fourth terms )

= x²(6x² - 5) + 2(6x² - 5) ← factor out (6x² - 5) from each term

= (6x² - 5)(x² + 2)

8 0
3 years ago
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Hello! i need 2 people to answer this question whoever is correct gets brainiest, if both are correct then ill go by whoever ans
mojhsa [17]

Answer:

-10b - 7

Step-by-step explanation:

    -11b - 4

  - (-b + 3)

-----------------

    -10b - 7

6 0
3 years ago
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Travis painted for 6 2/3 hours. He received $27 an hour for his work. How much was Travis paid for doing this painting job?
mars1129 [50]

Answer:

$180

Step-by-step explanation:

Well because it is $27 per hour, you can start with the whole hours:

6*27=162

then because you are working with fractions you can find what 1/3 of 27 is:

27* (1/3)=9

this tells us that one-third of an hour is worth 9 dollars but because it is 2/3 then you just add another 9. This means 2/3 of 27 is 18.

Then you just add it all up:

162+18=180

8 0
1 year ago
Air is being lost from a spherical balloon at a constant rate of 3/2cm3s-1. Find the rate at which the radius is decreasing at t
cricket20 [7]

Answer:

The rate at which the radius is decreasing when the radius is 6 cm is approximately 3.316 × 10⁻³ cm/s

Step-by-step explanation:

The rate at which air is being lost from the balloon = 3/2 cm³/s

The rate at which the radius is decreasing when the radius is 6 cm long is given as follows;

The rate at which air is being lost from the balloon = dV/dt = 3/2 cm³/s

dV/dt = dV/dr × dr/dt

Where;

dr/dt = The rate at which the radius is decreasing

dV/dr = d(4/3×π×r³)/dr = 4·π·r²

Therefore, we have;

dr/dt = (dV//dt)/(dV/dr) = (3/2 cm³/s)/(4·π·r²)

dr/dt = (3/2 cm³/s)/(4·π·r²)

When r = 6 cm, we have;

dr/dt = (3/2 cm³/s)/(4 × π × (6 cm)²) ≈ 3.316 × 10⁻³ cm/s

Therefore, the rate at which the radius is decreasing, dr/dt, when the radius is 6 cm long ≈ 3.316 × 10⁻³ cm/s.

7 0
3 years ago
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