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rusak2 [61]
3 years ago
15

one winter day the temperature ranged from a high of 20 f to a low of -25 f. by how many degrees did the temperature change?

Mathematics
1 answer:
Mamont248 [21]3 years ago
6 0

Answer:

45 degrees

Step-by-step explanation:

you would do 20-(-25) then that turns to 20+25 and you get 45

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I’m pretty sure it’s 75%
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Solve the quadratic equation by using a numeric approach.
Sonja [21]

The solution to the quadratic equation by using a numeric approach

are x = 6.3 and x = -31.3

Given the quadratic function expressed as;

0.02x²+ 0.5x - 4 = 0

Multiply through by 100 to have;

2x² + 50x - 400 = 0

Using the general formula to solve:

x = -50±√2500-4(2)(-400)/2(2)

x = -50±√2500+3200/4

x = -50±75.49/4

x = -31.3 or 6.3

Hence the solution to the quadratic equation by using a numeric approach

are x = 6.3 and x = -31.3

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2 years ago
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Veronika [31]
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For what values of x does -x2 +7x + 5 = 0?
uranmaximum [27]
Use the commutative property-to record the terms

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3 years ago
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Nhan and Phuong both left town A for town B at 5 am . After 5 hours , Nhan and Phuong were 102 km and 52 km away from town B res
Pepsi [2]

Answer:

Nhan reach town B at 13:24 pm

Step-by-step explanation:

Let

Speed of Nhan=3x km /hour

Speed of Phoung=4x km/hour

After 5 hours,

Distance covered by Nhan=3x(5)=15 xkm

Distance covered by Phuong=4x(5)=20x km

Total distance covered by Nhan to reach town B=15x+102 km

Total distance covered by Phuong to reach town B=20x+52 km

Distance between town A and town B for both person is same therefore,

15x+102=20x+52

102-52=20x-15x

50=5x

x=10

Speed of Nhan=3(10)=30 km/h

Speed of Phuong=4(10)=40 km/h

Total distance between town A and town B=102+15(10)=252 km

Time taken by Nhan to reach town B from town A=\frac{distance}{speed}

Time taken by Nhan to reach town B from town A=\frac{252}{30}=8.4 hours

8.4 hours=8+0.4\times 60

=8 hours 24 minutes

1 hour=60 minutes

Nhan reach town B at 13:24 pm

3 0
2 years ago
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