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NISA [10]
3 years ago
11

The average daily temperature in Denton went from –8°F in January to 43°F in March. What was the change in average daily tempera

ture from January to March?
Mathematics
1 answer:
11111nata11111 [884]3 years ago
7 0

Answer:

8

Step-by-step explanation:

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Describe the diagram below using correct geometric names
bagirrra123 [75]
Angles PQT and RQS are vertical angles and angles PQR and TQS are vertical angles.

angle PQS and TQR are straight angles

angles TQP and PQR form a linear pair. they are also adjacent by default. the other angles have these same features.

that should bout do it. hope it helped!
7 0
4 years ago
Drag each tile to the correct location. Not all tiles will be used.
Novosadov [1.4K]

Answer:

25:64

Step-by-step explanation:

surface : 4×pi×R^2

R1 = 5x

R2 =8x

4piR1^2 / 4piR2^2 = R1^2/R2^2 = 5x^2/8x^2 = 25x^2/64x^2 = 25/64

4 0
3 years ago
What is the equation of the line that passes through the point (5,-2) and has a
tresset_1 [31]

Answer:

y = 6/5x - 2

Step-by-step explanation:

use y = mx+ c

m = slope

c = y intercept

y = 6/5x - 2

3 0
3 years ago
Read 2 more answers
Find the least number must be added to 4015 to make it a perfect square. also,find the square root of the perfect square so obta
Vladimir [108]

To solve this, I used guess and check.

I started by finding 60²=3600 and 70²=4900. 70² is too much, so I then did 65²=4,225.

After 65², I did 62²=3,844. I knew that was pretty close, so I did 64. 64²=4,096 and 63²=3,969. So, 64² was it.

4,096-4015=81

So, you would need to add 81 to 4015 in order to receive a perfect square, which is 64²(4,096).

7 0
3 years ago
A central angle of a circle measures 1.5 radians. If the radius of the circle is 3 cm, what is the area of the related sector?
soldi70 [24.7K]

Answer:

Option B. 6.75\ cm^{2}

Step-by-step explanation:

we know that

The area of a circle is equal to

A=\pi r^{2}

we have

r=3\ cm

substitute

A=\pi (3^{2})=9 \pi\ cm^{2}

Remember that

2\pi radians subtends the complete circle of area 9 \pi\ cm^{2}

so

by proportion

Find the area of the related sector for a central angle of 1.5 radians

Let

x------> the area of the related sector

\frac{9 \pi}{2\pi}\frac{cm^{2}}{radians} =\frac{x}{1.5}\frac{cm^{2}}{radians}\\ \\x=9*1.5/2\\ \\x= 6.75\ cm^{2}

7 0
4 years ago
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