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Keith_Richards [23]
2 years ago
15

What is 122°F in °C?a.1 °Cb.10 °Cc.100 °Cd.110 °C​

Mathematics
2 answers:
frutty [35]2 years ago
8 0

Answer:

<h2>50° C</h2>

Step-by-step explanation:

<h2>By the Formula :-</h2>

= > (122°F − 32) × 5/9

= > 50°C...ans

<h2>Hope it Helps you!!</h2>
aksik [14]2 years ago
7 0

Answer:

None of the above ºC=50ºC

Step-by-step explanation:

122ºF=50ºC

Hope this helps :)

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Please please pleaseeeeeeeeeeeeeeeeeee I beg of your assistance
hoa [83]

Answer:

80/160

Step-by-step explanation:

so basically...

50 is half of 100

so then you  look at it and your like welp if its half then ur going to have to do half for 80

so 80 x 2= 160

and for A can you please show the options or at least tell me

7 0
3 years ago
Read 2 more answers
Triangle ABC is a right triangle.
mina [271]

Answer:

They are complementary.

Step-by-step explanation:

we know that

The sum of the interior angles of a triangle must be equal to 180 degrees

so

A+B+C=180°

where

A,B and C are the measures of the interior angles of triangle

In a right triangle

The measure of angle C ( I assume that the right angle is C) is equal to 90 degrees

so

A+B+90°=180°

A+B=180°-90°

A+B=90°

Remember that

If two angles are complementary, then their sum is equal to 90 degrees

therefore

A and B are complementary angles

3 0
3 years ago
Read 2 more answers
Erica is participating in a road race. The first part of the race is on a 5.2-mile-long straight road-oriented at an angle of 25
kap26 [50]

Answer:

6.31 mi

Step-by-step explanation:

The diagram below explains the solution better.

From the diagram,

C = starting point of the race.

A = end of the first part of the race.

B = end of the race.

Using Cosine rule, we can find the straight-line distance between the starting point and the end of the race.

Cosine rule states that:

a^2 = b^2 + c^2 - 2bc[cos(A)]

where A = angle A = <A

Given that

b = 5.2 miles

c = 2.0 miles

<A = 115° (from the diagram)

Hence,

a^2 = 5.2^2 + 2.0^2 - 2*5.2*2.0[cos(115)]\\\\a^2 = 27.04 + 4 - 20.8[cos(115)]\\\\a^2 = 31.04 + 8.79\\\\a^2 = 39.83\\\\a = \sqrt{39.83}\\ \\a = 6.31 mi

The straight-line distance between the starting point and the end of the race is 6.31 mi

4 0
3 years ago
Aaron has a digital scale. He puts a marshmallow on The digital scale and it reads 7.2 grams. How much would you except 10 marsh
madreJ [45]

Answer:

72 grams

Step-by-step explanation:

If one marshmallow weighs 7.2 all you have to do is take that times 10, which would give you 72 grams as the answer.

4 0
3 years ago
Help please ASAP thank you
jonny [76]

<u>Question 1</u>

If we let FE=x, then DE=x-5.

Also, as \overline{DF} bisects \overline{BC}, this means BE=EC=6.

Thus, by the intersecting chords theorem,

6(6)=x(x-5)\\\\36=x^2 - 5x\\\\x^2 - 5x-36=0\\\\(x-9)(x+4)=0\\\\x=-4, 9

However, as distance must be positive, we only consider the positive case, meaning FE=9

<u>Question 2</u>

If we let CE=x, then because AB bisects CD, CE=ED=x.

We also know that since FB=17, the radius of the circle is 17. So, this means that the diameter is 34, and as AE=2, thus means EB=32.

By the intersecting chords theorem,

2(32)=x^2\\\\64=x^2\\\\x=-8, 8

However, as distance must be positive, we only consider the positive case, meaning CE=8

4 0
2 years ago
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