All categories

3 years ago

. what is the difference between these 16 degrees farenheit and -15 degrees farenheit

Mathematics

1 answer:

Ivahew [28]3 years ago

7 0

Answer:

The answer 1 degrees Fahrenheit because you have to subtract 16-15. I subtracted 16-15 instead of putting the negative sign

You might be interested in

In the figure, find m∠RST if m∠RSU = 81° and m∠UST = 18°. Answers : A) 109° B) 81° C) 99° D) 63°

ohaa [14]

C because you add the angles together

6 0

3 years ago

Read 2 more answers

Use substitution to solve each system of equations<br> 2x -y = 6 <br> -3y = -6x + 18

vichka [17]

Hope this helps
Step are all below

3 0

2 years ago

You have 50$ to spend at the carnival. First, you must pay $9.20 for an entrance ticket. Each snack costs $4.25, including tax.

asambeis [7]

50 - 9.20 = 40.80 so 40.80/ 4.25 = 9.6 but you can’t buy .6 of a snack so it’s 9

5 0

3 years ago

Read 2 more answers

Lyn puts dvds on shelves. She has 236 dvds.each full shelf has 28 dvds.how many full shelves of dvds does Lyn have?

cluponka [151]

Answer:

8 shelves

Step-by-step explanation:

As 28 multiple by 2 is 224 and 28*9 is 252 so she will have 8 full shelves

3 0

3 years ago

Read 2 more answers

Find the measures of the angles of the triangle whose vertices are A = (-3,0) , B = (1,3) , and C = (1,-3).A.) The measure of ∠A

alekssr [168]

Answer:

$\theta_{CAB}=128.316$

$\theta_{ABC}=25.842$

$\theta_{BCA}=25.842$

Step-by-step explanation:

A = (-3,0) , B = (1,3) , and C = (1,-3)

We're going to use the distance formula to find the length of the sides:

$r= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$

$AB= \sqrt{(-3-1)^2+(0-3)^2}=5$

$BC= \sqrt{(1-1)^2+(3-(-3))^2}=9$

$CA= \sqrt{(1-(-3))^2+(-3-0)^2}=5$

we can use the cosine law to find the angle:

it is to be noted that:

the angle CAB is opposite to the BC.

the angle ABC is opposite to the AC.

the angle BCA is opposite to the AB.

to find the CAB, we'll use:

$BC^2 = AB^2+CA^2-(AB)(CA)\cos{\theta_{CAB}}$

$\dfrac{BC^2-(AB^2+CA^2)}{-2(AB)(CA)} =\cos{\theta_{CAB}}$

$\cos{\theta_{CAB}}=\dfrac{9^2-(5^2+5^2)}{-2(5)(5)}$

$\theta_{CAB}=\arccos{-\dfrac{0.62}}$

$\theta_{CAB}=128.316$

Although we can use the same cosine law to find the other angles. but we can use sine law now too since we have one angle!

To find the angle ABC

$\dfrac{\sin{\theta_{ABC}}}{AC}=\dfrac{\sin{CAB}}{BC}$

$\sin{\theta_{ABC}}=AC\left(\dfrac{\sin{CAB}}{BC}\right)$

$\sin{\theta_{ABC}}=5\left(\dfrac{\sin{128.316}}{9}\right)$

$\theta_{ABC}=\arcsin{0.4359}\right)$

$\theta_{ABC}=25.842$

finally, we've seen that the triangle has two equal sides, AB = CA, this is an isosceles triangle. hence the angles ABC and BCA would also be the same.

$\theta_{BCA}=25.842$

this can also be checked using the fact the sum of all angles inside a triangle is 180

$\theta_{ABC}+\theta_{BCA}+\theta_{CAB}=180$

$25.842+128.316+25.842$

$180$

6 0

3 years ago

Read 2 more answers