Will give brainiliest, 15 points!! Please help, due in 10 minutes.

Trishia is 4 years older than Caila now. The sum of their ages in 8 years from now will be 38. How old are Caila and Trisha now?

Nostrana [21]

Answer: 13 and 9

Step-by-step explanation:

because since there are two people 38-(8x2)=22

The only numbers that add up to 22 that have a difference of 4 are 13 and 9

Another way:

T=C+4

T+8+C+8=38

C+4+C+8=

4 0

2 years ago

Amanda left the theatre, drove toward the shop, got gas, and then continued driving to the shop. Let g be the amount of gas (in

Vedmedyk [2.9K]

Any time the car is driving, the amount of gas is decreasing.
The only time the amount of gas can increase is the short time when
the car is parked at the gas station and gas is being poured into it.

Amanda drover towards the shop, stopped and got gas, then
drove some more. So the amount of gas decreases for a while,
then increases for a short time, then decreases again.

The middle graph is the one doing that.

4 0

3 years ago

What's the area of the shaded region?

satela [25.4K]

Do you remember how to find the area of a circle ?

Big Fat Hint: The area of a circle is (pi) x (its radius)² .

Can you find the area of the big circle ... the one with radius = 18 m ?
Do that. The whole big circle, just as if there was nothing inside it, and
no shaded region.
I'll wait.

Now, can you find the area of the small circle ... the one with radius = 12 m ?
Do that.
I'll wait.

The area of the shaded region is just

(the area of the big circle) minus (the area of the small circle) !

I got 180 pi or roughly 565.2 m². I could be wrong. You should check it.

8 0

3 years ago

If sin A = 3/8, find the value of cosec A - sec A.

alexira [117]

Answer:

$\csc A - \sec A = \dfrac 83 + \dfrac{8}{\sqrt{55}}\\\\\csc A - \sec A = \dfrac 83 - \dfrac{8}{\sqrt{55}}$

Step by step explanation:

$\text{Given that,}\\\\~~~~~~\sin A = \dfrac 38 \\\\\implies \sin^2 A = \dfrac 9{64}\\\\\implies 1 - \cos^2 A = \dfrac{9}{64}\\\\\implies \cos ^2 A = 1 - \dfrac 9{64}\\\\\implies \cos^2 A = \dfrac{55}{64}\\\\\implies \cos A =\pm\sqrt{\dfrac{55}{64}}\\ \\\implies \cos A = \pm\dfrac{\sqrt{55}}8\\\\$

$\implies \dfrac 1{\cos A} = \pm\dfrac{8}{\sqrt{55}}$

$\text{Now,}\\\\\csc A - \sec A\\\\=\dfrac{1}{\sin A}- \dfrac{1}{\cos A}\\\\=\dfrac 83 -\left(\pm \dfrac 8{\sqrt {55}} \right)\\ \\\text{Hence,}\\\\\csc A - \sec A = \dfrac 83 + \dfrac{8}{\sqrt{55}}\\\\\csc A - \sec A = \dfrac 83 - \dfrac{8}{\sqrt{55}}$