I WILL GIVE YOU BRAINLIEST FOR THE CORRECT ANSWER WITH AN EXPLANATION!

Gas mileage is the number of miles you can drive on a gallon of gasoline. A test of a new car results in 440 miles on10 gallons

Paraphin [41]

Okay. So you can go 440 miles on 10 miles of gas. That’s 44 miles for 1 gallon of gas. For 60 miles of gas, you have to do 440 * 6. When you do that, your product should be 2,640. You can drive 2,640 miles on 69 gallons of gas and 44 miles per gallon of gas.

7 0

3 years ago

If an object accelerates at a rate of 3 m/s2, it will exert a force of 15 N. How much force will the same object exert if it acc

fenix001 [56]

Answer:

13 newtons

Step-by-step explanation:

6 0

2 years ago

PLEASE HELP

miss Akunina [59]

In the smallest of the three triangles (with sides $a,b,3$ ) we have

$a^2+3^2=b^2$

In the second-largest triangle (with sides $a,108$ ), the length of the missing side is $\sqrt{a^2+108^2}$ . Then in the largest triangle (with sides $b,111,\sqrt{a^2+108^2}$ ), we have

$b^2+\left(\sqrt{a^2+108^2}\right)^2=(108+3)^2=111^2$

$\implies b^2+a^2+108^2=111^2$

$\implies b^2+a^2=657$

Equivalently, we have

$(a^2+9)+a^2=657\implies 2a^2=648\implies a^2=324\implies a=18$

In turn, we find

$b^2=18^2+3^2=298\implies b=\sqrt{333}=3\sqrt{37}$

6 0

3 years ago

Please help? I’m super lost...

babunello [35]

Answer:

Step-by-step explanation:

In all of these problems, the key is to remember that you can undo a trig function by taking the inverse of that function. Watch and see.

a. $sin2\theta =-\frac{\sqrt{3} }{2}$

Take the inverse sin of both sides. When you do that, you are left with just 2theta on the left. That's why you do this.

$sin^{-1}(sin2\theta)=sin^{-1}(-\frac{\sqrt{3} }{2} )$

This simplifies to

$2\theta=sin^{-1}(-\frac{\sqrt{3} }{2} )$

We look to the unit circle to see which values of theta give us a sin of -square root of 3 over 2. Those are:

$2\theta =\frac{5\pi }{6}$ and

$2\theta=\frac{7\pi }{6}$

Divide both sides by 2 in both of those equations to get that values of theta are:

$\theta=\frac{5\pi }{12},\frac{7\pi }{12}$

b. $tan(7a)=1$

Take the inverse tangent of both sides:

$tan^{-1}(tan(7a))=tan^{-1}(1)$

Taking the inverse tangent of the tangent on the left leaves us with just 7a. This simplifies to

$7a=tan^{-1}(1)$

We look to the unit circle to find which values of <em>a</em> give us a tangent of 1. They are:

$7\alpha =\frac{5\pi }{4},7\alpha =\frac{\pi }{4}$

Dibide each of those equations by 7 to find that the values of alpha are:

$\alpha =\frac{5\pi}{28},\frac{\pi}{28}$

c. $cos(3\beta)=\frac{1}{2}$

Take the inverse cosine of each side. The inverse cosine and cosine undo each other, leaving us with just 3beta on the left, just like in the previous problems. That simplifies to:

$3\beta=cos^{-1}(\frac{1}{2})$