Raphael graphed the system of equations shown.

Need help on question please!!

Zepler [3.9K]

you'd use the pythagorean theorem.

a^2+b^2=c^2

a is the first leg, b is the second leg, and c is the third leg.

10^2+13^2=c^2

100+169=c^2

269=c^2

take the square root of both sides

16.4=c

16.4cm is your answer!

3 0

3 years ago

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Can some one explain how to do this?

Delvig [45]

Answer:

see explanation

Step-by-step explanation:

Using the tangent and sine ratios in the right triangle EFG

tan60° = $\frac{opposite}{adjacent}$ = $\frac{FG}{EG}$ = $\frac{28}{EG}$ ( multiply both sides by EG )

EG × tan60° = 28 ( divide both sides by tan60° )

EG = $\frac{28}{tan60}$ ≈ 16.2 in ( to the nearest tenth )

--------------------------------------------------------------

sin60° = $\frac{opposite}{hypotenuse}$ = $\frac{FG}{EF}$ = $\frac{28}{EF}$ ( multiply both sides by EF )

EF × sin60° = 28 ( divide both sides by sin60° )

EF = $\frac{28}{sin60}$ ≈ 32.3 in ( to the nearest tenth )

3 0

3 years ago

The bottom plzz help me

gladu [14]

The answer uses NMF for formatting, which you can find out about in my profile if needed:

the equation is this:
time = {total paid - 35}/{15}

Put in what we know:
time = {95 - 35}/{15}
time = {60}/{15}
time = 4 hours

7 0

3 years ago

Dwayne drove 18 miles to the airport to pick up his father and then returned home. On the return trip he was able to drive an av

Nesterboy [21]

Answer:

30 mph

Step-by-step explanation:

Let d = distance (in miles)

Let t = time (in hours)

Let v = average speed driving <u>to</u> the airport (in mph)

⇒ v + 15 = average speed driving <u>from</u> the airport (in mph)

Using: distance = speed x time

$\implies t=\dfrac{d}{v}$

Create two equations for the journey to and from the airport, given that the distance one way is 18 miles:

$\implies t=\dfrac{18}{v} \ \ \textsf{and} \ \ t=\dfrac{18}{v+15}$

We are told that the total driving time is 1 hour, so the sum of these expressions equals 1 hour:

$\implies \dfrac{18}{v} +\dfrac{18}{v+15}=1$

Now all we have to do is solve the equation for v:

$\implies \dfrac{18(v+15)}{v(v+15)} +\dfrac{18v}{v(v+15)}=1$

$\implies \dfrac{18(v+15)+18v}{v(v+15)}=1$

$\implies 18(v+15)+18v=v(v+15)$

$\implies 18v+270+18v=v^2+15v$

$\implies v^2-21v-270=0$

$\implies (v-30)(v+9)=0$

$\implies v=30, v=-9$

As v is positive, v = 30 only

So the average speed driving to the airport was 30 mph

(and the average speed driving from the airport was 45 mph)

8 0

1 year ago

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Youssef plans to get rid of cable and is comparing two high-speed internet providers. He narrows his search to two companies. Co

nadezda [96]

100 per setup fee. a.

4 0

3 years ago