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3 years ago

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The fifth grade raised $182 by selling tickets for a car wash. If each ticker cost $3.25, how many tickets did the fifth grade s

ell?

1 answer:

dimaraw [331]3 years ago

8 0

The fifth grade sold 56 tickets. You solve this by dividing 182/3.25

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Jason's bank account balance was $300. He paid of his bills. Now, his account balance is -$50. Which equation can be used to fin

Katyanochek1 [597]

Answer: -50= 300- X

Step-by-step explanation:

Solving the equation:

Subtract 300 from both sides, which gives you -350= -X. Divide by -1 and get x=350. His bills were 350 dollars, which can be stated using the equation I put in the answer.

I hope this helped.

6 0

3 years ago

Help SOS pretty please

Pepsi [2]

Answer:

x = 12

Step-by-step explanation:

Recall: the secant-tangent rule states that when a secant and a tangent meet at an external point of a circle, the product of the secant and the external segment is equal to the square of the tangent segment

(x)(3) = 6² (secant-tangent rule)

3x = 36

Divide both sides by 3

3x/3 = 36/3

x = 12

6 0

3 years ago

The answer for this question

Ivanshal [37]

I'm pretty sure it would be 4 5/6 - 3 2/3 = 1 1/6.

3 0

3 years ago

Two planes leave airport at the same time, one traveling south and one traveling west. If three hours later they are 500 miles a

Snezhnost [94]

It traveled 300 miles

4 0

3 years ago

from the top of a 120-foot high tower an air traffic controller observes an airplane on the runway at an angle of depression of

Crazy boy [7]

Data:
h (<span>Tower height) = 120 ft
d (</span><span>Distance from the base of the tower to the airplane) = ?
</span> $\alpha$ (<span>Angle formed from observation of the tower to the airplane)=19º

Data: tg 19º </span>≈ 0.34
<span>
Note: </span><span>The angle formed is tangent to the height of the tower and the distance from the base of the tower to the airplane.

</span><span>Formula:
</span> $tg \alpha = \frac{opposite\:leg}{adjacent\:leg}$
<span>
Solving:
</span> $tg \alpha = \frac{opposite\:leg}{adjacent\:leg}$
$tg 19^0 = \frac{h}{d}$
$tg 19^0 = \frac{120}{d}$
$0.34 = \frac{120}{d}$
$0.34*d = 120$
$0.34d = 120$
$d = \frac{120}{0.34}$
$\boxed{\boed{d \approx 352.9\:ft}}\end{array}}\qquad\quad\checkmark$

Answer:
<span>Distance from the base of the tower to the airplane is aproximately 352.9 ft</span>

6 0

3 years ago