All categories

3 years ago

Geometry !!! Pls help it’s timed!!

Mathematics

1 answer:

DENIUS [597]3 years ago

6 0

Answer:

the answer is the first choice .

You might be interested in

A jar contains a mixture of 10 black paperclips, 8 red paperclips, and 2 white paperclips all the same size. Determine the proba

Marizza181 [45]

There are 20 paper clips in total so

Black probability: 10/20 (50%)

Red probability: 8/20 (40%)

White probability: 2/20 (10%)

3 0

4 years ago

Translate and solve using proportions: What number is 115% of 95?<br> Provide your answer below:

Vanyuwa [196]

Answer:

Step-by-step explanation:

115% = 115/100 = 1.15

of means multiply

1.15 * 95 = 109.25

3 0

3 years ago

Suppose your car gets 28 miles per gallon of gasoline, and you are driving at 55 miles per hour. Using unit analysis, find the a

pentagon [3]

<h3>Answer:</h3>

1 27/28 ≈ 1.964 gallons/hour

<h3>Step-by-step explanation:</h3>

You want gallons in the numerator of your unit rate, but that unit is in the denominator of the mileage rate. So, the computation must involve division by 28 mpg. Hours is already in the denominator of 55 mph, so the computation will involve multiplication by that rate.

... (55 mi/h)/(28 mi/gal) = (55 mi/h)·(1 gal/(28 mi)) = 55/28 gal/h

... = 1 27/28 gal/h

7 0

4 years ago

Please help.<br> Is algebra.<br> PLEASE HELP NO LINKS OR FILES

user100 [1]

Answer:

Hello, the answer for your question is c<em> (the expression has a variable in the denominator of a fraction)</em>

Step-by-step explanation:

4 0

3 years ago

A boat travel 312 miles each way downstream and back. The trip downstream took 13 hours. the trip back took 26 hours. What is th

zaharov [31]

Remember that $Speed= \frac{distance}{time}$ . Since the distance of the complete trip is 312 miles, the distance of each way is $\frac{312}{2} =156$ miles.
Let $S _{b}$ be the speed of the boat in still water and $S_{c}$ the speed of the current.

Now, for the downstream trip the boat is traveling with the current, so:
$S_{b} +S_{c} = \frac{156}{13}$
$S_{b} +S_{c} =12$ this will be our equation (1)

For the trip back the boat is traveling against the current, so:
$S _{b} -S_{c} = \frac{156}{26}$
$S_{b} -S_{c} =6$ this will be our equation (2)

Next, lets add equation (1) and equation (2) to get rid of $S _{c}$ :
$\left \{ {{S_{b}+S_{c} =12} \atop {+(S_{b}-S_{c} =6})} \right.$
$2S_{b} =18$
$S _{b} = \frac{18}{2}$
$S _{b} =9$

Finally, now that we know the speed of the boat in still water, lets replace that value in our equation (1) to find the speed of the current:
$9+S_{c} =12$
$S_{c} =12-9$
$S_{c} =3$

We can conclude that the speed of the boath in still water is 9 $\frac{mi}{h}$ , and the current of the stream is 3 $\frac{mi}{h}$ .

7 0

4 years ago