Answer:

Step-by-step explanation:
<u>Theorem 1:</u> The height of right triangle drawn from the right angle to the hypotenuse divides the hypotenuse into two segments. The length of the altitude is the geometric mean of these two segments. Hence,

<u>Theorem 2:</u> In a right triangle, the altitude drawn from the right angle to the hypotenuse divides the hypotenuse into two segments. The length of each leg of the right triangle is the geometric mean of the lengths of the hypotenuse and the segment of the hypotenuse that is adjacent to the leg.
Thus,

Answer:
It will take Ellie 64 minutes to put as many boxes as possible.
Step-by-step explanation:
Let us work with meters.
The dimensions of Ellie's boxes in meters are: 0.45m by 0.40m by 0.35 ( <em>to convert from centimeters to meters we just divide by 100, because 1m =100cm).</em> therefore the volume of each box is:
<em>
</em>
Now the dimensions of the empty van are 3.6m by 1.6m by 2.1 m, therefore its volume
is:

So the amount of boxes that Ellie can put in the van is equal to the volume of the van
divided by the volume
of each box:

So 192 boxes can be put into the van.
Now Ellie can put 3 boxes in the van in 1 minute, therefore the amount of time it will take her to put 192 boxes into the van will be:

So it takes Ellie 64 minutes to put as many boxes into the van as she can.
The answer to the question above as to which system of equations can be used to model the situation of Graham and Hunter when a cable lifts Graham into the air at a speed of 1.5 ft/s and Hunter throws the ball from a position of 5 ft. above the ground with an initial velocity of 24 ft/s. the equation will be : H = 18t + 1.5t - 16t ^ 2
H = 5 + 24t - 16t ^ 2
Answer:
The area of the shaded region is 
Step-by-step explanation:
<u><em>The correct question is</em></u>
A circle with radius of 4cm sits inside a circle with a radius of 11cm. What is the area of the shaded region?
The shaded region is the area outside the smaller circle and inside the larger circle
we know that
To find out the shaded region subtract the area of the smaller circle from the area of the larger circle
so

simplify

where
r_a is the radius of the larger circle
r_b is the radius of the smaller circle
we have

substitute


assume

substitute
