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3 years ago

Please help me tysm!!!

Mathematics

2 answers:

Oksana_A [137]3 years ago

7 0

Answer:

GCF: 7x
First parenthesis: (x+4)
Second parenthesis: (x-4)

Explanation:

7 is the only thing that goes into both 7 and 112.

Also when you put the GCF and both parenthesis together you’ll end up getting 7x^3-112x.

I hope this helps! Have a good rest of your day!

ziro4ka [17]3 years ago

5 0

Answer: GCF would be 7x

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The distance Max's bike moves is directly proportional to how many rotations his bike's crank shaft has

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Answer:

187.5 feet

Step-by-step explanation:

we know that

Max's bike moves 25 feet after two rotations

using proportion

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Let

x ----> the distance in feet

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Answer:

Step-by-step explanation:

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The Better Baby Buggy Co. has just come out with a new model, the Turbo. The market research department predicts that the demand

dlinn [17]

Answer:

$68

Step-by-step explanation:

We have been given the demand equation for Turbos as $q=-4p+544$ , where q is the number of buggies the company can sell in a month if the price is $p per buggy.

Let us find revenue function by multiplying price of p units by demand function as:

Revenue function: $pq=p(-4p+544)$

$pq=-4p^2+544p$

Since revenue function is a downward opening parabola, so its maximum point will be vertex.

Let us find x-coordinate of vertex using formula $\frac{-b}{2a}$ .

$\frac{-b}{2a}=\frac{-544}{2\times -4}$

$\frac{-b}{2a}=\frac{-544}{-8}$

$\frac{-b}{2a}=68$

The maximum revenue would be the y-coordinate of vertex.

Let us substitute $x=68$ in revenue formula.

$pq=-4(68)^2+544(68)$

$pq=-4*4624+544(68)$

$pq=-18496+36992$

$pq=18496$

Therefore, the company should sell each buggy for $68 to get the maximum revenue of $18,496.

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3 years ago

In the pulley system shown in this figure, MQ = 30 mm, NP = 10 mm, and QP = 21 mm. Find MN.

ankoles [38]

<h2>Answer:</h2>

$\boxed{\overline{MN}=37.96}$

<h2>Step-by-step explanation:</h2>

For a better understanding of this problem, see the figure below. Our goal is to find $\overline{MN}$ . Since:

$\angle MRS=\angle MQP=90^{\circ} \\ \\ \overline{MQ}=\overline{MR}=30mm$

and $\overline{MN}$ is a common side both for ΔMRN and ΔMQN, then by SAS postulate, these two triangles are congruent and:

$\overline{RN}=\overline{QN}$

By Pythagorean theorem, for triangle NQP:

$\overline{QN}=\sqrt{\overline{NP}^2+\overline{QP}^2} \\ \\ \overline{QN}=\sqrt{10^2+21^2} \\ \\ \overline{QN}=\sqrt{541}$

Applying Pythagorean theorem again, but for triangle MQN:

$\overline{MN}=\sqrt{\overline{MQ}^2+\overline{QN}^2} \\ \\ \overline{MN}=\sqrt{30^2+(\sqrt{541})^2} \\ \\ \boxed{\overline{MN}=37.96}$

3 0

3 years ago

A triangle has one leg measuring 10 inches and the hypotenuse measuring 20 inches. the other leg measures 10\sqrt[]{3} 10 3 inch

stiv31 [10]

From the description given for the triangle above, I think the type of triangle that is represented would be a right triangle. This type of triangle contains a right angle and two acute angles. In order to say or prove that it is a right triangle, it should be able to satisfy the Pythagorean Theorem which relates the sides of the triangle. It is expressed as follows:

c^2 = a^2 + b^2

where c is the hypotenuse or the longest side and a, b are the two shorter sides.

To prove that the triangle is indeed a right triangle, we use the equation above.

c^2 = a^2 + b^2
c^2 = 20^2 = 10^2 + (10sqrt(3))^2
400 = 100 + (100(3))
400 = 400

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4 years ago