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wlad13 [49]
3 years ago
13

What is the simplified form of 5 - 4y + 2x - 3y - 2 + 5x?

Mathematics
1 answer:
leva [86]3 years ago
7 0
Combine like terms.

5-2=3

-4y-3y=-7y

2x+5x=7x

Put these combined values together:
3-7y+7x

Final answer: A
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9. Mark is filling up crates with watermelons for the food drive. He has large watermelons that weigh 10 pounds each and smaller
abruzzese [7]
9. 10x + 5y is the equation for larger watermelons + smaller watermelons (you did not provide information about the medium watermelons, so we must assume given the options you miswrote one of them). We can hold NO MORE than 500 pounds, so 10x + 5y must be smaller than 500. The best answer is C assuming that "5y" represents your "Medium Watermelons"- because smaller watermelons are stated to be 5 and medium ones should therefore be between 5 and 10, the option provided by A wouldn't make since because 3 pounded watermelons are not "medium" in comparison to the "small ones" that are heavier/bigger. Your best option is C, 10x + 5y < 500, the exact answer technically would be 10x + 5y <= 500.
4 0
3 years ago
A high-interest savings account pays 5.5% interest compounded annually. If $300 is deposited initially and again at the first of
inessss [21]
That is an annuity and use the attached formula.

Total = 300 * [(1.055)^11 -1] / .055 -300
Total = 300 * <span> <span> <span> 1.8020924036  </span> </span> </span> -1 /.055 -300
Total = 300 * <span>.8020924036 / .055 - 300
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4 0
3 years ago
If we inscribe a circle such that it is touching all six corners of a regular hexagon of side 10 inches, what is the area of the
Brrunno [24]

Answer:

\left(100\pi - 150\sqrt{3}\right) square inches.

Step-by-step explanation:

<h3>Area of the Inscribed Hexagon</h3>

Refer to the first diagram attached. This inscribed regular hexagon can be split into six equilateral triangles. The length of each side of these triangle will be 10 inches (same as the length of each side of the regular hexagon.)

Refer to the second attachment for one of these equilateral triangles.

Let segment \sf CH be a height on side \sf AB. Since this triangle is equilateral, the size of each internal angle will be \sf 60^\circ. The length of segment

\displaystyle 10\, \sin\left(60^\circ\right) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}.

The area (in square inches) of this equilateral triangle will be:

\begin{aligned}&\frac{1}{2} \times \text{Base} \times\text{Height} \\ &= \frac{1}{2} \times 10 \times 5\sqrt{3}= 25\sqrt{3} \end{aligned}.

Note that the inscribed hexagon in this question is made up of six equilateral triangles like this one. Therefore, the area (in square inches) of this hexagon will be:

\displaystyle 6 \times 25\sqrt{3} = 150\sqrt{3}.

<h3>Area of of the circle that is not covered</h3>

Refer to the first diagram. The length of each side of these equilateral triangles is the same as the radius of the circle. Since the length of one such side is 10 inches, the radius of this circle will also be 10 inches.

The area (in square inches) of a circle of radius 10 inches is:

\pi \times (\text{radius})^2 = \pi \times 10^2 = 100\pi.

The area (in square inches) of the circle that the hexagon did not cover would be:

\begin{aligned}&\text{Area of circle} - \text{Area of hexagon} \\ &= 100\pi - 150\sqrt{3}\end{aligned}.

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3 years ago
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