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Arlecino [84]
3 years ago
15

A Square yellow rug has a red square in the center. The side length of the red square is X inches. The width of the yellow band

that's surrounds the red square is 6 inches. What is the area of the yellow band?
Mathematics
1 answer:
Mariana [72]3 years ago
4 0

Answer: The area of yellow band is 24x+144 sq. inches.

Step-by-step explanation:

Let the side length of the red square be 'x' inches

Width of the yellow band surrounds the red square = 6 inches

So, Length of  band would be

6+x+6=x+12

Width of  band would be

6+x+6=x+12

So, Area of the yellow band would be

\text{Area of whole band}-\text{Area of red square}\\\\=(x+12)\times (x+12)-x^2\\\\=(x+12)^2-x^2\\\\=x^2+144+24x-x^2\\\\=24x+144

Hence, the area of yellow band is 24x+144 sq. inches.

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Serjik [45]

I think the answer is c

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The Miller family bought 4 crates of eggs last week. Each crate had 18 eggs. Since then, they have eaten 27 of the eggs. How man
Varvara68 [4.7K]

Answer:

45 eggs left

Step-by-step explanation:

if they had 4 crates and each crate has 18, then we have a total of 72 eggs

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4 0
3 years ago
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26. Law of sines, please give a step by step explanation
Stels [109]

Answer:

  174.6 ft

Step-by-step explanation:

It can be helpful to draw a diagram of the triangle we're concerned with. (See attached.)

We know the angle at the end of the shadow inside the triangle is 52°-22° = 30°. We assume the tree is growing straight up out of the hillside, so its angle with the hill inside the triangle is 90°+22° = 112°. Then the remaining angle between the shadow and the tree at the top of the tree is ...

  180° -30° -112° = 38°

Now, we have the angle opposite the tree, and the angle opposite the known side length of the triangle (215 feet along the hill, AC in the diagram). This is enough information to usefully use the Law of Sines.

  c/sin(C) = a/sin(A)

  c = a(sin(C)/sin(A)) = (215 ft)(sin(30°)/sin(38°)) ≈ 174.6 ft

The height of the tree is about 174.6 feet.

5 0
3 years ago
Try to sketch by hand the curve of intersection of the parabolic cylinder y = x2 and the top half of the ellipsoid x2 + 7y2 + 7z
vovikov84 [41]

Plug y=x^2 into the equation of the ellipsoid:

x^2+7(x^2)^2+7z^2=49

Complete the square:

7x^4+x^2=7\left(x^4+\dfrac{x^2}7+\dfrac1{14^2}-\dfrac1{14^2}\right)=7\left(x^2+\dfrac1{14}\right)^2+\dfrac1{28}

Then the intersection is such that

7\left(x^2+\dfrac1{14}\right)^2+7z^2=\dfrac{1371}{28}

\left(x^2+\dfrac1{14}\right)^2+z^2=\dfrac{1371}{196}

which resembles the equation of a circle, and suggests a parameterization is polar-like coordinates. Let

x(t)^2+\dfrac1{14}=\sqrt{\dfrac{1371}{196}}\cos t\implies x(t)=\pm\sqrt{\sqrt{\dfrac{1371}{196}}\cos t-\dfrac1{14}}

y(t)=x(t)^2=\sqrt{\dfrac{1371}{196}}\cos t-\dfrac1{14}

z=\sqrt{\dfrac{1371}{196}}\sin t

(Attached is a plot of the two surfaces and the intersection; red for the positive root x(t), blue for the negative)

4 0
3 years ago
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