A table is rectangular and its length is twice its width. Find the dimensions of the table if it occupies 18 ft2 of floor space.

Answers

Answer 1

Answer: the dimensions of the table width would be 2 feet and the length would be 4 feet.

Answer 2

Answer: The answer would be length is 6, and width would be 3.
How to find the answer:
First you make combinations that multiply to 18 (because area is length times width)
2 x 9
3 x 6
6 x 3
9 x 2
since the last two are basically the same as the first two, you can get rid of those.
2 x 9
3 x 6
since 3 is half of 6 (and 2 isn't half of 9),
3 x 6 is the answer.
Hope this helps.

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Determine the factors of 5x2 + 29x _ 6

Answers

The answer is (x + 6)(5x - 1).

5x² + 29x - 6 = 5x² + 30x - x - 6 =
= (5x · x + 5x · 6) - (x + 6) =
= 5x(x + 6) - 1(x + 6) =
= (x + 6)(5x - 1)

Answer:

The factors of given equation $5x^2+29x-6$ is $(5x-1)(x+6)$

Step-by-step explanation:

Given equation $5x^2+29x-6$

We have to determine the factors of given equation.

Quadratic equation is an equation with highest degree 2.

Consider the given quadratic equation $5x^2+29x-6$

We can solve the quadratic equation $5x^2+29x-6$ using middle term splitting method.

splitting middle term can be done by writing middle term in such a way that the product of coefficient is the product of remaining two terms.

29x can be written as 30x-1x

$\Rightarrow 5x^2+29x-6$

$\Rightarrow 5x^2+30x-x-6$

Taking 5x common from first two terms and -1 from last two terms , we have,

$\Rightarrow 5x(x+6)-1(x+6)$

$5x^2+29x-6=(5x-1)(x+6)$

Thus, the factors of given equation $5x^2+29x-6$ is $(5x-1)(x+6)$

What is the value of 3.5 (11) + 1.9 (11) + 1.6 (11)
40
77
4,096
9,317

Answers

The answer would be 77 :)

Step-by-step explanation:

What is the next letter in the sequence A Z B Y C X D

Answers

The answer is w !!!!!

W ..... because it is going backwards ZYXW then forward ABCD

Find the area of the figure to the nearest tenthA. 8.2 in.2
B. 74.2 in.2
C. 148.4 in.2
D. 23.6 in.2

Answers

That's a part of a circle that has a radius of 9 inches.

Area of any circle = (pi) x (radius)²

Area of the full circle that this is a piece of = (pi) x (9)² = (81 pi) square inches.

That's the whole circle. How much of the whole circle is in the picture ?

A whole circle has 360 degrees in the middle. This piece has 105 degrees in the middle.
This piece is ( 105/360 ) of the whole circle.

So the area of the piece is (105/360) x (81 pi) = 74.22 square inches (rounded)

Indra baked and frosted a rectangular cake to sell at a bake sale for the Model UN club. The cake, including frosting, was 4 inches high, 16 inches wide, and 12 inches long. She centered the cake on a circular tray. The circular tray had a radius of 10 inches. What is the area, to the nearest square inch, of the tray that is not covered by the cake?A. 100
B. 114
C. 122
D. 314
E. 192

Answers

Answer:

C. 122

Step-by-step explanation:

Given,

Dimension of the cake:

Length = 12 in Width = 16 in Height = 4 in

We have to find out the area of the tray that is not covered by the cake.

Indra centered the cake on a circular tray.

Radius of the tray = 10 in.

So area of the tray is equal to π times square of the radius.

Framing in equation form, we get;

Area of tray = $\pi* r^2=3.14*{10}^2=3.14* 100=314\ in^2$

Since the cake is placed in circular tray.

That means only base area of cake has covered the tray.

Base Area of cake = $length* width=12* 16=192\ in^2$

Area of the tray that is not covered by the cake is calculated by subtracting area of base of cake from area of circular tray.

We can frame it in equation form as;

Area of the tray that is not covered by the cake = Area of tray - Base Area of cake

Area of the tray that is not covered by the cake = $314-192=122\ in^2$

Hence The area of the tray that is not covered by the cake is 122 sq. in.

Prove that 1+cosA/sinA + sinA/1+cosA=2cosecA

Answers

$(1+cos\alpha)/(sin\alpha)+(sin\alpha)/(1+cos\alpha)=2cosec\alpha\n\nL=((1+cos\alpha)(1+cos\alpha)+sin\alpha\cdot sin\alpha)/(sin\alpha(1+cos\alpha))=(1+2cos\alpha+cos^2\alpha+sin^2\alpha)/(sin\alpha(1+cos\alpha))\n\n=(1+2cos\alpha+1)/(sin\alpha(1+cos\alpha))=(2+2cos\alpha)/(sin\alpha(1+cos\alpha))=(2(1+cos\alpha))/(sin\alpha(1+cos\alpha))\n\n=(2)/(sin\alpha)=2\cdot(1)/(sin\alpha)=2cosec\alpha=R$

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