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3 years ago

Need help finding the dimension to this

Mathematics

1 answer:

gladu [14]3 years ago

3 0

D. 2,400 sq. ft. is the closest
1) Convert mixed numbers to improper fractions...
6 1/12 = 73/121 and 11 5/6 = 71/6

2) To get area of one tile, multiply length x width...73/121 times 71/6 = 5,183/72 = 72 sq. in. or 6 sq. ft.

3) 6 sq. ft. times 408 tiles = 2,448 sq. ft.

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If you answer can you write how you solved it?

Liono4ka [1.6K]

Answer:

$145.3in^2$

Step-by-step explanation:

$A_{L} =2\pi rh=2*\pi *2.5*9.25\\=145.3in^2$

6 0

3 years ago

Martina made a block tower that was 2 1/3 ft tall. She had a toy giraffe that was 1 1/3 ft tall. She had another tower that was

Zarrin [17]

The answer is B. 2 1/3+ 3 1/4 equals5 7/12

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3 years ago

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Find the locus of a point such that the sum of its distance from the point ( 0 , 2 ) and ( 0 , -2 ) is 6.

jok3333 [9.3K]

Answer:

$\displaystyle \frac{x^2}{5}+\frac{y^2}{9}=1$

Step-by-step explanation:

We want to find the locus of a point such that the sum of the distance from any point P on the locus to (0, 2) and (0, -2) is 6.

First, we will need the distance formula, given by:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Let the point on the locus be P(x, y).

So, the distance from P to (0, 2) will be:

$\begin{aligned} d_1&=\sqrt{(x-0)^2+(y-2)^2}\\\\ &=\sqrt{x^2+(y-2)^2}\end{aligned}$

And, the distance from P to (0, -2) will be:

$\displaystyle \begin{aligned} d_2&=\sqrt{(x-0)^2+(y-(-2))^2}\\\\ &=\sqrt{x^2+(y+2)^2}\end{aligned}$

So sum of the two distances must be 6. Therefore:

$d_1+d_2=6$

Now, by substitution:

$(\sqrt{x^2+(y-2)^2})+(\sqrt{x^2+(y+2)^2})=6$

Simplify. We can subtract the second term from the left:

$\sqrt{x^2+(y-2)^2}=6-\sqrt{x^2+(y+2)^2}$

Square both sides:

$(x^2+(y-2)^2)=36-12\sqrt{x^2+(y+2)^2}+(x^2+(y+2)^2)$

We can cancel the x² terms and continue squaring:

$y^2-4y+4=36-12\sqrt{x^2+(y+2)^2}+y^2+4y+4$

We can cancel the y² and 4 from both sides. We can also subtract 4y from both sides. This leaves us with:

$-8y=36-12\sqrt{x^2+(y+2)^2}$

We can divide both sides by -4:

$2y=-9+3\sqrt{x^2+(y+2)^2}$

Adding 9 to both sides yields:

$2y+9=3\sqrt{x^2+(y+2)^2}$

And, we will square both sides one final time.

$4y^2+36y+81=9(x^2+(y^2+4y+4))$

Distribute:

$4y^2+36y+81=9x^2+9y^2+36y+36$

The 36y will cancel. So:

$4y^2+81=9x^2+9y^2+36$

Subtracting 4y² and 36 from both sides yields:

$9x^2+5y^2=45$

And dividing both sides by 45 produces:

$\displaystyle \frac{x^2}{5}+\frac{y^2}{9}=1$

Therefore, the equation for the locus of a point such that the sum of its distance to (0, 2) and (0, -2) is 6 is given by a vertical ellipse with a major axis length of 3 and a minor axis length of √5, centered on the origin.

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3 years ago

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4(3+2x)-8=8x+4<br><br> algebra

Rufina [12.5K]

Answer:

x= infinite

Step-by-step explanation:

12+8x-8=8x+4

4+8x=8x+4

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3 years ago

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Find the volume of the composite figure. please!

Irina-Kira [14]

Volume= Length * width ( aka Base ) * height

Break up the figure into easier figures, a small square and a bigger square.

Small square- 3*2*1 = 6 cm

Large Square- 7*6*1 = 42 cm

42 + 6 = 48 cm. But wait! You have to take one more step, which is minus-ing 6 from 48. Why? Notice that there is a little area where a side of the small square meets the bigger square. That little area is worth 3 cm ( length is 3, height is 1 cm ) times 2 ( 3 cm is one side, another 3 cm is the other side from the other square ) = 6 cm.

Your total answer should be 42. ( or 48, if your teacher doesn't count the area where the squares meet/join together ).

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3 years ago

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