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inysia [295]
3 years ago
13

The ratio of the areas of the two squares is 25 to 9. The side length of the smaller square is 30 meters. How long is the side l

ength of the larger square? Explain your reasoning
Mathematics
1 answer:
Ratling [72]3 years ago
6 0
Ratios of Area of the two squares = 25:9
So then let the areas be A1 = 25A and A2 = 9A, A is common element
Side of the smaller area S2 = 30 meters
Area of the smaller square A2 = 30 x 30 = 900
We have area of smaller square as 9A = 900 => A = 100
Area of the large square = 25A = 25 x 100 = 2500.
Hence S1^2 = 2500 => S1 = 50 meters which is 20 meters longer than the
side of the smaller square.
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What is sum of two mixed numbers of 5.815+6.021
Julli [10]
5.815+6.021= 11.836


good luck!
3 0
3 years ago
Is the given point interior , exterior, or on the circle k (x+2)2 + (y-3)2 =18 P (8,4)
Mnenie [13.5K]
One way would be to find the distance from the point to the center of the circle and compare it to the radius

for
(x-h)^2+(y-k)^2=r^2
the center is (h,k) and the radius is r

and the distance formula is
distance between (x_1,y_1) and (x_2,y_2) is
D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}


r=radius
D=distance form (8,4) to center

if r>D, then (8,4) is inside the circle
if r=D, then (8,4) is on the circle
if r<D, then (8,4) is outside the circle


so
(x+2)^2+(y-3)^2=18
(x-(-2))^2+(y-3)^2=(\sqrt{18})^2
(x-(-2))^2+(y-3)^2=(3\sqrt{2})^2

the radius is 3\sqrt{2}
center is (-2,3)

find distance between (8,4) and (-2,3)

D=\sqrt{(8-(-2))^2+(4-3)^2}
D=\sqrt{(8+2)^2+(1)^2}
D=\sqrt{10^2+1}
D=\sqrt{100+1}
D=\sqrt{101}




r=3\sqrt{2}≈4.2
D=\sqrt{101}≈10.04

do r<D

(8,4) is outside the circle

6 0
4 years ago
I need help! Will mark Brainliest for full answer!!!
Softa [21]

<u><em>Answer:</em></u>

Part a .............> x = 11

Part b .............> k = 57.2

Part c .............> y = 9.2

<u><em>Explanation:</em></u>

The three problems deal with inverse variation between two variables

An inverse variation relation between two variables means that when one of the variables increases, the other will decrease (and vice versa)

<u>Mathematically, an inverse variation relation is represented as follows:</u>

y = \frac{k}{x}

where x and y are the two variables and k is the constant of variation

<u><em>Now, let's check the givens:</em></u>

<u>Part a:</u>

We are given that y = 3 and k = 33

<u>Substitute in the original relation and solve for x as follows:</u>

y = \frac{k}{x}\\ \\3 = \frac{33}{x}\\ \\x=\frac{33}{3}=11

<u>Part b:</u>

We are given that y = 11 and x = 5.2

<u>Substitute in the original relation and solve for k as follows:</u>

y=\frac{k}{x}\\ \\11=\frac{k}{5.2}\\ \\k=11*5.2=57.2

<u>Part c:</u>

We are given that x=7.8 and k=72

<u>Substitute in the original relation and solve for y as follows:</u>

y=\frac{k}{x}=\frac{72}{7.8}=9.2 to the nearest tenth

Hope this helps :)

6 0
3 years ago
you are given the measure of angle 4. You also know the following angles are supplementary: Angles 1 and 2, angles 2 and 3, angl
Harrizon [31]
If you are given the measure of angle 4, then we know angle 2 is equal since it is a vertical angle. The angle of a line is 180 degrees, so subtract the measure of angle 4 (and/or 2) which  will give you the angle measure for both angle 1 and 3, since those are also equal due to vertical angles. Hope that helped! :)
4 0
3 years ago
Line segment BA is tangent to the circle. A circle is shown. Secant D B and tangent B A intersect at point B outside of the circ
Mama L [17]

Answer: 88 units

Step-by-step explanation:

4 0
4 years ago
Read 2 more answers
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