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Elan Coil [88]
2 years ago
12

If anyone know this can you help !

Mathematics
2 answers:
GrogVix [38]2 years ago
3 0

Answer:

It is 8 ft I think.

lidiya [134]2 years ago
3 0

Answer:

8

Step-by-step explanation:

so QM and OM are equal to each other. the difference between these is that one is 5 and the other is 10. put these over each other to see how different they are. 1/2 is what the relationship so you want the same relationship with the other triangle. 4/8 equals 1/2 so it would be 8!

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The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2). On a coordinate plane, line A B has points (4, 1) and (n
GarryVolchara [31]

Answer:

(-1,1),(4,-2)

Step-by-step explanation:

Given: The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2).

To find: coordinates of vertex of the right angle

Solution:

Let C be point (x,y)

Distance between points (x_1,y_1),(x_2,y_2) is given by \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

AC=\sqrt{(x-4)^2+(y-1)^2}\\BC=\sqrt{(x+1)^2+(y+2)^2}\\AB=\sqrt{(4+1)^2+(1+2)^2}=\sqrt{25+9}=\sqrt{34}

ΔABC is a right angled triangle, suing Pythagoras theorem (square of hypotenuse is equal to sum of squares of base and perpendicular)

34=\left [ (x-4)^2+(y-1)^2 \right ]+\left [ (x+1)^2+(y+2)^2 \right ]

Put (x,y)=(-1,1)

34=\left [ (-1-4)^2+(1-1)^2 \right ]+\left [ (-1+1)^2+(1+2)^2 \right ]\\34=25+9\\34=34

which is true. So, (-1,1) can be a vertex

Put (x,y)=(4,-2)

34=\left [ (4-4)^2+(-2-1)^2 \right ]+\left [ (4+1)^2+(-2+2)^2 \right ]\\34=9+25\\34=34

which is true. So, (4,-2) can be a vertex

Put (x,y)=(1,1)

34=\left [ (1-4)^2+(1-1)^2 \right ]+\left [ (1+1)^2+(1+2)^2 \right ]\\34=9+4+9\\34=22

which is not true. So, (1,1) cannot be a vertex

Put (x,y)=(2,-2)

34=\left [ (2-4)^2+(-2-1)^2 \right ]+\left [ (2+1)^2+(-2+2)^2 \right ]\\34=4+9+9\\34=22

which is not true. So, (2,-2) cannot be a vertex

Put (x,y)=(4,-1)

34=\left [ (4-4)^2+(-1-1)^2 \right ]+\left [ (4+1)^2+(-1+2)^2 \right ]\\34=4+25+1\\34=30

which is not true. So, (4,-1) cannot be a vertex

Put (x,y)=(-1,4)

34=\left [ (-1-4)^2+(4-1)^2 \right ]+\left [ (-1+1)^2+(4+2)^2 \right ]\\34=25+9+36\\34=70

which is not true. So, (-1,4) cannot be a vertex

So, possible points for the vertex are (-1,1),(4,-2)

7 0
3 years ago
Read 2 more answers
This is the full question
nekit [7.7K]
Given:  cos A = 5/13 = adj side / hypotenuse

The the opp side is given by  (hypo)^2 = (adj)^2 + (opp)^2.  Here,
                                                 13^2     =   5^2     + (opp)^2, so that 
(opp)^2 = 169 - 25 - 144.  Then opp = +12.  All of these lengths are in Q I.
                       opp        12
Then sin A = -------- = --------  (answer)
                       hyp         13
8 0
3 years ago
The sphere below has a radius of 2.5 inches and an approximate volume of 65.42 cubic inches.
Stells [14]

Part a: The radius of the second sphere is 5 inches.

Part b: The volume of the second sphere is 523.33 in³

Part c; The radius of the third sphere is 1.875 inches.

Part d: The volume of the third sphere is 27.59 in³

Explanation:

Given that the radius of the sphere is 2.5 inches.

Part a: We need to determine the radius of the second sphere.

Given that the second sphere has twice the radius of the given sphere.

Radius of the second sphere = 2 × 2.5 = 5 inches

Thus, the radius of the second sphere is 5 inches.

Part b: we need to determine the volume of the second sphere.

The formula to find the volume of the sphere is given by

V=\frac{4}{3}  \pi r^3

Substituting \pi=3.14 and r=5 , we get,

V=\frac{4}{3} (3.14)(125)

V=\frac{1580}{3}

V=523.3333 \ in^3

Rounding off to two decimal places, we have,

V=523.33 \ in^3

Thus, the volume of the second sphere is 523.33 in³

Part c: We need to determine the radius of the third sphere

Given that the third sphere has a diameter that is three-fourths of the diameter of the given sphere.

Hence, we have,

Diameter of the third sphere = \frac{3}{4} (5)=3.75

Radius of the third sphere = \frac{3.75}{2} =1.875

Thus, the radius of the third sphere is 1.875 inches

Part d: We need to determine the volume of the third sphere

The formula to find the volume of the sphere is given by

V=\frac{4}{3}  \pi r^3

Substituting \pi=3.14 and r=1.875 , we get,

V=\frac{4}{3} (3.14)(1.875)^3

V=\frac{4}{3} (3.14)(6.59)

V=27.5901 \ in^3

Rounding off to two decimal places, we have,

V=27.59 \ in^3

Thus, the volume of the third sphere is 27.59 in³

4 0
3 years ago
Erin buys a bag of peanuts that weighs 3/4 of a pound.Later that week, the bag is 2/3 full.How much does the bag of peanuts weig
dusya [7]
So he started with 3/4 lb and ended up with 2/3 of that so 2/3 of 3/4

'of' can be roughly translated as multiply

so 2/3 times 3/4 =6/12=1/2 lb left

or you know that there are 3 out of 4 parts pound left so 2 out of 3 parts of nuts is = 2/4 or 1/2 lb
6 0
3 years ago
Please help I don't understand this
xxTIMURxx [149]
3r^2 = 3 x r^2 = 3 x 1/6^2 = 3 x 1/36 = 3/36 = 1/12

Answer:
1/12
3 0
2 years ago
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