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4 years ago

9/12 + 6/12 A. 1^1/3 B. 4/5 C. 5/8 D. 1^/14

1 answer:

5 0

I don't think you have the correct options. 9/12 + 6/12 = 15/12. To simplify, you divide each value by 3 and get 5/4, which, as a mixed number, would be 1 1/4, and neither 5/4 or 1 1/4 is an option. I hope this helps!

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To win the recycling contest, you must collect a total average between 83 and 87 pop tabs for 4 weeks, inclusive. Suppose you co

Dahasolnce [82]

Answer:

Inequality:

83 ≤ (82 + 86 + 84 + x)/4 ≤ 87

Step-by-step explanation:

You must collect no less than 80 and no more than 96 pop tabs.

4 0

3 years ago

9m-19=3M+1. M=10/3 can you tell me the answer

lana [24]

Plug in 10/3

9(10/3)-19= 3(10/3)+1

30-19= 10+1

11=11

That is the solution.

I hope this helps!
~kaikers

7 0

3 years ago

(3-2sqrt-18)-(2+3sqrt-8)

Mama L [17]

Tell me the question and I’ll help you with it is it simplify or evalute etc.

7 0

3 years ago

What are the endpoint coordinates for the midsegment of △PQR that is parallel to PQ¯¯¯¯¯?

andriy [413]

Answer:

M(x₄ ,y₄) = (-3.5 , 0.5) and

N (x₅ ,y₅) = ( -1 , -0.5 )

Step-by-step explanation:

Let the endpoint coordinates for the mid segment of △PQR that is parallel to PQ be

M(x₄ ,y₄) and N(x₅ ,y₅) such that MN || PQ

point P( x₁ , y₁) ≡ ( -3 ,3 )

point Q( x₂ , y₂) ≡ (2 , 1 )

point R( x₂ , y₂) ≡ (-4 , -2)

To Find:

M(x₄ ,y₄) = ? and

N (x₅ ,y₅) = ?

Solution:

We have Mid Point Formula as

$Mid\ point(x,y)=(\frac{x_{1}+x_{2} }{2}, \frac{y_{1}+y_{2} }{2})$

As M is the mid point of PR and N is the mid point of RQ so we will have

$Mid\ pointM(x_{4} ,y_{4})=(\frac{x_{1}+x_{3} }{2}, \frac{y_{1}+y_{3} }{2})$

$Mid\ pointN(x_{5} ,y_{5})=(\frac{x_{2}+x_{3} }{2}, \frac{y_{2}+y_{3} }{2})$

Substituting the given value in above equation we get

$Mid\ pointM(x_{4} ,y_{4})=(\frac{-3+-4 }{2}, \frac{3+-2} }{2})$

∴ $Mid\ pointM(x_{4} ,y_{4})=(\frac{-7} }{2}, \frac{1}{2})$

∴ $Mid\ pointM(x_{4} ,y_{4})=(-3.5, 0.5)$

Similarly,

$Mid\ pointN(x_{5} ,y_{5})=(\frac{2+-4 }{2}, \frac{1+-2 }{2})$

∴ $Mid\ pointN(x_{5} ,y_{5})=(\frac{-2 }{2}, \frac{-1}{2})$

∴ $Mid\ pointN(x_{5} ,y_{5})=(-1, -0.5)$

∴ M(x₄ ,y₄) = (-3.5 , 0.5) and

N (x₅ ,y₅) = ( -1 , -0.5 )

3 0

3 years ago

Find the value of each variable.

Harrizon [31]

Step-by-step explanation:

b is per the identity of angles on parallel lines when intersected by one inclined line the same as the 40° angle.

so,

b = 40°

due to the parallel nature of the 2 lines there is a symmetry effect for such shapes inscribed a circle. the upper and the lower triangle must be similar. and when applying a vertical line through the central crossing point, everything to the left is mirrored by everything on the right.

so, angle c must be equal to angle b.

c = 40°

and as the sum of all angles in a triangle is always 180°, d is then

d = 180 - 40 - 40 = 100°

the interior angle of the arc angle a is the supplementary angle of d (together they are 180°), because together with d they cover the full down side of the top-left to bottom-right line.

interior angle to a = 180 - 100 = 80°

due to the symmetry again, the arc angle opposite to a is the same as a.

as we know, the interior angle to a pair of opposing arc angles is the mean value of the 2 angles.

so, we have

(a + a)/2 = 80

2a/2 = 80

a = 80°

there might (and actually should) be some more direct approaches for "a" out of the other pieces of information, but that was the most straight one right out of my mind, and I don't spend time on finding additional shortcuts, when I have already a working approach.

8 0

2 years ago