Answer:

Step-by-step explanation:
Well we can simplify the numerator, by multiplying the 4 by the 6 and the m^3 and m^4 (add the exponents, explained in one of my previous answers I think)
This gives us the fraction: 
We can now divide the m^7 by m^2 by subtracting the exponents, and the reason why this works, is you're simply cancelling out the m's, If we express this in expanded form we have the following fraction: 
Since there is two m's in the denominator and there is also two (more than two) m's in the numerator, we can cancel those two m's out, and we get the fraction:
which can be simplified in exponent form as:
, now all we have to do is divide the 24 by the 3, to get 8
This gives us the answer: 
9514 1404 393
Answer:
A. ∠EAG and ∠BAC
Step-by-step explanation:
Vertical angles share a vertex and have opposite rays for sides.
Rays AE and AB are opposite; rays AG and AC are opposite. These are the only pairs of opposite rays in the figure, so the vertical angles must be constructed from one of the rays of the first pair and one of the rays of the second pair. ∠EAG and ∠BAC are a pair of vertical angles.
The other pair is ∠EAC and ∠BAG.
Answer:
The option is C i.e 115°, 65°. proof is given below.
Step-by-step explanation:
Given:
ABCD is a quadrilateral.
m∠ A = 100 + 5x
m∠ B = 77 - 4y
m∠ C = 106 + 3x
m∠ D = 47 + 6y
To Prove:
ABCD is a parallelogram if opposing angles are congruent by finding the measures of angles.
m∠ A = m∠ C and
m∠ B = m∠ D
Proof:
ABCD is a quadrilateral and is a parallelogram if opposing angles are congruent.
∴ m∠ A = m∠ C
On substituting the given values we get
∴ 100 + 5x = 106 +3x
∴ 
m∠ A = 100 + 5x = 100 + 5 × 3 =100 + 15 = 115°
m∠ C = 106 + 3x = 106 + 3 ×3 =106 + 9 = 115°
∴ m∠ A = m∠ C = 115°
Similarly,
∴ m∠ B = m∠ D
77 - 4y = 47 + 6y
10y = 77 - 47
10y =30
∴
m∠ B = 77 - 4y =77 - 4 × 3 = 77 - 12 = 65°
m∠ D = 47 + 6y = 47 + 6 × 3 = 47 + 18 = 65°
∴ m∠ B = m∠ D = 65°
Therefore the option is C i.e 115°, 65°
Answer:

Step-by-step explanation:
Hello,
a and b are the zeros, we can say that

So we can say that

Now, we are looking for a polynomial where zeros are 2a+3b and 3a+2b
for instance we can write

and we can notice that
so
![(x-2a-3b)(x-3a-2b)=x^2-5(a+b)x+6[(a+b)2-2ab]+13ab\\= x^2-5(a+b)x+6(a+b)^2+ab](https://tex.z-dn.net/?f=%28x-2a-3b%29%28x-3a-2b%29%3Dx%5E2-5%28a%2Bb%29x%2B6%5B%28a%2Bb%292-2ab%5D%2B13ab%5C%5C%3D%20x%5E2-5%28a%2Bb%29x%2B6%28a%2Bb%29%5E2%2Bab)
it comes

multiply by 3

When you expand the equation, you will have
9x + 9=25 + x
Next you have to collect like terms
+x will go over the equality sign to become -x. The same applies to +9
9x - x=23 - 9
9x minus x is 8x
8x=14
Divide both sides by 8
x=14/8
You will have 1 6/8
6 is the remainder while 1 is the number of times 8 divided 14.
In decimal, it will be 1.75
Hope that helped. Good luck