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2 years ago

I need help with this question.

Mathematics

2 answers:

never [62]2 years ago

6 0

An exterior angle is equal to the two opposite inside angles.

You are given the exterior angle and the two opposite inside angles so you can set up an equation to solve for x, then replace x in the equation to solve for the angle.

7x-16 = 4x + x+8

Simplify:

7x-16 = 5x+8

Add 16 to each side:

7x = 5x+24

Subtract 5x from each side:

2x = 24

Divide both sides by 2:

x = 24/2

x = 12

Now replace x with 12:

7(12)-16

Simplify:

84 -16

Subtract:

Answer is 68 degrees.

mina [271]2 years ago

4 0

Answer:

Step-by-step explanation:

The answer is 82 because itś not a right angle but it is very close to being one.

Step 1: (7x + 16) The first step would be to figre out what x equals in this case x = 14

Step 2: 7(14) if you multiply you would get 98

Step 3: 98 - 16 = 82

Step 4: Apply 82 into the inner angle

Step 5: Find what the inner angles are because when you add a triangle together you should get 180 degrees

Step 6: (x + 8) x = 30 so the new prblem would be 30 + 8 = 38

Step 7: 4x Find out what x equals x = 15 so 4(15) = 60

Step 8: 82 + 38 + 60 = 120 + 60 = 180

This explains why the answer is 82 hope this helps.

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1 year ago

Find the least common multiple (LCM) of 8y^6+ 144y^5+ 640y^4 and 2y^4 + 40y^3 + 200y^2.

Mice21 [21]

Answer:

LCM = $8y^{4}(y+ 10)^{2}(y + 8)$

Step-by-step explanation:

Making factors of $8y^{6}+ 144y^{5}+ 640y^{4}$

Taking $8y^{4}$ common:

$\Rightarrow 8y^{4} (y^{2}+ 18y+ 80)$

Using factorization method:

$\Rightarrow 8y^{4} (y^{2}+ 10y + 8y + 80)\\\Rightarrow 8y^{4} (y (y+ 10) + 8(y + 10))\\\Rightarrow 8y^{4} (y+ 10)(y + 8))\\\Rightarrow \underline{2y^{2}} \times 4y^{2} \underline{(y+ 10)}(y + 8)) ..... (1)$

Now, Making factors of $2y^{4} + 40y^{3} + 200y^{2}$

Taking $2y^{2}$ common:

$\Rightarrow 2y^{2} (y^{2}+ 20y+ 100)$

Using factorization method:

$\Rightarrow 2y^{2} (y^{2}+ 10y+ 10y+ 100)\\\Rightarrow 2y^{2} (y (y+ 10) + 10(y + 10))\\\Rightarrow \underline {2y^{2} (y+ 10)}(y + 10) ............ (2)$

The underlined parts show the Highest Common Factor(HCF).

i.e. HCF is $2y^{2} (y+ 10)$ .

We know the relation between LCM, HCF of the two numbers 'p' , 'q' and the numbers themselves as:

$HCF \times LCM = p \times q$

Using equations (1) and (2): $\Rightarrow 2y^{2} (y+ 10) \times LCM = 2y^{2} \times 4y^{2}(y+ 10)(y + 8) \times 2y^{2} (y+ 10)(y + 10)\\\Rightarrow LCM = 2y^{2} \times 4y^{2}(y+ 10)(y + 8) \times (y + 10)\\\Rightarrow LCM = 8y^{4}(y+ 10)^{2}(y + 8)$