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2 years ago

Help please...... 28, 29, 30 please show work

Mathematics

1 answer:

love history [14]2 years ago

5 0

(28)

∠WAE = ∠WAM + ∠MAE

Since M bisects ∠WAE then ∠WAM = ∠MAE, hence

8x = 5x + 15 ( subtract 5x from both sides )

3x = 15 ( divide both sides by 3 )

x = 5

∠WAM = 8 × 3 = 24°

∠MAE = (5 × 3 ) + 15 = 30°

(29)

∠ABC = 90°, hence

∠ABD + ∠DBC = 90, that is

6x + 5 + 5x - 3 = 90

11x + 2 = 90 ( subtract 2 from both sides )

11x = 88 ( divide both sides by 11 )

x = 8

∠ABD = (6 × 8 ) + 5 = 53°

∠DBC = (5 × 8 ) - 3 = 37°

(30)

EH = EF + FG + GH = 28, hence

2x - 1 + x + 5 + x = 28

4x + 4 = 28 ( subtract 4 from both sides )

4x = 24 ( divide both sides by 4 )

x = 6

EF = (2 × 6 ) - 1 = 11

FG = 6 + 5 = 11

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2 years ago

What is the completely factored form of f(x)=x^3+4x^2+7x+6?

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we are given

$f(x)=x^3+4x^2+7x+6$

We will use rational root theorem to find factors

We can see that

Leading coefficient =1

constant term is 6

so, we will find all possible factors of 6

$6=\pm 1,\pm 2,\pm 3,\pm 6$

now, we will check each terms

At x=-2:

We can use synthetic division

we get

$f(-2)=0$

so, x+2 will be factor

and we can write our expression from synthetic division as

$f(x)=x^3+4x^2+7x+6=(x+2)(x^2+2x+3)$

$f(x)=(x+2)(x^2+2x+3)$

now, we can find factor of remaining terms

$x^2+2x+3=0$

we can use quadratic formula

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

we can compare our equation with quadratic equation

we get

$a=1,b=2,c=3$

now, we can plug these values

$x=\frac{-2+\sqrt{2^2-4\cdot \:1\cdot \:3}}{2\cdot \:1}$

$x=-1+\sqrt{2}i$

$x=\frac{-2-\sqrt{2^2-4\cdot \:1\cdot \:3}}{2\cdot \:1}$

$x=-1-\sqrt{2}i$

so, we get

$x=-1+\sqrt{2}i,\:x=-1-\sqrt{2}i$

so, we can write factor as

$x^2+2x+3=(x-(-1+\sqrt{2}i))(x-(-1-\sqrt{2}i))$

so, we get completely factored form as

$f(x)=(x+2)(x-(-1+\sqrt{2}i))(x-(-1-\sqrt{2}i))$ ...............Answer

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