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eimsori [14]
3 years ago
9

(20 points) Find AC, BC, m Pleasee

Mathematics
1 answer:
Alexxandr [17]3 years ago
6 0
Givens
====
DF = 8
FE = 2
======
Find BF 
By similar triangles 
FC/FE = DF/FC
FC^2 = 8*2
FC = sqrt(16)
FC = 4
Find AC
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Suppose M is the midpoint of FG. Find the missing measure.<br><br><br><br> FM=12x−4, MG=5x+10
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Answer:

FG = 17x - 6

Step-by-step explanation:

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The size of a TV is measured by its diagonal. You want to buy a TV that has a 32
ANEK [815]

The width of the TV is 27.9 inches

Step-by-step explanation:

Given,

Diagonal of TV = 32 inch

Height of TV = 15.7 inch

The diagonal makes the TV into a right triangle,

Therefore,

Diagonal = hypotenuse = c = 32 inch

Height = perpendicular = a = 15.7 inch

Width = base = b

Using pythagoras theorem

a^2+b^2=c^2\\(15.7)^2+b^2=(32)^2\\246.49+b^2=1024\\b^2=1024-246.49\\b^2=777.51

Taking square root on both sides

\sqrt{b^2}=\sqrt{777.51}\\b=27.88

Rounding off to nearest tenth

b = 27.9 inches

The width of the TV is 27.9 inches

Keywords: Pythagoras theorem, square root

Learn more about square root at:

  • brainly.com/question/2154850
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3 years ago
Five workers dig a trench 5 yards long in 5 hours. How many workers are needed to dig a 100 yards long in 100 hours?
Pavlova-9 [17]

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​Find all roots: x^3 + 7x^2 + 12x = 0 <br> Show all work and check your answer.
Aliun [14]

The three roots of x^3 + 7x^2 + 12x = 0 is 0,-3 and -4

<u>Solution:</u>

We have been given a cubic polynomial.

x^{3}+7 x^{2}+12 x=0

We need to find the three roots of the given polynomial.

Since it is a cubic polynomial, we can start by taking ‘x’ common from the equation.

This gives us:

x^{3}+7 x^{2}+12 x=0

x\left(x^{2}+7 x+12\right)=0   ----- eqn 1

So, from the above eq1 we can find the first root of the polynomial, which will be:

x = 0

Now, we need to find the remaining two roots which are taken from the remaining part of the equation which is:

x^{2}+7 x+12=0

we have to use the quadratic equation to solve this polynomial. The quadratic formula is:

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

Now, a = 1, b = 7 and c = 12

By substituting the values of a,b and c in the quadratic equation we get;

\begin{array}{l}{x=\frac{-7 \pm \sqrt{7^{2}-4 \times 1 \times 12}}{2 \times 1}} \\\\{x=\frac{-7 \pm \sqrt{1}}{2}}\end{array}

<em><u>Therefore, the two roots are:</u></em>

\begin{array}{l}{x=\frac{-7+\sqrt{1}}{2}=\frac{-7+1}{2}=\frac{-6}{2}} \\\\ {x=-3}\end{array}

And,

\begin{array}{c}{x=\frac{-7-\sqrt{1}}{2}} \\\\ {x=-4}\end{array}

Hence, the three roots of the given cubic polynomial is 0, -3 and -4

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3 years ago
Help me, please<br> Which statement is true about the following pair of rectangles?
miv72 [106K]

Answer:

C

Step-by-step explanation:

\frac{27}{33} =  \frac{9}{11}

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2 years ago
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