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Korvikt [17]
4 years ago
10

Solve for x. Round to the nearest hundredth

Mathematics
1 answer:
pashok25 [27]4 years ago
6 0

Answer:

D

Step-by-step explanation:

We name the corners of the triangle as the attached image.

As we can see, triangle ABC is a right-angled triangle with Angle B equal to 90 degree.

As ABC is a right triangle, so that total measures of angle A and angle C is 90 degree

=> m∠A + m∠C = 90

=> m∠A + 45 = 90

=> m∠A = 90 - 45 = 45

=> m∠A = m∠C

So that ABC is also an isosceles triangle.

=> AB = AC  = x

As ABC is a right triangle, according to Pythagoras theorem, we have the following equation:

AB^2 + BC^2 = AC^2

=> x^2 + x^2 = 20^2

=> 2.x^2 = 400

=> x^2 = 400/2 = 200

=> x = 10\sqrt{2} ≈ 14.14

So that x ≈ 14.14, the correct answer is D

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Answer:

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Step-by-step explanation:

Start by finding the area of the whole thing with the missing triangle:

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Then find the area of the triangle. You can do this by first finding out the side lengths of the triangle:

33 - 24 = 9 and 22 - 12 = 10

Then multiply both sides and divide by two because it is a triangle:

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Then subtract the area of the triangle from the rectangle:

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zheka24 [161]

Answer:

x=11 (read below)

Step-by-step explanation:

7 + x = 18

<em>Subtract 7 from both sides</em>

x=11

It's quite simple because if you do something to one side of the equation, you need to do it to the other because otherwise the equation won't be equal. This is how you need to simplify most problems, by taking something from one side and taking from the other as well.

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Consider functions f and g.1 + 12f(1) = 12 + 4. – 12for * # 2 and 7 -64.2 – 16. + 1641 +48for a # -12 Which expression is equal
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Given the following functions below,

\begin{gathered} f(x)=\frac{x+12}{x^2+4x-12}\text{ and} \\ g(x)=\frac{4x^2-16x+16}{4x+48} \end{gathered}

Factorising the denominators of both functions,

Factorising the denominator of f(x),

\begin{gathered} f(x)=\frac{x+12}{x^2+4x-12}=\frac{x+12}{x^2+6x-2x-12}=\frac{x+12}{x(x+6)-2(x+6)}=\frac{x+12}{(x-2)(x+6)} \\ f(x)=\frac{x+12}{(x-2)(x+6)} \end{gathered}

Factorising the denominator of g(x),

\begin{gathered} g(x)=\frac{4x^2-16x+16}{4x+48}=\frac{4(x^2-4x+4)}{4(x+12)} \\ \text{Cancel out 4 from both numerator and denominator} \\ g(x)=\frac{x^2-4x+4}{x+12}=\frac{x^2-2x-2x+4}{x+12}=\frac{x(x-2)-2(x-2)}{x+12}=\frac{(x-2)^2}{x+12} \\ g(x)=\frac{(x-2)^2}{x+12} \end{gathered}

Multiplying both functions,

undefined

4 0
1 year ago
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