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Eduardwww [97]
3 years ago
14

HURRY ILL GIVE U BRAINLIEST!!! Find the area of each figure. Which do these belong in?

Mathematics
1 answer:
nordsb [41]3 years ago
4 0

Answer:

Area of rectangle PQRS is 14 square units.

Area of triangles UVW and EFG and square ABCD are 16 square units each.

Step-by-step explanation:

Let us find the area of each of the figures shown in the graph.

Area of rectangle PQRS is given as the product of PQ and QS.

From the graph,

Length of PQ = 7 units

Length of QS = 2 units

Now, area of rectangle PQRS = PQ\times QS=7\times 2=14 square units.

Now, area of square ABCD is given as the square of any of its side.

From the graph, AB = 4 units.

So, area of ABCD = AB^{2}=4^2=16 square units.

Area of triangle UVW is given as the half of the product of its base VW and height UW.

From the graph, UW = 4 units, VW = 8 units

Therefore, area of triangle UVW = \frac{1}{2}\times UW\times VW=\frac{1}{2}\times 4\times 8=16 square units.

For triangle EFG, EF = 4 units and FG = 8 units.

Area =  \frac{1}{2}\times EF\times FG=\frac{1}{2}\times 4\times 8=16 square units.

Area of rectangle PQRS is 14 square units. The remaining figures have areas each equal to 16 square units.

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Step-by-step explanation:

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The sides of those triangles are in ratio:

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3 years ago
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What is the quotient <br> 1/2 divided by 8
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Two problems here I need solved! I need every step, so please have that with your answers!!
Free_Kalibri [48]
QUESTION 1

We want to solve,

\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-6x+8}

We factor the denominator of the fraction on the right hand side to get,

\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-4x - 2x+8}.

This implies
\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x(x-4) - 2(x - 4)}.

\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{(x-4)(x - 2)}

We multiply through by LCM of
(x-4)(x - 2)

(x - 2) + x(x-4) = 2

We expand to get,

x - 2 + {x}^{2} - 4x= 2

We group like terms and equate everything to zero,

{x}^{2} + x - 4x - 2 - 2 = 0

We split the middle term,

{x}^{2} + - 3x - 4 = 0

We factor to get,

{x}^{2} + x - 4x- 4 = 0

x(x + 1) - 4(x + 1) = 0

(x + 1)(x - 4) = 0

x + 1 = 0 \: or \: x - 4 = 0

x = - 1 \: or \: x = 4

But
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It is an extraneous solution.

\therefore \: x = - 1
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QUESTION 2

\sqrt{x+11} -x=-1

We add x to both sides,

\sqrt{x+11} =x-1

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x + 11 = (x - 1)^{2}

We expand to get,

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This implies,

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We solve this quadratic equation by factorization,

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x = - 2 \: or \: x = 5

But
x = - 2
is an extraneous solution

\therefore \: x = 5
7 0
3 years ago
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