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3 years ago

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If ABC=PQR, then BC=?

1 answer:

olya-2409 [2.1K]3 years ago

4 0

If ABC=PQR, then
BC=QR

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Solving equations (algebra). Thank you!

LenKa [72]

Answer:

$\large\boxed{\dfrac{1}{x^2}+x^2=23}$

Step-by-step explanation:

$\left(\dfrac{1}{x}+x\right)^2=25\qquad\text{use}\ (a+b)^2=a^2+2ab+b^2\\\\\left(\dfrac{1}{x}\right)^2+2(x\!\!\!\!\diagup)\left(\dfrac{1}{x\!\!\!\!\diagup}\right)+x^2=25\\\\\dfrac{1}{x^2}+2+x^2=25\qquad\text{subtract 2 from both sides}\\\\\dfrac{1}{x^2}+x^2=23$

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aev [14]

It depends on the equation that it's used in

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3 years ago

Solving 2 step equations 3x + 2=5

julsineya [31]

Subtract 2 from both sides

3x+2-2=5-2

3x=3

Divide both sides by 3

3x/3= 3/3

The answer is ×=1

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3 years ago

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The dotted line represents the mid segment of the trapezoid. Find the value of X

Ganezh [65]

<u>Given</u>:

Given that the bases of the trapezoid are 21 and 27.

The midsegment of the trapezoid is 5x - 1.

We need to determine the value of x.

<u>Value of x:</u>

The value of x can be determined using the trapezoid midsegment theorem.

Applying the theorem, we have;

$Midsegment = \frac{b_1+b_2}{2}$

where b₁ and b₂ are the bases of the trapezoid.

Substituting Midsegment = 5x - 1, b₁ = 21 and b₂ = 27, we get;

$5x-1=\frac{21+27}{2}$

Multiplying both sides of the equation by 2, we have;

$2(5x-1)=(21+27)$

Simplifying, we have;

$10x-2=48$

Adding both sides of the equation by 2, we get;

$10x=50$

Dividing both sides of the equation by 10, we have;

$x=5$

Thus, the value of x is 5.

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Jlenok [28]

Adults, 18 and older

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