Solve for X 7(X-3)+3(4-X)=-8

1 answer:

7 0

$7(x-3)+3(4-x)=-8\\\\7x-21+12-3x=-8\\\\4x-9=-8\ \ \ \ |add\ 9\ to\ both\ sides\\\\4x=1\ \ \ \ |divide\ both\ sides\ by\ 4\\\\\boxed{x=0.25}$

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PLS HELO IM MARKING BRAINLEIST!!

Musya8 [376]

∠O ≅ ∠U is third statement which is congruent .

What is congruent triangle?

Two figures are said to be "congruent" if they can be placed perfectly over one another. Both of the bread slices are the same size and shape when stacked one on top of the other.
Congruent refers to having precisely the same form and size.
If all three sides of two triangles are the same, they are said to be congruent.
If we have a side, an angle between the sides, and then a second side that is congruent, we know they are congruent. In other words, side, angle, side.

ΔONP and ΔUTV are congruent .

By using ASA Theorem ,

NP ≅ TV

∠P ≅ ∠V

∠O ≅ ∠U

Learn more about congruent

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4 0

10 months ago

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Someone can help??

oee [108]

Answer:

Step-by-step explanation:

let cos^{-1}x=t

cos t=x

when x=1,cos t=1=cos 0

t \rightarrow 0

$\lim_{x \to 1} \frac{1-\sqrt{x}}{(cos ^{-1}x)^2 } \\= \lim_{t \to 0} \frac{1-\sqrt{cos~t}}{t^2} \times \frac{1+\sqrt{cos~t}}{1+\sqrt{cos ~t}} \\= \lim_{t \to 0} \frac{1-cos~t}{t^2(1+\sqrt{cos~t})}} \\= \lim_{t \to 0 }\frac{2 sin^2~\frac{t}{2}}{t^2(1+\sqrt{cos~t})}} \\= 2\lim_{t \to 0 }(\frac{sin~t/2}{\frac{t}{2} })^2 \times \frac{1}{4} \times \lim_{t \to 0 }\frac{1}{1+\sqrt{cos~t}} \\=\frac{2}4} \times 1^2 \times \frac{1}{1+\sqrt{cos~0}} \\=\frac{1}{2} \times \frac{1}{1+1} \\=\frac{1}{4}$