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3 years ago

Please can you guy help for this questiong please

Mathematics

1 answer:

MrRissso [65]3 years ago

6 0

Answer:

Step-by-step explanation:

From the figure attached,

Given: ∠P ≅ ∠S

TQ ≅ RQ

To Prove : ΔQRS ≅ ΔQTP

Statements Reasons

1). ∠P ≅ ∠S, TQ ≅ RQ Given

2). ∠RQS ≅ ∠TQP Reflexive property

3). ΔQRS ≅ ΔQTP AAS

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Find the value of n in 2m-3n = 6 and 2m + 6n = 12.

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Answer:

n = 2/3

Step-by-step explanation:

2m - 3n = 6

– 2m + 6n = 12

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n = 2/3

Hence, the value of n in the given equations is 2/3.

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Find the value of x please!! ASAP

cricket20 [7]

Answer:

x = 10

Step-by-step explanation:

By looking at the picture, I can tell the angle of the 3x is also 30 degrees.

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2 years ago

CALC- limits<br> please show your method

gladu [14]

A. Factor the numerator as a difference of squares:

$\displaystyle\lim_{x\to9}\frac{x-9}{\sqrt x-3}=\lim_{x\to9}\frac{(\sqrt x-3)(\sqrt x+3)}{\sqrt x-3}=\lim_{x\to9}(\sqrt x+3)=6$

c. As $x\to\infty$ , the contribution of the terms of degree less than 2 becomes negligible, which means we can write

$\displaystyle\lim_{x\to\infty}\frac{4x^2-4x-8}{x^2-9}=\lim_{x\to\infty}\frac{4x^2}{x^2}=\lim_{x\to\infty}4=4$

e. Let's first rewrite the root terms with rational exponents:

$\displaystyle\lim_{x\to1}\frac{\sqrt[3]x-x}{\sqrt x-x}=\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}$

Next we rationalize the numerator and denominator. We do so by recalling

$(a-b)(a+b)=a^2-b^2$
$(a-b)(a^2+ab+b^2)=a^3-b^3$

In particular,

$(x^{1/3}-x)(x^{2/3}+x^{4/3}+x^2)=x-x^3$
$(x^{1/2}-x)(x^{1/2}+x)=x-x^2$

so we have

$\displaystyle\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}\cdot\frac{x^{2/3}+x^{4/3}+x^2}{x^{2/3}+x^{4/3}+x^2}\cdot\frac{x^{1/2}+x}{x^{1/2}+x}=\lim_{x\to1}\frac{x-x^3}{x-x^2}\cdot\frac{x^{1/2}+x}{x^{2/3}+x^{4/3}+x^2}$

For $x\neq0$ and $x\neq1$ , we can simplify the first term:

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So our limit becomes

$\displaystyle\lim_{x\to1}\frac{(1+x)(x^{1/2}+x)}{x^{2/3}+x^{4/3}+x^2}=\frac{(1+1)(1+1)}{1+1+1}=\frac43$

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3 years ago

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gavmur [86]

2 hours is also 120 minutes

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3 years ago