3 years ago

9263/88 and what is the remainder

Mathematics

2 answers:

WITCHER [35]3 years ago

5 0

The remainder is 23.
105 R23

Simora [160]3 years ago

4 0

105 with a remainder of 23

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Explain your answer !! Have a nice day Will give braisnlt

liubo4ka [24]

Answer:

$x=3$

$y=-4$

Step-by-step explanation:

$4x+5y=-8$ ----------- (1)

$-4x-4y=4$ -----------(2)

Adding equation (1) and (2)

$4x+5y-4x-4y=-8+4$

$y=-4$

Substitute y=-4 in equation (1) or (2)

$4x+5(-4)=-8$

$4x-20=-8\\4x=-8+20\\4x=12\\x=3$

8 0

3 years ago

Select the correct answer(s) from the given options:

Olegator [25]

Answer:

Step-by-step explanation:

All you have to is multiply 5x20 = 100 and 5 x 10 = 50 Then add and you get 150.

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3 years ago

lim x rightarrow 0 1 - cos ( x2 ) / 1 - cosx The limit has to be evaluated without using l'Hospital'sRule.

zaharov [31]

Answer with Step-by-step explanation:

Given

$f(x)=\frac{1-cos(2x)}{1-cos(x)}\\\\\lim_{x \rightarrow 0}f(x)=\lim_{x\rightarrow 0}(\frac{1-(cos^2{x}-sin^2{x})}{1-cos(x)})\\\\(\because cos(2x)=cos^2x-sin^2x)\\\\\lim_{x \rightarrow 0}f(x)=\lim_{x\rightarrow 0}(\frac{1-cos^2x}{1-cos(x)}+\frac{sin^2x}{1-cosx})\\\\=\lim_{x\rightarrow 0}(\frac{(1-cosx)(1+cosx)}{1-cosx}+\frac{sin^2x}{1-cosx})\\\\=\lim_{x\rightarrow 0}((1+cosx)+\frac{sin^2x}{1-cosx})\\\\\therefore \lim_{x \rightarrow 0}f(x)=1$