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2 years ago

How do you find angles on a rhombus

Mathematics

1 answer:

Naily [24]2 years ago

6 0

Answer:

you can try adding the adding angles up to 180

or if its graphed you can count the units

Step-by-step explanation:

and know that all four sides are congruent and diagonals are perpendicular

i tried :)

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Julie buys a pair of earrings for $7. Now she would like to the same earrings for 2 friends. How much will she spend for all 3 p

ankoles [38]

The easiest way to do this is to multiple 7 and 3, which gives you 21. Julie will spend $21 on the gifts.

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2 years ago

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What is 3 4/15 divided by ( 8 times 6 3/10

Karolina [17]

The answer is 49/216

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2 years ago

What is 2 plus 2 add 14 subtract 6

olasank [31]

What is 2 plus 2 add 14 subtract 6

2+2+14-6=

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2 years ago

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Plz help ASAP I don’t know just tell me the Awnser plz ASAP

Ann [662]

Answer:

The experimental probability is 1/6

Step-by-step explanation:

WIN = 8/48

= 1/6 or 0.1667 = 16.67% chance of winning

LOSE = 40/48

= 5/6 or 0.8333 = 83.33% chance of losing.

I hope this helps. ;)

5 0

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:

$\dfrac{x-x^3}{x-x^2}=\dfrac{x(1-x^2)}{x(1-x)}=\dfrac{x(1-x)(1+x)}{x(1-x)}=1+x$

So our limit becomes

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

3 0

2 years ago