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3 years ago

Asap answer pls ----------------

Mathematics

2 answers:

sineoko [7]3 years ago

7 0

Answer is D because square-roots cannot have negative domains.
Hope it helps (:

satela [25.4K]3 years ago

6 0

Answer:

Step-by-step explanation:

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I SPENT 50 POINTS ON THIS, PLEASE HELP

Anestetic [448]

In the figure, we can consider that the base is the side that mesures 12 in and that the height is the side that measures 15 in, since that sides are perpendicular. So, we just need to use the given formula:

$A = \dfrac{bh}{2}\Longrightarrow A = \dfrac{12\cdot15}{2}\Longrightarrow A=6\cdot15\Longrightarrow \boxed{A = 90~\text{in}^2}$

Hence, the area of the triangle is 90 in² (B).

5 0

3 years ago

Read 2 more answers

Norma bought a plastic bag that contains 100 ballons the lable says that there are 20 each of blue,yellow, red, ,green and orang

Volgvan

Answer:

It is 120

Step-by-step explanation:

7 0

3 years ago

(x+2)(2x+1)+(x+2)(2x-3) factorizar

koban [17]

Answer:

(x + 2)(4x - 2).

Step-by-step explanation:

(x+2)(2x+1)+(x+2)(2x-3)

Note that (x + 2) is common to 2 parts of the expression. So we have:

(x + 2)(2x + 1 + 2x - 3)

= Ix + 2)(4x - 2)

8 0

3 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

3 years ago

May someone help me with this? thank you, sorry to waste any time.

omeli [17]

Answer:

man that's hard good luck

7 0

3 years ago