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3 years ago

10

The length of a rectangular frame is 15 inches , and the width of the frame is 8 inches . What is the length of a diagonal of th

is frame in inches ?

1 answer:

sp2606 [1]3 years ago

6 0

Answer:

Step-by-step explanation:

d=17in

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A computer training institute has 625 students that are paying a course fee of $400. Their research shows that for every $20 red

Anastasy [175]

1.
If no changes are made, the school has a revenue of :

625*400$/student=250,000$

2.
Assume that the school decides to reduce n*20$.

This means that there will be an increase of 50n students.

Thus there are 625 + 50n students, each paying 400-20n dollars.

The revenue is:

(625 + 50n)*(400-20n)=12.5(50+n)*20(20-n)=250(n+50)(20-n)

3.

check the options that we have,

a fee of $380 means that n=1, thus

250(n+50)(20-n)=250(1+50)(20-1)=242,250 ($)

a fee of $320 means that n=4, thus

250(n+50)(20-n)=250(4+50)(20-4)=216,000 ($)

the other options cannot be considered since neither 400-275, nor 400-325 are multiples of 20.

Conclusion, neither of the possible choices should be applied, since they will reduce the revenue.

6 0

4 years ago

Plz answer will be marked BRAINLIEST

Anastaziya [24]

A=90 degrees divide by 5
=18 degrees
2c+a=90
2c+18=90
2c=90-18
2c=72
divide each side by 2
c=36
therefore:
2c=72

b=4a+c
b=4(18)+36
b=72+36
b=108

3 0

3 years ago

Read 2 more answers

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

Factor the polynomial. 3x3 – 12x2 + 27^x

Lynna [10]

12

Simplify 3x3 to 9

9- 12 x 2 + 27

Simplify 12 x 2 to 24

9- 24 + 27

8 0

4 years ago

How to solve (112+3/4(252-112),62+3/4(142-62)

Ilia_Sergeevich [38]

(112+3/4(140),62+(80)
(112+105)62+60
(217)122

5 0

3 years ago