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1 year ago

CAN SOMEONE PLEASE HELP I NEED THIS IN BY 11:59 PLEASE IM GUVING BRAINLIEST!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics

2 answers:

scoray [572]1 year ago

5 0

Elise did the problem wrong. She needed to find out the kids surveyed total. Her answer was a smaller number than the number of kids that liked math the most.

i’m not so good at explaining things :/
hope this helps a little!!

ludmilkaskok [199]1 year ago

5 0

Error Analysis:

The mistake Elise of multiplying the revealed part (over whole) by the percentage given. She calculated 25% of the math students surveyed rather than finding the students in whole.

Correct Work:

20 students / .25 = 80 students surveyed

20 students * 4 = 80 students surveyed.

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Pre Calculus:

Gwar [14]

Let n be a number of friends at the perty. If dinner costs $30 per person, then it cost $30n for all friends.

Robert has booked a banquet hall which costs $200, then his total expenses are $200+30n.

If the final bill is p, then p=200+30n.

Answer: p=200+30n.

3 0

3 years ago

Read 2 more answers

Find the measure of the arc. MFD=

Mila [183]

Answer:

124

Step-by-step explanation:

6 0

1 year ago

How do you add 323+412 in base 5

Alenkasestr [34]

Answer:

Addition of 323+412 = 1240 at base 5

Step-by-step explanation:

We have to add 323+412 on base 5

As we have to use base 5 for addition so we can use only numbers 0,1,2,3.4 and 5 as we use 0,1,2,3,4,5,6,7,8,9,10 in decimal when base is 10

Means in addition of sum is greater than 5 then we have to take carry as in base 10 we take carry if the sum is greater than 10

So sum 323+412 = 1240

8 0

2 years ago

PLEASE HURRY

kiruha [24]

The answer (I believe) Is 6.

bc you do this...

x+(5x-6)+(5x-6)=54

then solve

8 0

2 years ago

CALC- limits 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