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3 years ago

Please help me.........

Mathematics

1 answer:

natali 33 [55]3 years ago

3 0

Answer:

No acceleration, 4 m/s

Step-by-step explanation:

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two classmates have decided to read all the volumes in a popular series of books. Katie has already read 4 volumes and will cont

scoray [572]

To solve this we can make it into an equation, which looks like this

y = 2x (for Rebecca)
y = x + 10 (for Alice)

The we subtract one equation from the other to solve

0 = x - 10
10 = x
x = 10

Now we replace x in whichever equation and we get 20

So, each girl will read 20 books for them to be at the same book.

Hope this helps, remember that Jesus loves you

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2 years ago

Write a quadratic equation to represent the situation. Solve the equation by factoring. Use the table and graphing method to che

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Answer:

Step-by-step explanation:

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2 years ago

Find the greatest common factor of 26 and 14.

Molodets [167]

Answer:

Step-by-step explanation:

A factor is a number that can evenly divide into another number.

Let’s list all the factors of both numbers.

26: 1, 2, 13, 26

14: 1,2, 7, 14

Now find the GCF, or greatest common factor. The biggest number that is a factor of both 26 and 14 is 2.

Therefore, the GCF of 26 and 14 is 2.

6 0

3 years ago

Factor Completely:<br> 4x²+8x

Sidana [21]

The answer is 4x(x+2).

8 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