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2 years ago

Be Precise What is the difference between a terminating decimal and a repeating decimal?

Mathematics

1 answer:

matrenka [14]2 years ago

5 0

Answer:

Terminating decimals have a finite number of digits after the decimal point. Repeating decimals have one or more repeating numbers or sequences of numbers after the decimal point, which continue infinitely.

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SVETLANKA909090 [29]

Answer:

The point of intersection is at (2,3)

Step-by-step explanation:

7 0

3 years ago

Look at the expression 1/2 (c + 8). Tell whether each statement about the expression is True or False. 50 points if right and if

Alika [10]

Answer:

a)true becuase ×1/2 and ÷2 are doing the same thing

b) true bc they simplified the equation

c)true bc a term is a number or letter and 1/2 is the coefficient

d) true bc thats what the equation is telling you to do

Step-by-step explanation:

I hope that helped

4 0

3 years ago

Read 2 more answers

3+4=4+___? I need help for the math assignment. Thank you.

Montano1993 [528]

Answer:

Step-by-step explanation:

because that's the communicative property

6 0

3 years ago

Read 2 more answers

The equation for the cost in dollars of producing automobile tires is c = 0.000015x2 - 0.03x + 35, where x is the number of tire

statuscvo [17]

The cost function is
c = 0.000015x² - 0.03x + 35
where x = number of tires.

To find the value of x that minimizes cost, the derivative of c with respect to x should be zero. Therefore
0.000015*2x - 0.03 = 0
0.00003x = 0.03
x = 1000

Note:
The second derivative of c with respect to x is positive (= 0.00003), so the value for x will yield the minimum value.

The minimum cost is
Cmin = 0.000015*1000² - 0.03*1000 + 35
= 20

Answer:
Number of tires = 1000
Minimum cost = 20

3 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