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3 years ago

Evaluate |x-13| when x=5. A.8 B.-8 C.21 D.-21

1 answer:

8 0

5 - 13 = -8

But remember it is in the ||. This means the answer will, no matter what, be positive.

So the answer is |-8|, or +8.

A is the answer.

Now for an explanation of why the || makes it positive.

They are know as "absolute values." Absolute values will always, no matter what, be a positive number. These can be very tricky to deal with, but just remember to answer what is in the || first, and THEN do the absolute value. Remember that on a number line, the middle is always 0. And because -8 and 8 are both 8 spaces away from 0, the absolute value will be 8. This is why it is positive. You cannot be negative spaces away from something.

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2x^2-3x-9 please help

Romashka-Z-Leto [24]

Im assuming you mean 2x^2-3x-9=0 ?
If so, you have to factorise the equation
(2x-3) (x+3) = 0

2x - 3 = 0
2x = 3
x = 1.5

x + 3 = 0
x = -3

x is either 1.5 or -3

8 0

3 years ago

Read 2 more answers

Find T5(x) : Taylor polynomial of degree 5 of the function f(x)=cos(x) at a=0 . (You need to enter function.) T5(x)= Find all va

Burka [1]

Answer:

$\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}$

The polynomial is an approximation with an error less than or equals to 0.002652 for x in the interval

[-1.113826815, 1.113826815]

Step-by-step explanation:

According to Taylor's theorem

$\bf f(x)=f(0)+f'(0)x+f''(0)\displaystyle\frac{x^2}{2}+f^{(3)}(0)\displaystyle\frac{x^3}{3!}+f^{(4)}(0)\displaystyle\frac{x^4}{4!}+f^{(5)}(0)\displaystyle\frac{x^5}{5!}+R_6(x)$

with

$\bf R_6(x)=f^{(6)}(c)\displaystyle\frac{x^6}{6!}$

for some c in the interval (-x, x)

In the particular case f

f(x)=cos(x)

we have

$\bf f'(x)=-sin(x)\\f''(x)=-cos(x)\\f^{(3)}(x)=sin(x)\\f^{(4)}(x)=cos(x)\\f^{(5)}(x)=-sin(x)\\f^{(6)}(x)=-cos(x)$

therefore

$\bf f'(x)=-sin(0)=0\\f''(0)=-cos(0)=-1\\f^{(3)}(0)=sin(0)=0\\f^{(4)}(0)=cos(0)=1\\f^{(5)}(0)=-sin(0)=0$

and the polynomial approximation of T5(x) of cos(x) would be

$\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}$

In order to find all the values of x for which this approximation is within 0.002652 of the right answer, we notice that

$\bf R_6(x)=-cos(c)\displaystyle\frac{x^6}{6!}$

for some c in (-x,x). So

$\bf |R_6(x)|\leq|\displaystyle\frac{x^6}{6!}|=\displaystyle\frac{|x|^6}{6!}$

and we must find the values of x for which

$\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652$

Working this inequality out, we find

$\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652\Rightarrow |x|^6\leq1.90944\Rightarrow\\\\\Rightarrow |x|\leq\sqrt[6]{1.90944}\Rightarrow |x|\leq1.113826815$

Therefore the polynomial is an approximation with an error less than or equals to 0.002652 for x in the interval

[-1.113826815, 1.113826815]

8 0

2 years ago

Ben needs to buy orange juice. He can buy 64 fl oz at the grocery store for $2.00, or he can buy 256 fl oz at the wholesale club

maw [93]

2. unit price = $ ÷ oz
a) 64 fl oz for $2 = 2/64 ≈ $0.031 per ounce 
b) 256 fl oz for $7.5 = 7.5/256 = $0.029 per ounce 
c) buy wholesale because 0.029 < 0.031 

3. unit rate means 1 for the miles 
3/10 in ÷ 7/8 mi 
= 12/35 inches per mile ...

8 0

3 years ago

Can someone plz explain to me when to use an open and closed dot

balandron [24]

closed dot is if the number is part of the solution so if it is equal to

and

use open if it is not part of solution as in if it is greater than or less than

so for your picture x greater or equal to -2 it would be a closed dot

5 0

3 years ago

What is the value of y in the equation 4 + y = −3?

omeli [17]

4 + y = -3

subtract 4 from each side

y = -3-4

y = -7

4 0

3 years ago