Help ASAP!!!!!!! Plz

The original price of a shirt at Walmart is $25. It goes on sale for $20. What is the

Katena32 [7]

The answer is 20% off

7 0

2 years ago

Your teacher is selecting class partners by drawing names out of a bag. There are 12 girl's names and 10 boy's names. What is th

sasho [114]

The answer would be A)20/77.
The reason is that without replacement it would go (10/22)*(12/21)=0.2597
You can do the easy route I did then do (20/77) which would also give you 0.2597.

8 0

2 years ago

Read 2 more answers

1) 5.10-2.8.16. 10-1

const2013 [10]

Answer: -2511 (if the periods mean multiplication)

Step-by-step explanation:

I’m not sure whether you’re trying to have the periods be multiplication, but if that’s what they are, this is the answer

5x10-2x8x16x10-1

50-16x160-1

50-2560-1

-2511

Again, I’m not sure if that’s what you mean but I can’t really see what else they would mean since you can’t have a 3 part decimal

5 0

3 years ago

What is the remainder when (3x4 + 2x3 − x2 + 2x − 19) ÷ (x + 2)? 0 5 10 15

Elanso [62]

Answer: Second option is correct.

Explanation:

Since we have given that

$f(x)=3x^4+2x^3-x^2+2x-19$

And $g(x)=x+2$

To find the remainder, we use the Remainder Theorem, which states that when f(x) is divided by (x-c) then the f(c) is the required remainder.

So,

Here, we have,

$x+2=0\\\\x=-2$

So, we will find f(-2) which will give us the required remainder,

$f(-2)=3(-2)^4+2(-2)^3-(-2)^2+2(-2)-19\\\\f(-2)=(3\times 16)-(2\times 8)- 4-4-19\\\\f(-2)=48-16-36-4-19\\\\f(-2)=5$

Hence, Second option is correct.

7 0

3 years ago

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