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3 years ago

I’m taking a test plz anyone help!

Mathematics

1 answer:

Mandarinka [93]3 years ago

7 0

Ok so, basically reflect the shape horizontally across the y - intercept. In the exact same position but reflected across. I’m not sure because I can’t really see the points. Does it give you options? Or do you have to type it?

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The original price of a car is entered into spreadsheet cell A1 and the annual depreciation amount in cell B1.

Reil [10]

Answer:

A1 / B1 ;

(D1 - A1) / B1 ;

(A1 - E1*A1) / B1

Step-by-step explanation:

A1 = original price of car

B1 = annual Depreciation amount

Number of years it will take for the car to depreciate totally :

Using the straight line Depreciation relation :

y = mx + c

c = intercept = initial or original value of car

m = annual Depreciation amount

x = number of years

y = value after x years

For total Depreciation, final value, y = 0

0 = mx + c

mx = - c

x = - c / m

Hence, x = A1 / B1

B.)

D1 = car value

Length it will take for car to depreciate to value in D1 :

y = mx + c

y = D1; m = B1 ; c = A1

D1 = B1x + A1

B1x = D1 - A1

x = (D1 - A1) / B1

C.)

E1 = decrease percentage

Time it takes for car to decrease by percentage in E1

y = E1 * A1

E1 * A1 = B1x + A1

(A1 - E1*A1) = B1x

x = (A1 - E1*A1) / B1

4 0

2 years ago

What is the product?

earnstyle [38]

-9 -8
2. 0

Just multiply

3 0

3 years ago

I need help it’s for a very hard test

Mumz [18]

Answer:

Step-by-step explanation

This has a factor of 6% or 0.06. But, you are adding and not subtracting so it is 1.06. ( If you multiply by a number less than 1 then it is actually dividing) So C is the only on that is correct.

6 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

3 years ago

3(4-2x)=-2x will make brainiest to the first two people that answer correctly with explanation

ASHA 777 [7]

Answer:

x = 3

Step-by-step explanation:

First, distribute 3 into the parenthesis:

3(4 - 2x) = -2x

12 - 6x = -2x

Next, combine your x variables by adding +6x to both side:

12 - 6x = -2x (-6x and +6x cancel out)

+6x +6x

12 = 4x (divide 12 by 4 to get x by itself)

/4 /4

x = 3

3 0

3 years ago