GONZALEZ

PLEASE HELP ASAP!!!!

OleMash [197]

Answer:

y = -2x + 8

Step-by-step explanation:

The point slope form of an equation is written as

y = mx + c ...............(i)

Where m is the slope and c is the constant

Now we Know that the equation is

y + 2 = -2(x-5)

and the given points are

m= -2

x = 5

and y= -2

Putting these values in equation (i) to find the value of c

y = mx + c

it becomes

-2 = -2(5) + c

-2 = -10 + c

Adding 10 on both sides

-2 + 10 = -10 + 10 + c

8 = c

or c=8

Now we have the values of m and c

where m= -2 and c = 8

Point slope form of an equation is

y = mx + c

putting the values of m and c to get equation in slope intercept form is

y = (-2)x + 8

or

y = -2x + 8

Other method:

The given point slope form is

y + 2 = -2(x-5)

We have to change it in y= mx + c form

so solving it

y + 2 = -2x + 10

Subtracting 2 from both sides

y + 2 -2 = -2x + 10 -2

y = -2x + 8

which is same is

y=mx + c

so the required equation is

y = -2x + 8

7 0

3 years ago

Mr solomon can run 2 1/2 in 2 hours. how many miles can he run in 6 hour PLZ HELP!!

Marat540 [252]

Answer:

7 1/2

Step-by-step explanation:

2 1/2 times 3 is 7 1/2

6 0

3 years ago

Read 2 more answers

Find the measures of the labeled angles

Andrei [34K]

Answer:

135 degrees \

Step-by-step explanation:

5x= x+108

Subtract x from the right

4x= 108

divide by 4

x= 27

Substitute for x in the original equation

(x+108)

(27)+(108)

= 135 degrees

4 0

3 years ago

Ameekah is working with consecutive integers

Irina-Kira [14]

Consecutive numbers: 1, 2, 3, 4, 5.

As we can see it's the latest plus 1

So,

x, x + 1, x + 1 + 1, x + 1 + 1 + 1

x, x + 1, x + 2, x + 3 and so on

So it's x + 1.

7 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

3 years ago