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3 years ago

Which set of ordered pairs could be generated by an exponential function?

Mathematics

2 answers:

AleksandrR [38]3 years ago

5 0

Answer:

(0, 0), , (2, 1)

Step-by-step explanation:

Leya [2.2K]3 years ago

3 0

Answer:

(3,1) +(1,6)

Step-by-step explanation:

first yiu

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What’s the equation of the blue line??

Sholpan [36]

Answer:

y = -1/2 + 2 (Slope intercept)

Step-by-step explanation:

Y-intercept is when the x-value is 0.

So the y-intercept is (0, 2)

To find slope, find two random points:

(0, 2) , (2, 1)

Slope (m) =ΔY/ΔX = -1/2 = -0.5

Equation of the line:

y = -0.5x + 2

y = -1/2 + 2

When x=0, y = 2

When y=0, x = 4

4 0

2 years ago

A triangle has side lengths of (8d – 3) centimeters, (6d + 2) centimeters, and

alisha [4.7K]

The perimeter of a triangle is the sum of its side lengths.

The perimeter of the triangle is: $14d + 7f +7$

The sides of the triangle are:

Side 1 = (8d - 3) cm

Side 2 = (6d + 2) cm

Side 3 = (7f + 8) cm

The perimeter (P) is calculated as:

P = Side 1 + Side 2 + Side 3

So, we have:

$P = (8d - 3) + (6d + 2) + (7f + 8)$

Remove brackets

$P = 8d - 3 + 6d + 2 + 7f + 8$

Collect like terms

$P = 8d + 6d + 7f - 3 + 2 + 8$

$P = 14d + 7f +7$

Hence, the perimeter of the triangle is: $14d + 7f +7$

Read more about perimeters at:

brainly.com/question/6465134

3 0

3 years ago

Megan wants to retile her bathroom floor. If the home improvment store sel

Pavel [41]

79.68 dollars...........

8 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

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