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3 years ago

Help plz. I'll mark brainliest

Mathematics

1 answer:

Vera_Pavlovna [14]3 years ago

6 0

1. rule = r - 12 (number = 25)
2. rule = a + 34 (number = 43)
3. rule = s - 5 (number = 20)
4. rule = b + 16 (number = 32)
5. rule = w + 3 (number = 9)
6. rule = n + 9 (number = 42)

7. 15 - n when n = 9 would be 15 - 9, and your answer would be 6.

8. "32 more than a number d" implies add d to 32, so your answer is C. 32 + d

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Help please?!?!?!?!?!?!?

nadezda [96]

Answer:

Choice D is correct answer.

Step-by-step explanation:

We have given two function.

f(x) =2ˣ+5x and g(x) = 3x-5

We have to find the addition of given two function.

(f+g)(x) = ?

The formula to find the addition, we have

(f+g)(x) = f(x) + g(x)

Putting given values in above formula, we have

(f+g)(x) = (2ˣ+5x)+(3x-5)

(f+g)(x) = 2ˣ+5x+3x-5

Adding like terms, we have

(f+g)(x) = 2ˣ+8x-5 which is the answer.

6 0

3 years ago

In a survey 9 out of 15 named math as their favorite class. express this rate as a decimal

ZanzabumX [31]

0.6 is 9/15 as a decimal.

3 0

3 years ago

Each problem below gives the endpoints of a segment. Find the coordinates of the midpoints of each segment. A.(5, 2) and (11, 14

kvv77 [185]

The midpoints are (8,3) and (6.5,6).

<u>Step-by-step explanation</u>:

Midpoint formula = ((x1+x2)/2 , (y1+y2)/2)

(x1,y1) = (5,2)

(x2,y2) = (11,4)

Midpoint = ((5+11)/2 , (2+4)/2)

⇒ ((16/2) , (6/2))

⇒ (8,3)

(x1,y1) = (3,8)

(x2,y2) = (10,4)

Midpoint = ((3+10)/2 , (8+4)/2)

⇒ ((13/2) , (12/2))

⇒ (6.5,6)

8 0

3 years ago

1. Approximate the given quantity using a Taylor polynomial with n3.

Jet001 [13]

Answer:

See the explanation for the answer.

Step-by-step explanation:

Given function:

$f(x) = x^{1/4}$

The n-th order Taylor polynomial for function f with its center at a is:

$p_{n}(x) = f(a) + f'(a) (x-a)+\frac{f''(a)}{2!} (x-a)^{2} +...+\frac{f^{(n)}a}{n!} (x-a)^{n}$

As n = 3 So,

$p_{3}(x) = f(a) + f'(a) (x-a)+\frac{f''(a)}{2!} (x-a)^{2} +...+\frac{f^{(3)}a}{3!} (x-a)^{3}$

$p_{3}(x) = f(a) + f'(a) (x-a)+\frac{f''(a)}{2!} (x-a)^{2} +...+\frac{f^{(3)}a}{6} (x-a)^{3}$

$p_{3}(x) = a^{1/4} + \frac{1}{4a^{ 3/4} } (x-a)+ (\frac{1}{2})(-\frac{3}{16a^{7/4} } ) (x-a)^{2} + (\frac{1}{6})(\frac{21}{64a^{11/4} } ) (x-a)^{3}$

$p_{3}(x) = 81^{1/4} + \frac{1}{4(81)^{ 3/4} } (x-81)+ (\frac{1}{2})(-\frac{3}{16(81)^{7/4} } ) (x-81)^{2} + (\frac{1}{6})(\frac{21}{64(81)^{11/4} } ) (x-81)^{3}$