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3 years ago

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Mathematics

2 answers:

Arte-miy333 [17]3 years ago

8 0

There is an infinite number of values that are in both the domain and range.

arsen [322]3 years ago

7 0

Answer:

100% sure There is an infinite number of values that are in both the domain and range.

Step-by-step explanation:

You might be interested in

5x – 12y – 17 = 0<br> 1*9(x + 2) - (y + 1) + 3 = 0

valentina_108 [34]

Answer: $x=\frac{12y+17}{5}$

$y=\frac{-5x+17}{-12}$

$x=\frac{y-20}{9}$

y=9x+20

Step-by-step explanation:

5x-12y-17=0 (plus 17 from both sides)

5x-12y=17 (plus 12y from both sides)

5x=12y+17 (divide 5 from everything)

$x=\frac{12y+17}{5}$

5x-12y-17=0 (plus 17 from both sides)

5x-12y=17 (minus 5x from both sides)

-12y=-5x+17 (divide everything by -12)

$y=\frac{-5x+17}{-12}$

1*9(x+2)-(y+1)+3=0 (distribute 9 to x and to 2) (distribute -1 to y and to 1)

1*9x+18-y-1+3=0 (times 9x by 1)

9x+18-y-1+3=0 (combine like terms)

9x-y+20=0 (minus 20 from both sides)

9x-y=-20 (plus y to both sides)

9x=y-20 (divide 9 from both sides)

$x=\frac{y-20}{9}$

1*9(x+2)-(y+1)+3=0 (distribute 9 to x and to 2) (distribute -1 to y and to 1)

1*9x+18-y-1+3=0 (times 9x by 1)

9x+18-y-1+3=0 (combine like terms)

9x-y+20=0 (minus 20 from both sides)

9x-y=-20 (minus 9x from both sides)

-y=-9x-20 (divide -1 from everything)

y=9x+20

3 0

3 years ago

Louann and Carla received equal scores on a test made up of multiple choice questions and an essay. Louann got 14 multiple choic

Gnoma [55]

Louann= 14x+24
Carla= 16x+14
So basically this is just like solving equations. The way I did this was just put random numbers for x and then solve it.
First attempt:
14(7)+24=16(7)+14
98+24=112+14
122=126
Okay the first one didn't work so I use 6 for x.
Second attempt:
14(6)+24=16(6)+14
84+24=96+14
108=110

So 6 didn't work and you already should know that the answers should be between 1 and 5.

So now I going to do 5 for x
Third attempt:
14(5)+24=16(5)+14
70+24=80=14
94=94

And so the answer is 5! :D The multiply choice questions were worth 5 points and Louann and Carla had 94 for their scores! I hope this help you out and clear for you! :D

7 0

4 years ago

Read 2 more answers

Which of the following solves the equation t/8 =<br> 2

JulsSmile [24]

Answer:

I don't know what the options are, but to solve t/8 = 2, you have to multiply by 8.

Hope this helps :)

6 0

3 years ago

The common ratio of a geometric series is 3 and the sum of the first 8 terms is 3280.

Pavlova-9 [17]

Answer:

The first term of the geometric series is 1

Step-by-step explanation:

In this question, we are tasked with calculating the first term of a geometric series, given the common ratio, and the sum of the first 8 terms.

Mathematically, the sum of terms in a geometric series can be calculated as;

S = a(r^n-1)/( r-1)

where a is the first term that we are looking for

r is the common ratio which is 3 according to the question

n is the number of terms which is 8

S is the sum of the number of terms which is 3280 according to the question

Plugging these values, we have

3280 = a(3^8 -1)/(3-1)

3280 = a( 6561-1)/2

3280 = a(6560)/2

3280 = 3280a

a = 3280/3280

a = 1

6 0

3 years ago

Find the nth taylor polynomial for the function, centered at c. f(x) = ln(x), n = 4, c = 5

masha68 [24]

The nth taylor polynomial for the given function is

P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

Given:

f(x) = ln(x)

n = 4

c = 3

nth Taylor polynomial for the function, centered at c

The Taylor series for f(x) = ln x centered at 5 is:

$P_{n}(x)=f(c)+\frac{f^{'} (c)}{1!}(x-c)+ \frac{f^{''} (c)}{2!}(x-c)^{2} +\frac{f^{'''} (c)}{3!}(x-c)^{3}+.....+\frac{f^{n} (c)}{n!}(x-c)^{n}$

Since, c = 5 so,

$P_{4}(x)=f(5)+\frac{f^{'} (5)}{1!}(x-5)+ \frac{f^{''} (5)}{2!}(x-5)^{2} +\frac{f^{'''} (5)}{3!}(x-5)^{3}+.....+\frac{f^{n} (5)}{n!}(x-5)^{n}$

Now

f(5) = ln 5

f'(x) = 1/x ⇒ f'(5) = 1/5

f''(x) = -1/x² ⇒ f''(5) = -1/5² = -1/25

f'''(x) = 2/x³ ⇒ f'''(5) = 2/5³ = 2/125

f''''(x) = -6/x⁴ ⇒ f (5) = -6/5⁴ = -6/625

So Taylor polynomial for n = 4 is:

P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

Hence,

The nth taylor polynomial for the given function is

P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

Find out more information about nth taylor polynomial here

brainly.com/question/28196765

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3 0

2 years ago