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avanturin [10]
2 years ago
10

Help me with this please im not sure how i passed my last test-

Mathematics
2 answers:
MaRussiya [10]2 years ago
7 0

Answer:

-2/3n + 7 = 15

-2/3n = 15 + 7

-2/3n = 22

n = 2/3 + 22

n = 24/3

n = 8

Alex2 years ago
6 0
Answer:
-2/3n + 7 = 15

Good luck on your test :D
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What is the value of the function at x=−2? <br> A) y=−2<br> B) y = 0<br> C) y = 2<br> D) y = 3
VashaNatasha [74]
It’s A for the value of the function u asking for
5 0
3 years ago
Solve the given linear Diophantine equation. Show all necessary work. A) 4x + 5y=17 B)6x+9y=12 C) 4x+10y=9
aliina [53]

Answer:

A) (-17+5k,17-4k)

B)  (-4+3k,4-2k)

C) No integer pairs.

Step-by-step explanation:

To do this, I'm going to use Euclidean's Algorithm.

4x+5y=17

5=4(1)+1

4=1(4)

So going backwards through those equations:

5-4(1)=1

-4(1)+5(1)=1

Multiply both sides by 17:

4(-17)+5(17)=17

So one integer pair satisfying 4x+5y=17 is (-17,17).

What is the slope for this equation?

Let's put it in slope-intercept form:

4x+5y=17

Subtract 4x on both sides:

     5y=-4x+17

Divide both sides by 5:

      y=(-4/5)x+(17/5)

The slope is down 4 and right 5.

So let's show more solutions other than (-17,17) by using the slope.

All integer pairs satisfying this equation is (-17+5k,17-4k).

Let's check:

4(-17+5k)+5(17-4k)

-68+20k+85-20k

-68+85

17

That was exactly what we wanted since we were looking for integer pairs that satisfy 4x+5y=17.

Onward to the next problem.

6x+9y=12

9=6(1)+3

6=3(2)

Now backwards through the equations:

9-6(1)=3

9(1)-6(1)=3

Multiply both sides by 4:

9(4)-6(4)=12

-6(4)+9(4)=12

6(-4)+9(4)=12

So one integer pair satisfying 6x+9y=12 is (-4,4).

Let's find the slope of 6x+9y=12.

6x+9y=12

Subtract 6x on both sides:

      9y=-6x+12

Divide both sides by 9:

       y=(-6/9)x+(12/9)

Reduce:

       y=(-2/3)x+(4/3)

The slope is down 2 right 3.

So all the integer pairs are (-4+3k,4-2k).

Let's check:

6(-4+3k)+9(4-2k)

-24+18k+36-18k

-24+36

12

That checks out since we wanted integer pairs that made 6x+9y=12.

Onward to the last problem.

4x+10y=9

10=4(2)+2

4=2(2)

So the gcd(4,10)=2 which means this one doesn't have any solutions because there is no integer k such that 2k=9.

6 0
3 years ago
Find the taylor series for f(x) centered at the given value of a. [assume that f has a power series expansion. do not show that
Bond [772]

Taylor series is f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2}  }

To find the Taylor series for f(x) = ln(x) centering at 9, we need to observe the pattern for the first four derivatives of f(x). From there, we can create a general equation for f(n). Starting with f(x), we have

f(x) = ln(x)

f^{1}(x)= \frac{1}{x} \\f^{2}(x)= -\frac{1}{x^{2} }\\f^{3}(x)= -\frac{2}{x^{3} }\\f^{4}(x)= \frac{-6}{x^{4} }

.

.

.

Since we need to have it centered at 9, we must take the value of f(9), and so on.

f(9) = ln(9)

f^{1}(9)= \frac{1}{9} \\f^{2}(9)= -\frac{1}{9^{2} }\\f^{3}(x)= -\frac{1(2)}{9^{3} }\\f^{4}(x)= \frac{-1(2)(3)}{9^{4} }

.

.

.

Following the pattern, we can see that for f^{n}(x),

f^{n}(x)=(-1)^{n-1}\frac{1.2.3.4.5...........(n-1)}{9^{n} }  \\f^{n}(x)=(-1)^{n-1}\frac{(n-1)!}{9^{n}}

This applies for n ≥ 1, Expressing f(x) in summation, we have

\sum_{n=0}^{\infinite} \frac{f^{n}(9) }{n!} (x-9)^{2}

Combining ln2 with the rest of series, we have

f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2}  }

Taylor series is f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2}  }

Find out more information about taylor series here

brainly.com/question/13057266

#SPJ4

3 0
2 years ago
What is the equation and soultion to x times 3 plus 4 equals 16
Zielflug [23.3K]

Hello!

Let's write this as an equation below.

3x+4=16

We subtract 4 from both sides.

3x=12

Divide both sides by 3

x=4

I hope this helps!

8 0
3 years ago
What is the place value of 3 in 5.349
Lisa [10]
The place value is 300.
5 0
3 years ago
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