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nalin [4]
1 year ago
10

Find the slope of the line that passes through the following points: (-4,-2) and (-3,5)

Mathematics
1 answer:
stepan [7]1 year ago
6 0
Formula: y2-y1/x2-x1
label your variables by x and y.
subtract y2 from y1, then x2 from x1.
simplify your answer
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Find the coordinates of the midpoint HX. H (5 1/2, -4 3/4), X (2 1/4, 1 1/4)
Fiesta28 [93]

Answer:

  (3 7/8, -1 3/4)

Step-by-step explanation:

The midpoint is the average of the end point coordinates:

  ((5 1/2, -4 3/4) +(2 1/4, 1 1/4))/2

  = (7 3/4, -3 1/2)/2 = (3 7/8, -1 3/4)

6 0
3 years ago
How is the definition of property similar in science and math​
sdas [7]

Answer:

Properties in math are used as general rules to solve problems.

I hope this helped!

Step-by-step explanation:

6 0
3 years ago
Find the quotient -14/7
AVprozaik [17]

The quotient is -2.


4 0
4 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
RUDIKE [14]

The taylor series for the f(x)=8/x centered at the given value of a=-4 is -2+2(x+4)/1!-24/16 (x+4)^{2}/2!+...........

Given a function f(x)=9/x,a=-4.

We are required to find the taylor series for the function f(x)=8/x centered at the given value of a and a=-4.

The taylor series of a function f(x)=f(a)+f^{1}(a)(x-a)/1!+ f^{11}(a)(x-a)^{2} /2! +f^{111}(a)(x-a)a^{3}/3!+..........

Where the terms in f prime f^{1}(a) represent the derivatives of x valued at a.

For the given function.f(x)=8/x and a=-4.

So,f(a)=f(-4)=8/(-4)=-2.

f^{1}(a)=f^{1}(-4)=-8/(-4)^{2}

=-8/16

=-1/2

The series of f(x) is as under:

f(x)=f(-4)+f^{1}(-4)(x+4)/1!+  f^{11}(-4)(x+4)^{2}/2!.............

=8/(-4)-8/(-4)^{2} (-4)(x+4)/1!+  24/(-4)^{3} (-4)(x+4)^{2}/2!.............

=-2+2(x+4)/1!-24/16 (x+4)^{2}/2!+...........

Hence the taylor series for the f(x)=8/x centered at the given value of a=-4 is -2+2(x+4)/1!-24/16 (x+4)^{2}/2!+...........

Learn more about taylor series at brainly.com/question/23334489

#SPJ4

3 0
1 year ago
Help
Art [367]

\huge\fbox{Answer ☘}

\bold{3( \frac{27}{8} ) {}^{ \frac{2}{3}  \times  - 1}  = 2( \frac{32}{243} ) {}^{2x} }\\  \\3(( \frac{3}{2} ) {}^{3} ) {}^{ \frac{2}{3} \times  - 1 }  = 2(( \frac{2}{3} ) {}^{5} ) {}^{2x}  \\\\ 3( \frac{3}{2} ) {}^{2 \times  - 1}  = 2( \frac{2}{3} ) {}^{10x}  \\\\ 3( \frac{2}{3} ) {}^{2}  = 2( \frac{2}{3} )  {}^{10x}  \\\\ 3( \frac{4}{9} ) = 2( \frac{4}{9} ) {}^{5x}  \\\\\bold\pink{ comparing \: powers }\\\\5x = 1 \\\\ \bold\blue{x =  \frac{1}{5} }

hope helpful~

5 0
2 years ago
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