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3 years ago

Write an equation in slope intercept form for the line that passes through (-1, -2) and (3, 4)

Mathematics

1 answer:

Vera_Pavlovna [14]3 years ago

3 0

Answer:

i do not no i do not no

Step-by-step explanation:

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Susan buys her lunch in the cafeteria each day for a week. On Monday she spends $1.50, on Tuesday $2.25, on Wednesday $1.76, on

olga_2 [115]

Answer:

The answer to your question is $11.17

Step-by-step explanation:

Data

Monday $1.50

Tuesday $2.25

Wednesday $1.76

Thursday $2.24

Friday $3.42

Process

1.- Sum up all the data, the result will be the answer

1.50

2.25

+ 1.76

2.24

$ 11.17

2.- Conclusion

Susan spent $11.17 on lunch this week

7 0

4 years ago

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ginny needs to leave the house by 9;00 she woke up at 5;00 how many minutes does she have to get ready

jok3333 [9.3K]

Answer:

240minutes

Step-by-step explanation:

She has 4 hours and 1 hour = 60minutes

so 4x60=240minutes

7 0

2 years ago

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The base of a triangular tent has angles of 45 degrees and 75 degrees. What is the measurement of the third angle?

vekshin1

Answer:i think its 180

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Billy knows that she can run .4 miles in 2 1/2 how minutes and .6 miles in 3 3/4 minutes how long would it take to walk 2 miles

wel

It is 12 1/2 minutes it would take

7 0

3 years ago

Use the appropriate substitutions to write down the first four nonzero terms of the Maclaurin series for the binomial (1+3x)^(-1

PolarNik [594]

Answer:

First term=1

Second term=-x

Third term= $2x^2$

Fourth term = $-\frac{28}{3!}x^3$

Step-by-step explanation:

We are given that function

$f(x)=(1+3x)^{-1/3}$

We have to find the first four non zero terms of the Maclaurin series for the binomial.

Maclaurin series of function f(x) is given by

$f(x)=f(0)+f'(0)x+\frac{1}{2!}f''(0)x^2+\frac{1}{3!}f'''(0)x^3+....$

$f(0)=(1+3x)^{\frac{-1}{3}}=1$

$f'(x)=-\frac{1}{3}(1+3x)^{-\frac{4}{3}}(3)=-(1+3x)^{-\frac{4}{3}}$

$f'(0)=-1$

$f''(x)=\frac{4}{3}\times 3 (1+3x)^{-\frac{7}{3}}$

$f''(0)=4$

$f'''(x)=-4\times \frac{7}{3}\times 3(1+3x)^{-\frac{10}{3}}$

$f'''(0)=-28$

Substitute the values we get

$(1+3x)^{-\frac{1}{3}}=1-x+\frac{4}{2!}x^2+\frac{-28}{3!}x^3+...$

$(1+3x)^{-\frac{1}{3}}=1-x+2x^2+\frac{-28}{3!}x^3+...$

First term=1

Second term=-x

Third term= $2x^2$

Fourth term = $-\frac{28}{3!}x^3$

5 0

3 years ago