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3 years ago

Help please... dont answer if u dont know bc this is for a grade

Mathematics

2 answers:

emmasim [6.3K]3 years ago

6 0

The answer is A !!!

salantis [7]3 years ago

3 0

The answer is a !!!!!

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Pls help will give brainlist

hjlf

Answer:

WELL I DONT SEE THE PICTUE BUT WHATS THE PROBLEM?

Step-by-step explanation:

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

Find an equation of the line having the given slope and containing the given point. m=8 (-4,-3)

notsponge [240]

Answer:

$y=8x+29$

Step-by-step explanation:

Since we know the slope and a point, we can use the point-slope form.

The point-slope form is:

$y-y_1=m(x-x_1)$

Substitute 8 for m, and let (-4,-3) be x₁ and y₁, respectively. Thus:

$y-(-3)=8(x-(-4))$

Simplify:

$y+3=8(x+4)$

Distribute:

$y+3=8x+32$

Subtract 3 from both sides:

$y=8x+29$

And we're done!

7 0

3 years ago

-1/2x>-2x - (-3/4x +1)

Sergeu [11.5K]

I neeeeeeeed it skkyvbnj

5 0

3 years ago

In the pair equations bx - 6y = 18 and 2x - 3y = 9, b is a constant. The system of two equations has infinitely many solutions.

sukhopar [10]

Well, if

{bx - 6y = 18 and {2x - 3y = 9, you might want to try elimination

Multiply second equation by -2: bx - 6y = 18 and -2(2x - 3y = 9)

Simplified: bx - 6y = 18 and -4x + 6y = -18

Now if you add them you get: -4x + bx = 0

Then simplified: bx = 4x

divide by x: b = 4

Your final solution is b = 4.

Hope that helps!

7 0

3 years ago