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3 years ago

Algebra 2

Mathematics

1 answer:

olga55 [171]3 years ago

4 0

Answer:

Step-by-step explanation:

-7-6(-2)

-7+12

sorry didn't read it right

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What is the equation of (4,10),(6,11) in standard form?

SIZIF [17.4K]

ANSWER

The standard form is:

$x - 2y = - 16$

EXPLANATION

The given line passes through (4,10)and (6,11).

The slope of this line is

$m = \frac{11 - 10}{6 - 4}$

$m = \frac{1}{2}$

The equation is given by the formula,

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

We plug in the first point and the slope to get.

$y-10= \frac{1}{2} (x-4)$

$2y - 20 = x - 4$

$x - 2y = - 20 + 4$

$x - 2y = - 16$

The standard form is

$x - 2y = - 16$

7 0

3 years ago

2. Mrs. Siebenaller bought a “bus” for $25,000 with a 7% interest rate. Mrs. S. gets a loan to pay off in 5 years. How much inte

PSYCHO15rus [73]

Answer: the correct option is A

Step-by-step explanation:

Cost of the bus is $25000. She took a loan of $25000 to be paid of in 5 years at an interest rate of 7%.

The formula for simple interest is expressed as

I = PRT/100

Where

I = interest

P = principal or initial amount borrowed

T = time in years

R = interest rate in percentage

From the information given,

P = 25000

R = 7%

T = 5 years

Therefore

I = (25000 × 7 × 5)/100 = 875000/100

I = $8750

3 0

3 years ago

Read 2 more answers

What is the best estimate for the percent of students scoring greater than 92 on at test?

son4ous [18]

Answer:

Step-by-step explanation:

7 0

3 years ago

10) The binary notation of 231) is

siniylev [52]

Answer: 1110 0111

Step-by-step explanation:

3 0

2 years ago

Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n

aliya0001 [1]

Answer:

$f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}$

Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

$f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...$

We first compute the n-th derivative of $f(x)=\ln(1+2x)$ , note that

$f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\$

Now, if we compute the n-th derivative at 0 we get

$f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!$

and so the Maclaurin series for f(x)=ln(1+2x) is given by

$f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2 \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}$

3 0

3 years ago