All categories

3 years ago

Complete the square for the expression. Also, identify the resulting expression as a binomial squared.

Mathematics

2 answers:

lakkis [162]3 years ago

4 0

Answer:

20x?

Step-by-step explanation:

Ede4ka [16]3 years ago

4 0

Answer:

13 is the answer

Step-by-step explanation:

You might be interested in

Find the area of the shape shown below<br><br> please any ty:)

Vedmedyk [2.9K]

Answer:

240

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

I'LL GIVE BRAINLIEST TO WHO EVER ANSWERS!!!!

Nookie1986 [14]

Answer:

Answer is Step 2

6 0

3 years ago

Solve the following systems of equations.<br> 2^(2x+y−1) = 32<br> 4^(x−2y) = 2

Rina8888 [55]

Answer:(x,y)=(2.5,1)

Step-by-step explanation:

Given

$2^{2x+y-1}=32$

$2^{2x+y-1}=2^5$

thus on comparing

2x+y=6------1

$4^{x-2y}=2$

$2^{2(x-2y)}=2$

On comparing exponent of 2

$x-2y=\frac{1}{2}$ ------2

we get x=2.5

substitute x in 1

y=1

(x,y)=(2.5,1)

6 0

3 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

Write the standard equation of the circle with the given center and passes through the given point. Center (7,-2); Point= (1,-6)

vodomira [7]

Answer:

(x - 7)² + (y + 2)² = (2√13)²

Step-by-step explanation:

We already know that the center is at (7, -2), but must find the radius. The radius is the distance between the points (7, -2) and (1, -6):

r = √ [6² + 4²] ) = √(36 + 16) = √52 = √4√13 = 2√13.

Then the desired equation is (x - 7)² + (y + 2)² = (2√13)²

6 0

3 years ago