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3 years ago

Draw a number line . Locate each integer on the number line.

Mathematics

2 answers:

Mumz [18]3 years ago

8 0

Sorry for my messy handwriting. Here you go :)

kvv77 [185]3 years ago

7 0

-5, -4, 0, 2, 3. From smallest to largest

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What is the side length of the largest square that can fit into a circle with a radius of 5 units?

Mariana [72]

So 1st consider that it's a square! That's very important. So for a square, all 4 sides are equal.

And now considering that the given information is the diameter. So any angle made at the circle extended from the 2 points of diameter gives an angle of 90°

Now consider one triangle. So we already know that 2 sides of the triangle are equal (because they are 2 sides of a square) , has a side of 10 (diameter) and and angle of 90°. So remaining 2 angles are 45°

Now solve it by applying

$\sin(45) \: \: \: \: = x \div 10 \\ (1 \div \sqrt{2} ) = (x \div 10) \\ 10 \div \sqrt{2} \: = x$

5 0

3 years ago

Write the point-slope form of the line that passes through (-8, 2) and is perpendicular to a line with a slope of -8

Schach [20]

Answer:

y - 2 = (1/8)(x + 8)

Step-by-step explanation:

The slope of the given line is -8. That means the slope of any line perpendicular to the given line will be the negative reciprocal of -8, which would be (1/8).

Then the desired equation is

y - 2 = (1/8)(x + 8)

4 0

4 years ago

Help.... Other question got deleted.

Helen [10]

-2/25 is the slope
x1=4
hope this helps!

7 0

3 years ago

Write the explicit formula which can be used to represent the sequence 5,9,13,17

Oliga [24]

Answer:

4n+1

Step-by-step explanation:

Term 1: 4(1)+1 = 5

Term 2: 4(2)+1 = 9

Term 3: 4(3)+1 = 13

Term 4: 4(4)+1 = 17

4 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