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2 years ago

Anyone help pls? No links! :)

Mathematics

1 answer:

Lunna [17]2 years ago

4 0

Answer:

6. D

7. F

8.A

9. B

10 C

13.Question- Write an expression using division and subtraction with a difference of 3.

answer- (12 divide 3) -1

Twelve divided by three is four. Four minus 1 is three.

12. Question- Write an expression using multiplication and addition with a sum of 16.

answer- There are many answers for this and this is one of them:

Y=4(3+1)

Y=16

Y=2(6+2)

Y=16

11. Question- Sam bought two CDs for $13 each. Sales tax for both CDs was $3. Write an expression to show how much Sam paid in all.

answer- Expression- (13*2)(3*2)=$32

This is the answer because Sam bought two CD's for 13, 13*2, and the sales tax was $ 3, so 3*2=6

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A cylinder has a radius of 4x + 3 and a height of 3x + 6. Which polynomial in standard form best

MrRa [10]

Answer:

Step-by-step explanation:

R=4x+3

H=3x+6

Volume of a cylinder= Πr²h

inserting the parameters

v=π(4x+3)²(3x+6)

v=π(16x²+24x+9)(3x+6)

V=π[3x(16x²+24x+9) +6(16x²+24x+9)]

v=Π(48x³+72x²+27x+96x²+144x+54)

v=π(48x³+168x²+171x+54)m³

8 0

3 years ago

If f(x) is parallel to the line y

meriva

It’s the first one since parallel means same slope

6 0

2 years ago

Read 2 more answers

Place the quadratic y=2x^2+24x+79 into vertex form by using the method of completing the square and then state the coordinates o

katrin2010 [14]

Answer:

$y=2(x+6)^2+7$

$(-6,7)$

Step-by-step explanation:

$y=2x^2+24x+79$

Completing the square is a process of converting a quadratic equation in standard form into vertex form.

The first step in completing the square is grouping the quadratic and linear terms of the quadratic equation.

$y=(2x^2+24x)+79$

Factor out the coefficient of the quadratic term,

$y=2(x^2+12x)+79$

Now complete the square, add a term to make the grouped part of the equation a complete square, then balance the equation.

$y=2(x^2+12x+36-36)+79$

Simplify,

$y=2(x^2+12x+36)+79+(2)(-36)$

$y=2(x+6)^2+79-72$

$y=2(x+6)^2+7$

The x-coordinate of the vertex of the equation is equal to (-1) times the numerical part of the quadratic term, and the y-coordinate is equal to the constant.

$(-6,7)$

7 0

3 years ago

Sketch the graph of and identify the axis of symmetry

kaheart [24]

The axis of symmetry is found within the set of parenthesis with the x. If our h value of the vertex is -4, then the axis of symmetry is x = -4. D is that choice. Cannot graph it here, but your vertex is sitting at (-4, 4), it's an upside down parabola, and some other points on this graph are (-5, 0), (-3, 0), (-6, -12), (-2, -12). You could graph it using those points and the vertex without a problem, I'm sure.

6 0

3 years ago

Read 2 more answers

Use the Taylor series you just found for sinc(x) to find the Taylor series for f(x) = (integral from 0 to x) of sinc(t)dt based

Marina CMI [18]

In this question (brainly.com/question/12792658) I derived the Taylor series for $\mathrm{sinc}\,x$ about $x=0$ :

$\mathrm{sinc}\,x=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}$

Then the Taylor series for

$f(x)=\displaystyle\int_0^x\mathrm{sinc}\,t\,\mathrm dt$

is obtained by integrating the series above:

$f(x)=\displaystyle\int\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}\,\mathrm dx=C+\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}$

We have $f(0)=0$ , so $C=0$ and so

$f(x)=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}$

which converges by the ratio test if the following limit is less than 1:

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(-1)^{n+1}x^{2n+3}}{(2n+3)^2(2n+2)!}}{\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}}\right|=|x^2|\lim_{n\to\infty}\frac{(2n+1)^2(2n)!}{(2n+3)^2(2n+2)!}$

Like in the linked problem, the limit is 0 so the series for $f(x)$ converges everywhere.

7 0

3 years ago