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1 year ago

Create a equation and label the graph

Mathematics

1 answer:

Alona [7]1 year ago

5 0

A. The linear equation is: y = 60x + 45

B. The graphed equation is shown in the attachment below.

<h3>What is a Linear Equation?</h3>

A linear equation is defined as, y = mx + b, where the initial value = b; unit rate = m; x and y are the variables.

A. In this situation given, we have:

Total cost = y
Number of months = x
Monthly charges = unit rate (m)= $60
Initial value (b) = $45

Substituting the values into y = mx + b, we would have the equation as:

y = 60x + 45

B. The graph of the equation is shown in the image attached below.

Learn more about linear equation on:

brainly.com/question/15602982

#SPJ1

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Given h(x) = |x+3| -5

statuscvo [17]

Answer:

The parent function f(x) is equal to $f\left(x\right)=\left|x\right|$

The translations is 3 units to the left and 5 units down

Step-by-step explanation:

we have

$h\left(x\right)=\left|x+3\right|-5$

The vertex of the function h(x) is the point (-3,-5)

we know that the parent function f(x) is equal to

$f\left(x\right)=\left|x\right|$

The vertex of the function f(x) is the point (0,0)

The rule of the transformation of f(x) to h(x) is equal to

(x,y) -----> (x-3,y-5)

That means ----> The translations is 3 units to the left and 5 units down

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3 years ago

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.

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3 years ago

Rewrite the following linear equation in slope intercept form.<br> Y-5=3(x+1)

jarptica [38.1K]

Y=3x+8

Step-by-step explanation:

The slope is 3

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2 years ago

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ella [17]

Answer: $x=\sqrt{17}-5$ and $x = -\sqrt{17}-5$

Step-by-step explanation:

Alright, lets get started.

The given equation is :

$x^{2} +10x = -8$

Adding 8 in both sides, it will become

$x^{2} +10x + 8 = -8 +8$

$x^{2} +10x + 8 = 0$

To make it perfect square, we need to add 25 and subtract 25

$x^{2} +10x + 8 +25-25= 0$

$x^2+10x+25+8-25=0$

$(x+5)^2+8-25=0$

$(x+5)^2-17=0$

Adding 17 in both sides

$(x+5)^2=17$

taking square root

$x+5=\sqrt{17}$

So,

$x=\sqrt{17}-5$ and $x = -\sqrt{17}-5$ .. Answer

Hope it will help :)

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