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2 years ago

Are the lines y=4x+4 and y=-4x-2 parallel

Mathematics

2 answers:

skelet666 [1.2K]2 years ago

7 0

Answer:

No these are not parallel

Step-by-step explanation:

Parallel lines have the same slope. 4 and -4 are not the same slope, which means they will intersect. A tool to check your answers such as this is Desmos. It really helps. Search Desmos graphing calculator. Hope this helps.

Alex Ar [27]2 years ago

6 0

Answer:

No they are not, they intersect at some point.

Step-by-step explanation:

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A B What is the domain and range?

posledela

Answer:

the domain is 5 and the range is 10

Step-by-step explanation:

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2 years ago

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there are 6 buses transporting students to a baseball game with 32 students on each bus.each row at the baseball stadium seats 8

quester [9]

Amount of students x Amount of buses
= 32 x 6
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You need 192 seats since there are 192 students. To find how many rows you need you divide 192 by 8 which gives you 24

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3 years ago

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Suppose that Y varies directly with X andY equals 12 when X equals -2 what is X when Y equals -6

VLD [36.1K]

Answer:

$x=1$

Step-by-step explanation:

A direct variation equation has the following format:

$y=kx$

Where k is a constant.

We know that y is 12 when x is -2. Thus, substitute:

$12=-2k$

Divide both sides by -2:

$k=-6$

So, our direct variation equations is:

$y=-6x$

To find what x is when y equals -6, substitute in -6 for y:

$-6=-6x$

Divide both sides by -6:

$x=1$

So, our answer is 1 :)

4 0

3 years ago

Find the surface area of the cylinder

STatiana [176]

<h2>Solution:-</h2>

• Surface area of a cylinder = 2πr(r + h) sq. units.

In the above diagram,

Radius (r) of the cylinder = 13 cm.

Height (h) of the cylinder = 39 cm.

<h3>• Taking value of π = 3.14</h3>

Hence, Surface area of the cylinder = 2 × 3.14 × 13(13 + 39)

Surface area of the cylinder = (81.64 × 52) cm²

Surface area of the cylinder = 4245.28 cm²

The surface area of the cylinder is 4245.28 cm². [Answer]

6 0

2 years ago

Use the definition of a Taylor series to find the first three non zero terms of the Taylor series for the given function centere

Ket [755]

Answer:

$e^{4x}=e^4+4e^4(x-1)+8e^4(x-1)^2+...$

$\displaystyle e^{4x}=\sum^{\infty}_{n=0} \dfrac{4^ne^4}{n!}(x-1)^n$

Step-by-step explanation:

Taylor series expansions of f(x) at the point x = a

$\text{f}(x)=\text{f}(a)+\text{f}\:'(a)(x-a)+\dfrac{\text{f}\:''(a)}{2!}(x-a)^2+\dfrac{\text{f}\:'''(a)}{3!}(x-a)^3+...+\dfrac{\text{f}\:^{(r)}(a)}{r!}(x-a)^r+...$

This expansion is valid only if $\text{f}\:^{(n)}(a)$ exists and is finite for all $n \in \mathbb{N}$ , and for values of x for which the infinite series converges.

$\textsf{Let }\text{f}(x)=e^{4x} \textsf{ and }a=1$

$\text{f}(x)=\text{f}(1)+\text{f}\:'(1)(x-1)+\dfrac{\text{f}\:''(1)}{2!}(x-1)^2+...$

$\boxed{\begin{minipage}{5.5 cm}\underline{Differentiating $e^{f(x)}$}\\\\If $y=e^{f(x)}$, then $\dfrac{\text{d}y}{\text{d}x}=f\:'(x)e^{f(x)}$\\\end{minipage}}$