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3 years ago

Your help would be greatly appreciated

Mathematics

1 answer:

vitfil [10]3 years ago

5 0

Answer:

12/13

Step-by-step explanation:

Sin, remember it as SOH. SOH is Sin=Opposite/hypotenuse.

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Is 9r+16=5 a literal equation? Explain.

Butoxors [25]

This is not a literal equation but you can divide 3.14 divide by 5

Step-by-step explanation:

5 0

3 years ago

<img src="https://tex.z-dn.net/?f=%7C%5Cfrac%7Bx%2B1%7D%7Bx-1%7D%2B1%7C%5C%20%20%5Ctextgreater%20%5C%20%5Cfrac%7Bx%2B1%7D%7Bx-1%

VMariaS [17]

The inequality boils down to

|y| > y

By definition of absolute value, we have

• |y| = y if y ≥ 0

• |y| = -y if y < 0

So if y ≥ 0, we have

y > y

but this is a contradiction.

On the other hand, if y < 0, we have

-y > y ==> 2y < 0 ==> y < 0

and no contradiction.

Now replace y with (x + 1)/(x - 1) + 1. Then you're left with solving

(x + 1)/(x - 1) + 1 < 0

(x + 1 + x - 1)/(x - 1) < 0

2x/(x - 1) < 0

The left side is negative if either 2x > 0 and x - 1 < 0, or 2x < 0 and x - 1 > 0. The first case reduces to x > 0 and x < 1, or 0 < x < 1. In the second case, we get x < 0 and x > 1, but x cannot satisfy both conditions, so we throw this case out.

7 0

3 years ago

Is 69 a prime or composite number

Alchen [17]

Composite number. 69=3×23

8 0

4 years ago

A. The slope is 55, representing the trip fee.

Morgarella [4.7K]

Answer:

c. the slope is 70, representing the old cost

4 0

3 years ago

Consider the curve defined by the equation y=6x2+14x. Set up an integral that represents the length of curve from the point (−2,

torisob [31]

Answer:

32.66 units

Step-by-step explanation:

We are given that

$y=6x^2+14x$

Point A=(-2,-4) and point B=(1,20)

Differentiate w.r. t x

$\frac{dy}{dx}=12x+14$

We know that length of curve

$s=\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^2}dx$

We have a=-2 and b=1

Using the formula

Length of curve= $s=\int_{-2}^{1}\sqrt{1+(12x+14)^2}dx$

Using substitution method

Substitute t=12x+14

Differentiate w.r t. x

$dt=12dx$

$dx=\frac{1}{12}dt$

Length of curve= $s=\frac{1}{12}\int_{-2}^{1}\sqrt{1+t^2}dt$

We know that

$\sqrt{x^2+a^2}dx=\frac{x\sqrt {x^2+a^2}}{2}+\frac{1}{2}\ln(x+\sqrt {x^2+a^2})+C$

By using the formula

Length of curve= $s=\frac{1}{12}[\frac{t}{2}\sqrt{1+t^2}+\frac{1}{2}ln(t+\sqrt{1+t^2})]^{1}_{-2}$

Length of curve= $s=\frac{1}{12}[\frac{12x+14}{2}\sqrt{1+(12x+14)^2}+\frac{1}{2}ln(12x+14+\sqrt{1+(12x+14)^2})]^{1}_{-2}$

Length of curve= $s=\frac{1}{12}(\frac{(12+14)\sqrt{1+(26)^2}}{2}+\frac{1}{2}ln(26+\sqrt{1+(26)^2})-\frac{12(-2)+14}{2}\sqrt{1+(-10)^2}-\frac{1}{2}ln(-10+\sqrt{1+(-10)^2})$

Length of curve= $s=\frac{1}{12}(13\sqrt{677}+\frac{1}{2}ln(26+\sqrt{677})+5\sqrt{101}-\frac{1}{2}ln(-10+\sqrt{101})$

Length of curve= $s=32.66$

5 0

3 years ago

Your help would be greatly appreciated ​

Your help would be greatly appreciated