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3 years ago

Graph the following equation

Mathematics

1 answer:

Shalnov [3]3 years ago

7 0

Answer:

Graphic attached

Step-by-step explanation:

The oblique asymptote of the function has the shape of a line of the form

$y = mx + b$

We need to find the slope m and the intercept b.

The oblique asymptote is found by these two limits:

$m = \lim_{x \to \infty}\frac{f(x)}{x}\\\\b = \lim_{x \to \infty}[f(x) - mx]$

If $f(x) = \frac{(2x+3)(x-6)}{(x+2)(x-1)}$ then:

$m = \lim_{x \to \infty} \frac{\frac{(2x+3)(x-6)}{(x+2)(x-1)}}{x}\\\\m = \lim_{x \to \infty} \frac{2x^2-9x-18}{x^3 +x^2 -2x}\\\\m = 0$

The slope is 0. Therefore the function has no oblique asymptote.

Horizontal asymptote:

$y = 2$

Vertical asymptote

$x = -2\\x = 1$

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Graph the line with the equation y=1/3x+1

devlian [24]

Answer:

See photo

Step-by-step explanation:

$\frac{1}{3}$ = $\frac{rise}{run}$

1 = slope intercept

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3 years ago

jake makes 9 loaves of olive bread. he uses 30 grams of olives in each loaf.he started with 1 kilogram of olives.how many grams

8090 [49]

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3 years ago

Michael drinks 355 milliliters of orange juice and puts 6 centiliters of syrup on his pancakes for breakfast. How many millilite

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2 years ago

Read 2 more answers

Find sin(a)&cos(B), tan(a)&cot(B), and sec(a)&csc(B).

Reil [10]

Answer:

Part A) $sin(\alpha)=\frac{4}{7},\ cos(\beta)=\frac{4}{7}$

Part B) $tan(\alpha)=\frac{4}{\sqrt{33}},\ tan(\beta)=\frac{4}{\sqrt{33}}$

Part C) $sec(\alpha)=\frac{7}{\sqrt{33}},\ csc(\beta)=\frac{7}{\sqrt{33}}$

Step-by-step explanation:

Part A) Find $sin(\alpha)\ and\ cos(\beta)$

we know that

If two angles are complementary, then the value of sine of one angle is equal to the cosine of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

$sin(\alpha)=cos(\beta)$

Find the value of $sin(\alpha)$ in the right triangle of the figure

$sin(\alpha)=\frac{8}{14}$ ---> opposite side divided by the hypotenuse

simplify

$sin(\alpha)=\frac{4}{7}$

therefore

$sin(\alpha)=\frac{4}{7}$

$cos(\beta)=\frac{4}{7}$

Part B) Find $tan(\alpha)\ and\ cot(\beta)$

we know that

If two angles are complementary, then the value of tangent of one angle is equal to the cotangent of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

$tan(\alpha)=cot(\beta)$

Find the value of the length side adjacent to the angle alpha

Applying the Pythagorean Theorem

Let

x ----> length side adjacent to angle alpha

$14^2=x^2+8^2\\x^2=14^2-8^2\\x^2=132$

$x=\sqrt{132}\ units$

simplify

$x=2\sqrt{33}\ units$

Find the value of $tan(\alpha)$ in the right triangle of the figure

$tan(\alpha)=\frac{8}{2\sqrt{33}}$ ---> opposite side divided by the adjacent side angle alpha

simplify

$tan(\alpha)=\frac{4}{\sqrt{33}}$

therefore

$tan(\alpha)=\frac{4}{\sqrt{33}}$

$tan(\beta)=\frac{4}{\sqrt{33}}$

Part C) Find $sec(\alpha)\ and\ csc(\beta)$

we know that

If two angles are complementary, then the value of secant of one angle is equal to the cosecant of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

$sec(\alpha)=csc(\beta)$

Find the value of $sec(\alpha)$ in the right triangle of the figure

$sec(\alpha)=\frac{1}{cos(\alpha)}$

Find the value of $cos(\alpha)$

$cos(\alpha)=\frac{2\sqrt{33}}{14}$ ---> adjacent side divided by the hypotenuse

simplify

$cos(\alpha)=\frac{\sqrt{33}}{7}$

therefore

$sec(\alpha)=\frac{7}{\sqrt{33}}$

$csc(\beta)=\frac{7}{\sqrt{33}}$

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3 years ago

Find the Area of the figure below,

murzikaleks [220]

Answer:

169.1 unit^2 to the nearest tenth.

Step-by-step explanation:

Area of the parallelogram = base * height

= 24 * 6 unit^2.

Area of the semicircle = 1/2 pi r^2.

Here r = 1/2 * 8 = 4.

So the area is 1/2 * pi * 4^2

= 8pi unit^2.

The area of the whole figure = 24*6 + 8pi

= 169.13 unit^2.

3 0

2 years ago