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3 years ago

12

What is the equation of the line that passes through the point (-8,2) and has an undefined slope?

1 answer:

astra-53 [7]3 years ago

4 0

Answer: x = -8

Step-by-step explanation: Since this line has <em>no slope</em>, it must be <em>vertical</em>.

The equation of a vertical line through a given point can

always be written x = the x-coordinate of the given point.

So the answer to this problem is x = -8.

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tekilochka [14]

Answer:

$6.03

Step-by-step explanation:

Let's begin by listing out the given variables:

height (h) = 12 cm, diameter (d) = 8 cm, r = d ÷ 2 ⇒ r = 4 cm, cost of coconut oil (c) = $0.01 /cm³

The formula of cylinder is given by:

V = πr²h = π * 4² * 12 = <u>603.1858 cm³</u>

cost of filling the tin can with coconut oil = cost of coconut oil * Volume of cylinder

Cost = c * V = 0.01 * 603.1858

Cost = $<u>6.03</u>

5 0

3 years ago

5 of what percent is 3.5

UNO [17]

If you take 3.5 and divide it by 0.05, or times 20, that would make it 70.

7 0

3 years ago

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What is the proper equation <br> (-5cd^-4)(2cd^2)^2

mojhsa [17]

Answer:

-20c^3

Step-by-step explanation:

((−5c)(d−4))((2cd2)2)

= −20c3d4 /d4

=-20c^3

7 0

3 years ago

Please solve 5 f <br> (Trigonometric Equations)<br> #salute u if u solved it

Zanzabum

Answer:

$\beta=45\degree\:\:or\:\:\beta=135\degree$

Step-by-step explanation:

We want to solve $\tan \beta \sec \beta=\sqrt{2}$ , where $0\le \beta \le360\degree$ .

We rewrite in terms of sine and cosine.

$\frac{\sin \beta}{\cos \beta} \cdot \frac{1}{\cos \beta} =\sqrt{2}$

$\frac{\sin \beta}{\cos^2\beta}=\sqrt{2}$

Use the Pythagorean identity: $\cos^2\beta=1-\sin^2\beta$ .

$\frac{\sin \beta}{1-\sin^2\beta}=\sqrt{2}$

$\implies \sin \beta=\sqrt{2}(1-\sin^2\beta)$

$\implies \sin \beta=\sqrt{2}-\sqrt{2}\sin^2\beta$

$\implies \sqrt{2}\sin^2\beta+\sin \beta- \sqrt{2}=0$

This is a quadratic equation in $\sin \beta$ .

By the quadratic formula, we have:

$\sin \beta=\frac{-1\pm \sqrt{1^2-4(\sqrt{2})(-\sqrt{2} ) } }{2\cdot \sqrt{2} }$

$\sin \beta=\frac{-1\pm \sqrt{1^2+4(2) } }{2\cdot \sqrt{2} }$

$\sin \beta=\frac{-1\pm \sqrt{9} }{2\cdot \sqrt{2} }$

$\sin \beta=\frac{-1\pm3}{2\cdot \sqrt{2} }$

$\sin \beta=\frac{2}{2\cdot \sqrt{2} }$ or $\sin \beta=\frac{-4}{2\cdot \sqrt{2} }$

$\sin \beta=\frac{1}{\sqrt{2} }$ or $\sin \beta=-\frac{2}{\sqrt{2} }$

$\sin \beta=\frac{\sqrt{2}}{2}$ or $\sin \beta=-\sqrt{2}$

When $\sin \beta=\frac{\sqrt{2}}{2}$ , $\beta=\sin ^{-1}(\frac{\sqrt{2} }{2} )$

$\implies \beta=45\degree\:\:or\:\:\beta=135\degree$ on the interval $0\le \beta \le360\degree$ .

When $\sin \beta=-\sqrt{2}$ , $\beta$ is not defined because $-1\le \sin \beta \le1$

4 0

3 years ago

Solve for x, angle A, and Angle B<br> A x+50 B x+80 c72

jolli1 [7]

I’m assuming you man degrees? Are these angles of a triangle?

This doesn’t really make sense because the angles in a triangle add up to 180° so your equation would be x+50+x+80+72=180

Combine like terms to get
2x+202=180

Then subtract 202 from both sides
2x=-22

So x would be -11 which doesn’t seem right lol

7 0

3 years ago

Read 2 more answers