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2 years ago

12

My sister needs help with this

1 answer:

Oliga [24]2 years ago

4 0

Sailing ships should be correct

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Daredevil Danny was cliff jumping . He jumped off a cliff that was 14.2 meters above the surface . He jumped 2.3 meters below wa

Novay_Z [31]

Answer:

that he is trying to die

Step-by-step explanation:

i don't know just stating the obvious.

6 0

2 years ago

Read 2 more answers

Y varies directly with x and y = 12 when x =5 what is the value of y when x = 8

torisob [31]

Y and x are directly proportional, as stated by hypothesis. Let's use the rule of three to find y when x = 8.

x=5 --> y=12
x=8 --> y=?

? = (8*12)/5 = 96/5 = 19.2

So y directly varies with x, and y=12 when x = 5, then y=19.2 when x=8.

Hope this Helps! :D

3 0

3 years ago

I WILL GIVE 20 POINTS TO THOSE WHO ANSWER THIS QUESTION RIGHT NOOOO SCAMS AND EXPLAIN WHY THAT IS THE ANSWER

luda_lava [24]

Answer:

Step-by-step explanation:

7 0

2 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

The point (2, 3) is on the line given by which equation below?

love history [14]

Plug in the points and see if y and x are equal to each other.

(2,3)
y= x-1
3= 2-1
3= 1 (not the solution)

y=3x
3=3*3
3=9 (not the solution)

y=x+1
3=2+1
3=3 (solution) We can check to make sure.

y=-3x
3=-3*3
3=-9 (not the solution)

y=x+1 <em>is the answer.
</em>
I hope this helps!
~kaikers

5 0

3 years ago

Read 2 more answers