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3 years ago

Can somebody help me on number 2 or 3? Please explain as well. HELP ASAP :)

Mathematics

1 answer:

harina [27]3 years ago

7 0

What grade a you in?

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<img src="https://tex.z-dn.net/?f=%5Cfrac%7B%5Csec%5Cleft%28x%5Cright%29%7D%7B%5Ccos%5Cleft%28x%5Cright%29%7D-%5Cfrac%7B%5Csin%5

DanielleElmas [232]

Answer:

Step-by-step explanation:

First, convert all the secants and cosecants to cosine and sine, respectively. Recall that $csc(x)=1/sin(x)$ and $sec(x)=1/cos(x)$ .

Thus:

$\frac{sec(x)}{cos(x)} -\frac{sin(x)}{csc(x)cos^2(x)}$

$=\frac{\frac{1}{cos(x)} }{cos(x)} -\frac{sin(x)}{\frac{1}{sin(x)}cos^2(x) }$

Let's do the first part first: (Recall how to divide fractions)

$\frac{\frac{1}{cos(x)} }{cos(x)}=\frac{1}{cos(x)} \cdot \frac{1}{cos(x)}=\frac{1}{cos^2(x)}$

For the second term:

$\frac{sin(x)}{\frac{cos^2(x)}{sin(x)} } =\frac{sin(x)}{1} \cdot\frac{sin(x)}{cos^2(x)}=\frac{sin^2(x)}{cos^2(x)}$

So, all together: (same denominator; combine terms)

$\frac{1}{cos^2(x)}-\frac{sin^2(x)}{cos^2(x)}=\frac{1-sin^2(x)}{cos^2(x)}$

Note the numerator; it can be derived from the Pythagorean Identity:

$sin^2(x)+cos^2(x)=1; cos^2(x)=1-sin^2(x)$

Thus, we can substitute the numerator:

$\frac{1-sin^2(x)}{cos^2(x)}=\frac{cos^2(x)}{cos^2(x)}=1$

Everything simplifies to 1.

7 0

3 years ago

What‘s 1+1? ;);) have a good day :)

galben [10]

Answer: 2 :))

Step-by-step explanation: hope you also have a good day!!

4 0

3 years ago

Read 2 more answers

I can’t figure this out at all.

Liula [17]

The fact that this triangle is a right angle triangle makes you now have 2 angles and the 1 side given, so it should be solvable.
First, You know that A=46 and C=90 as it is the right angle, and you know that the sum of any triangle's angles is 180. so now B=180-(90-46)=44
Now to the sides,
sin(B)=opp./hyp.=b/c=8/c=sin(44)
so, c=8/sin(44) which is approximately 11.52 unit length
now, use Pythagoras to find a,
a=√c²-b² =√11.52²-8² which is approximately 8.3 unit length.
Hope this helps.

6 0

3 years ago

Which graph represents this equation?

zheka24 [161]

The graph in the bottom left is correct

5 0

3 years ago

-1/225x^2+2/3x what is the vertex of this quadratic function?

Julli [10]

$\bf \textit{vertex of a parabola}\\ \quad \\\\

\begin{array}{lccclll}
y=&-\frac{1}{225}x^2&+\frac{2}{3}x\\\\
y=&-\frac{1}{225}x^2&+\frac{2}{3}x&+0\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad 
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)$

6 0

3 years ago