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goldfiish [28.3K]
3 years ago
12

Solve for c when,

iddle" class="latex-formula">
​
Mathematics
2 answers:
solong [7]3 years ago
5 0
C=4. hope this helps
allochka39001 [22]3 years ago
4 0

Answer:

c = 4

Step-by-step explanation:

hope this helps

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About 2.43 kg of pepper
Divide 17 by 7
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How is 7/6 the same as 21/18 please how do i show my work
Oduvanchick [21]

Answer:

Multiply 7/6 by 3/3

Step-by-step explanation:

Multiplying 7/6 by 3/3 will give you a result of 21/18.

Hope this helps!

6 0
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Write an expression for the sequence of operations described below.
Sveta_85 [38]
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6 0
3 years ago
Find a polynomial f(x) of degree 3 with real coefficents and following zeros. -4,4i
ElenaW [278]

Answer:

f(x)=x^3 +4x^2 +16x +64

Step by step explanation:

\text{Given that,  two roots are}~ -4~ \text{and}~  4i.\\\\\text{Let,}\\\\~~~~~~~x = 4i\\\\\implies  x^2 = 16i^2~~~~~~~;[\text{Square on both sides}]\\\\\implies x^2 = -16~~~~~~~~;[i^2 = -1]\\\\\implies x^2 +16 = 0\\\\\text{So,}~ x^2 +16~ \text{ is a factor of the 3 degree polynomial}.\\ \\ \text{The polynomial is ,}\\\\ f(x) = (x+4)(x^2 +16)\\\\~~~~~~~=x^3 +16x +4x^2 +64\\\\~~~~~~~=x^3 +4x^2 +16x +64

7 0
2 years ago
If sinx = p and cosx = 4, work out the following forms :<br><br><br>​
Kay [80]

Answer:

$\frac{p^2 - 16} {4p^2 + 16} $

Step-by-step explanation:

I will work with radians.

$\frac {\cos^2 \left(\frac{\pi}{2}-x \right)+\sin(-x)-\sin^2 \left(\frac{\pi}{2}-x \right)+\cos \left(\frac{\pi}{2}-x \right)} {[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)]}$

First, I will deal with the numerator

$\cos^2 \left(\frac{\pi}{2}-x \right)+\sin(-x)-\sin^2 \left(\frac{\pi}{2}-x \right)+\cos \left(\frac{\pi}{2}-x \right)$

Consider the following trigonometric identities:

$\boxed{\cos\left(\frac{\pi}{2}-x \right)=\sin(x)}$

$\boxed{\sin\left(\frac{\pi}{2}-x \right)=\cos(x)}$

\boxed{\sin(-x)=-\sin(x)}

\boxed{\cos(-x)=\cos(x)}

Therefore, the numerator will be

$\sin^2(x)-\sin(x)-\cos^2(x)+\sin(x) \implies \sin^2(x)- \cos^2(x)$

Once

\sin(x)=p

\cos(x)=4

$\sin^2(x)-\cos^2(x) \implies p^2-4^2 \implies \boxed{p^2-16}$

Now let's deal with the numerator

[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)]

Using the sum and difference identities:

\boxed{\sin(a \pm b)=\sin(a) \cos(b) \pm \cos(a)\sin(b)}

\boxed{\cos(a \pm b)=\cos(a) \cos(b) \mp \sin(a)\sin(b)}

\sin(\pi -x) = \sin(x)

\sin(2\pi +x)=\sin(x)

\cos(2\pi-x)=\cos(x)

Therefore,

[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)] \implies [\sin(x)+\cos(x)] \cdot [\sin(x)\cos(x)]

\implies [p+4] \cdot [p \cdot 4]=4p^2+16p

The final expression will be

$\frac{p^2 - 16} {4p^2 + 16} $

8 0
3 years ago
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