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3 years ago

Find sin D sin E cos D and cos E

Mathematics

1 answer:

Andrew [12]3 years ago

5 0

9514 1404 393

Answer:

sin(D) = cos(E) = (√3)/2

cos(D) = sin(E) = 1/2

Step-by-step explanation:

The mnemonic SOH CAH TOA is intended to remind you of the relationships between trig functions and right triangle sides.

Sin = Opposite/Hypotenuse

Cos = Adjacent/Hypotenuse

For this diagram, this means ...

sin(D) = cos(E) = (13√3)/26 = (√3)/2

cos(D) = sin(E) = 13/26 = 1/2

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Find the x= to the answer

pickupchik [31]

Answer:

17°

Step-by-step explanation:

I'm pretty sure because it has the arrow pointing up the angle x labeled 17 degrees...

(hope this helps!)

6 0

3 years ago

What is the determinant of the coefficient matrix of the system {4x+3y+2z=0 -3x+y+5z=0 -x-4y+3z=0

slava [35]

Answer:

130

Step-by-step explanation:

You want the determinant of the matrix ...

$\left[\begin{array}{ccc}4&3&2\\-3&1&5\\-1&-4&3\end{array}\right]$

One way to figure it is as the difference between the sum of products of the down-diagonals and the sum of products of the up-diagonals:

D = (4)(1)(3) +(3)(5)(-1) +(2)(-3)(-4) -(-1)(1)(2) -(-4)(5)(4) -(3)(-3)(3)

= 12 -15 +24 +2 +80 +27

D = 130

The determinant of the coefficient matrix is 130.

_____

Many scientific and graphing calculators and web sites can perform this calculation for you.

7 0

3 years ago

Find the slope and Y-intercept of the graph of this equation <br><br> y-9x=1/2

natita [175]

Hello,

Your answer would be:

Y=9x+1/2

Plz mark me brainliest

4 0

3 years ago

Devin has started a lawn care service. He will charge a mowing fee of $10 plus $3 per square yard of lawn. He also offers trimmi

pychu [463]

Answer is D because he charge 10 $ mowing free +3 per square yard of lawn
lawn has dimensions:
x and y
Area= x*y=xy squar yard
so charge is
10 +3*Area=10+3xy

7 0

3 years ago

Read 2 more answers

Save me the headache

maxonik [38]

$(9\sin2x+9\cos2x)^2=81$

Taking the square root of both sides gives two possible cases,

$9\sin2x+9\cos2x=9\implies\sin2x+\cos2x=1$

$9\sin2x+9\cos2x=-9\implies\sin2x+\cos2x=-1$

Recall that

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

If $\alpha=2x$ and $\beta=\dfrac\pi4$ , we have

$\sin\left(2x+\dfrac\pi4\right)=\dfrac{\sin2x+\cos2x}{\sqrt2}$

so in the equations above, we can write

$\sin2x+\cos2x=\sqrt2\sin\left(2x+\dfrac\pi4\right)=\pm1$

Then in the first case,

$\sqrt2\sin\left(2x+\dfrac\pi4\right)=1\implies\sin\left(2x+\dfrac\pi4\right)=\dfrac1{\sqrt2}$

$\implies2x+\dfrac\pi4=\dfrac\pi4+2n\pi\text{ or }\dfrac{3\pi}4+2n\pi$

(where $n$ is any integer)

$\implies2x=2n\pi\text{ or }\dfrac\pi2+2n\pi$

$\implies x=n\pi\text{ or }\dfrac\pi4+n\pi$

and in the second,

$\sqrt2\sin\left(2x+\dfrac\pi4\right)=-1\implies\sin\left(2x+\dfrac\pi4\right)=-\dfrac1{\sqrt2}$

$\implies2x+\dfrac\pi4=-\dfrac\pi4+2n\pi\text{ or }-\dfrac{3\pi}4+2n\pi$

$\implies2x=-\dfrac\pi2+2n\pi\text{ or }-\pi+2n\pi$

$\implies x=-\dfrac\pi4+n\pi\text{ or }-\dfrac\pi2+n\pi$

Then the solutions that fall in the interval $[0,2\pi)$ are

$x=0,\dfrac\pi4,\dfrac\pi2,\dfrac{3\pi}4,\pi,\dfrac{5\pi}4,\dfrac{3\pi}2,\dfrac{7\pi}4$

5 0

3 years ago

Read 2 more answers