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xxMikexx [17]
3 years ago
6

I NEED HELP PLEASE, THANKS! :)

Mathematics
2 answers:
Y_Kistochka [10]3 years ago
4 0
(I’m currently on mobile so sorry for possible bad formatting)

First, recall the trig identity: tan(A-B)=(tan(A)-tan(B))/(1+tan(A)tan(B))

We see that the given equation directly corresponds with this, with A=9x and B=5x.

Thus we can equate the given equation to tan(9x-5x)=tan(4x)
This cannot be simplified further.
KIM [24]3 years ago
3 0

Answer:  tan(14x)

<u>Step-by-step explanation:</u>

Consider the Sum Formula for tan:

tan(A + B)=\dfrac{tan(A)+tan(B)}{1-tan(A)(tan(B)}\\\\\\tan(9x+5x)=\dfrac{tan(9x)+tan(5x)}{1-tan(9x)(tan(5x)}\\\\\\\large\boxed{tan(14x)}=\dfrac{tan(9x)+tan(5x)}{1-tan(9x)(tan(5x)}

You might be interested in
Solve the inequality 2x - 5 \le -x +12. Give your answer as an interval.
ludmilkaskok [199]

Answer: \begin{bmatrix}\mathrm{Solution:}\:&\:x\le \frac{17}{3}\:\\ \:\mathrm{Decimal:}&\:x\le \:5.66666\dots \\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:\frac{17}{3}]\end{bmatrix}

Step-by-step explanation:

2x-5\le \:-x+12

2x\le \:-x+17\\

2x+x\le \:-x+17+x

3x\le \:17

\frac{3x}{3}\le \frac{17}{3}

x\le \frac{17}{3}

6 0
3 years ago
Please help! It would be very much appreciated :)
Trava [24]

Answer:

  1. Cos 45=6/x

x=6/cos 45=6√2

x=6√2

tan 45=y/6

y=tan45 ×6

y=6

2.

sin 60=y/10

y=sin60×10

y=5√3

again

Cos 60=x/10

x=½×10

x=5

3.

Sin 45=x/5

x=1/√2×5 or

x=5√2/2

again

y=5√2/2 or 5/√2[base angle of right angled isosceles triangle]

4.

Sin 30=8/x

½×x=8

x=8*2

x=16

again

cos 30= y/x

√3/2×16=y

y=8√3

5.

sin 60=7/x

x=7/[√3/2]

x=14/√3 or 14√3/3

again

cos 60=y/x

½×14√3/3=y

7√3/2=y

y=7√3/2

<u>N</u><u>o</u><u>t</u><u>e</u><u>:</u>

<u>S</u><u>i</u><u>n</u><u> </u><u>a</u><u>n</u><u>g</u><u>l</u><u>e</u><u> </u><u>=</u><u>p</u><u>e</u><u>r</u><u>p</u><u>e</u><u>n</u><u>d</u><u>i</u><u>c</u><u>u</u><u>l</u><u>a</u><u>r</u><u>/</u><u>h</u><u>y</u><u>p</u><u>o</u><u>t</u><u>e</u><u>n</u><u>u</u><u>s</u><u>e</u>

<u>C</u><u>o</u><u>s</u><u> </u><u>a</u><u>n</u><u>g</u><u>l</u><u>e</u><u> </u><u>=</u><u>b</u><u>a</u><u>s</u><u>e</u><u>/</u><u>h</u><u>y</u><u>p</u><u>o</u><u>t</u><u>e</u><u>n</u><u>u</u><u>s</u><u>e</u>

<u>T</u><u>a</u><u>n</u><u> </u><u>a</u><u>n</u><u>g</u><u>l</u><u>e</u><u> </u><u>=</u><u>p</u><u>e</u><u>r</u><u>p</u><u>e</u><u>n</u><u>d</u><u>i</u><u>c</u><u>u</u><u>l</u><u>a</u><u>r</u><u> </u><u>/</u><u>b</u><u>a</u><u>s</u><u>e</u>

7 0
3 years ago
I’ve done this at least 40 times and keep getting it wrong??
aleksandrvk [35]

Answer:

4 times it is tails

Step-by-step explanation:

Sample Space: 1H, 1T, 2H, 2T, 3H, 3T, 4H, 4T

H= heads

T= tails

Event is is tails; 1T, 2T, 3T, 4T

4 times or 50% chance

5 0
3 years ago
Read 2 more answers
If j and k are nonzero integers, which pair of points must lie in the same quadrant?
Marizza181 [45]
<h3>Answer:  Choice D.   (3j, 3k)  and (3/j, 3/k)</h3>

The slash indicates a fraction.

=============================================

Proof:

We'll need to consider 4 different cases.

-----------------------

Case (1): j > 0 and k > 0

If j > 0, then 3j > 0 and 3/j > 0

If k > 0, then 3k > 0 and 3/k > 0

The two points (3j, 3k) and (3/j, 3/k) are both in quadrant 1.

-----------------------

Case (2): j > 0 and k < 0

If j > 0, then 3j > 0 and 3/j > 0

If k < 0, then 3k < 0 and 3/k < 0

Points (3j, 3k) and (3/j, 3/k) are both in quadrant 4.

------------------------

Case (3): j < 0 and k > 0

If j < 0, then 3j < 0 and 3/j < 0

If k > 0, then 3k > 0 and 3/k > 0

Points (3j, 3k) and (3/j, 3/k) are in quadrant 3.

------------------------

Case (4): j < 0 and k < 0

If j < 0, then 3j < 0 and 3/j < 0

If k < 0, then 3k < 0 and 3/k < 0

Points (3j, 3k) and (3/j, 3/k) are in quadrant 4.

------------------------

For nonzero integers j and k, we've shown that Points (3j, 3k) and (3/j, 3/k) are in the same quadrant. This concludes the proof.

8 0
3 years ago
Read 2 more answers
Qaudriladrial ABCD had cordinates A (3, 5) B(5, 2) C(8, 4) D(6, 7) qaudriladrial ABCD is a what?
Lorico [155]

The quadrilateral is a square

<h3>What are quadrilaterals?</h3>

Quadrilaterals are polygons, that have four sides and four corners

The coordinates are given as:

  • A = (3, 5)
  • B = (5, 2)
  • C = (8, 4)
  • D = (6, 7)

Start by calculating the adjacent side lengths using:

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

So, we have:

AB = \sqrt{(3- 5)^2 + (5- 2)^2}

AB = \sqrt{13}

BC = \sqrt{(8- 5)^2 + (4- 2)^2}

BC = \sqrt{13}

The side lengths are equal.

Hence, the quadrilateral is a square

Read more about quadrilaterals at:

brainly.com/question/5715879

8 0
2 years ago
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