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3 years ago

Can anyone help with this question?

Mathematics

1 answer:

Fittoniya [83]3 years ago

4 0

Answer:

Step-by-step explanation:

tan 36° = 16/x

x = 16/(tan 36°)

x ≈ 22.0

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A phone company offers two monthly plans. Plan A costs $18 plus an additional $0.10 for each minute of calls. Plan B costs $22 p

Serggg [28]

Answer:

Step-by-step explanation:

18 + .10x = 22 + .08x

.04x=5

x = 22/0.08 = 275 minutes, is the number of minutes the two plans cost the same

18 + .10*125 = $30.5 is the cost when the two plans cost the same

8 0

3 years ago

Estimate 15 -(-5.03) -7.88

MissTica

(15 + 5) - 8... I get approximately 12. Hope this helps ;)

4 0

3 years ago

Read 2 more answers

Periodic grab some phase shifts

nikitadnepr [17]

Answer:

The sine function is:

$y(x)=\frac{2}{3}sin[3\pi (x-\frac{2\pi}{3})]-2$

Step-by-step explanation:

We need to recall that a sine function can write as:

$y(x)=Asin[B(x+C)]+D$

Where:

A is the amplitude of the function
B is the period
C is the phase shift left
D is the vertical shift

So we just need to use the listed attributes

Therefore, the sine function is:

$y(x)=\frac{2}{3}sin[3\pi (x-\frac{2\pi}{3})]-2$

I hope it helps you!

8 0

3 years ago

I need help with this please

vova2212 [387]

The coordinates of the vertices of LMN acquired a -ve sign to form the image L'M'N'. The line passes through the origin.

Step-by-step explanation:

When a point is rotated through 180° about the origin clockwise/anticlockwise, the coordinates of the object point T (h,k) takes a new position (-h,-k) to form the image T'. A line joining an object point to its image point will pass through the origin that is the center of rotation.

Learn More

Rotation about the origin through 180°: brainly.com/question/7747099

Keywords : Rotate, origin,graph, triangle

#LearnwithBrainly

7 0

4 years ago

Solve the equation 2 sec² x = 3 - tan x for the domain 0° ≤ x ≤ 360°

vodka [1.7K]

$2 \sec^2 x = 3- \tan x \\\\\implies 2(1 +\tan^2 x) = 3- \tan x \\\\\implies 2 + 2 \tan^2 x +\tan x -3 =0\\\\\implies 2 \tan^2 x + \tan x -1 =0\\\\\implies 2u^2 + u -1 =0~~~ ;[\text{set} \tan x = u]\\\\\implies u = \dfrac{-1\pm \sqrt{1-4\cdot 2 \cdot (-1)}}{2(2)}\\\\\implies u =\dfrac{-1 \pm\sqrt{9}}{4}\\\\\implies u = \dfrac{-1 \pm 3}{4}\\\\\implies u = \dfrac{2}{4}=\dfrac 12~~ \text{or} ~~u =\dfrac{-4}{4} =-1\\\\$

$\implies \tan x = \dfrac 12 ~~ \text{or} ~~ \tan x =-1~~~ ;[\text{Substitute back u =tan x}]\\\\\text{Now,}~ \\\\\tan x = -1,\\\\\implies x = n\pi - \dfrac{\pi}4\\\\\\\text{For interval,}~ [0,2\pi) \\\\x = \dfrac{3\pi}4, \dfrac{7\pi}4\\\\\text{In degrees,}~ x = 135^{\circ}, x =315^{\circ}\\\\\tan x = \dfrac 12\\\\\implies x = n\pi + \tan^{-1} \left(\dfrac 12 \right)\\\\\text{For interval} ~[0,2\pi),\\\\$

$x=\tan^{-1} \left(\dfrac 12 \right), \left[\pi +\tan^{-1} \left( \dfrac 12 \right)\right]\\\\\text{in degrees,} ~~x=\tan^{-1} \left(\dfrac 12 \right), \left[180^{\circ} +\tan^{-1} \left( \dfrac 12 \right)\right]\\\\\\\\\text{Combine all solutions:}\\\\x=\dfrac{3\pi}4, \dfrac{7\pi}4, \tan^{-1} \left(\dfrac 12 \right), \left[\pi +\tan^{-1} \left( \dfrac 12 \right)\right]\\\\\\$

$\text{In degrees,}~ x = 135^{\circ},~315^{\circ},~\tan^{-1} \left(\dfrac 12 \right), \left[180^{\circ} +\tan^{-1} \left( \dfrac 12 \right)\right]$

4 0

2 years ago

Can anyone help with this question?​

Can anyone help with this question?