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alexgriva [62]
3 years ago
9

12

Mathematics
1 answer:
Setler [38]3 years ago
6 0
Around 17000-30000 it’s different for each person
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Solve the equation x/4+3=-6
charle [14.2K]
X/4 = -9

x/4 = -9/1

cross multiply 

x = -36
3 0
3 years ago
What is the length of side c, given a = 10, b = 8, and c = 105°?
posledela
We can calculate using cosinus method in triangle
c² = a² + b² - 2ab cos c

Plug in the number to the formula
c² = a² + b² - 2ab cos c
c² = 10² + 8² - 2(10)(8) cos 105°
c² = 100 + 64 - 160 cos 105°
c² = 164 - 160 (-0.26)
c² = 164 + 41.6
c² = 205.6
c = √205.6
c =14.34

C is 14.34 unit length
5 0
2 years ago
which of the following statements best describes the effect of replacing the graph of y = f(x) with the graph of y = f(x) − 11?
gtnhenbr [62]
The graph of y = f(x) will shift down 11 units.
3 0
3 years ago
Read 2 more answers
A student is given that point P(a, b) lies on the terminal ray of angle Theta, which is between StartFraction 3 pi Over 2 EndFra
Harman [31]

Answer:

<em>A.</em>

<em>The student made an error in step 3 because a is positive in Quadrant IV; therefore, </em>

<em />cos\theta = \frac{a\sqrt{a^2 + b^2}}{a^2 + b^2}

Step-by-step explanation:

Given

P\ (a,b)

r = \± \sqrt{(a)^2 + (b)^2}

cos\theta = \frac{-a}{\sqrt{a^2 + b^2}} = -\frac{\sqrt{a^2 + b^2}}{a^2 + b^2}

Required

Where and which error did the student make

Given that the angle is in the 4th quadrant;

The value of r is positive, a is positive but b is negative;

Hence;

r = \sqrt{(a)^2 + (b)^2}

Since a belongs to the x axis and b belongs to the y axis;

cos\theta is calculated as thus

cos\theta = \frac{a}{r}

Substitute r = \sqrt{(a)^2 + (b)^2}

cos\theta = \frac{a}{\sqrt{(a)^2 + (b)^2}}

cos\theta = \frac{a}{\sqrt{a^2 + b^2}}

Rationalize the denominator

cos\theta = \frac{a}{\sqrt{a^2 + b^2}} * \frac{\sqrt{a^2 + b^2}}{\sqrt{a^2 + b^2}}

cos\theta = \frac{a\sqrt{a^2 + b^2}}{a^2 + b^2}

So, from the list of given options;

<em>The student's mistake is that a is positive in quadrant iv and his error is in step 3</em>

3 0
2 years ago
How does tan 135 degrees compare to tan -135 degrees
Anna007 [38]
Evaluating the tangent function for the two values we obtain the following:
tan 135=tan(135+180)=tan 315=-1

next:
tan -135=tan(180-135)=tan 45=1
thus comparing the two we see that tan 135 is negative while tan -135 is positive
3 0
3 years ago
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