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3 years ago

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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:

A.

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

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

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;

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

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