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larisa86 [58]
3 years ago
8

NEED HELP MARK CROWN ASAP

Mathematics
1 answer:
m_a_m_a [10]3 years ago
5 0

Answer:

The answer is <em><u>0.48</u></em>!

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Select the functions that have a value of -1. sin180° cos180° tan180° csc180° sec180° cot180°
kupik [55]

We have to break each degree in terms of 90

A) sin180^\circ=sin(90\times2+0)

Which is in third quadrant, therefore sine is negative hence

sin(90\times2+0)= -sin0 ^\circ = 0


B) cos180^\circ =cos(90\times2+0)

Which is in third quadrant, therefore cosine is negative hence

cos(90\times2+0)= -cos0^\circ  = -1


C) tan180^\circ=tan(90\times2+0)

Which is in third quadrant, therefore tangent is positive hence

tan(90\times2+0)= tan0^\circ  = 0


D) csc180^\circ=csc(90\times2+0)

Which is in third quadrant, therefore cosec is negative hence

cosec(90\times2+0)= -csc0^\circ  =not defined


E)sec180^\circ=sec(90\times2+0)

Which is in third quadrant, therefore secant is negative hence

sec(90\times2+0)= -sec0^\circ  = -1


F) cot180^\circ=cot(90\times2+0)

Which is in third quadrant, therefore tangent is positive hence

cot(90\times2+0)= cot0^\circ = not defined


Hence only cos 180^\circ and

sec180^\circ have value -1

Hope this will help

7 0
3 years ago
How many integers from 1 through a
AysviL [449]

Answer:

sorry but I don't understand

Step-by-step explanation:

please forgive me

comment if I am forgiven

8 0
3 years ago
How can you solve the equation 8x = 2x + 2 graphically?
ehidna [41]

subtract 2x from both sides... 6x=2 divide by 6 = .3

3 0
3 years ago
Find the equation of the line that has a slope of -3 and passes through (4, 4)
sergey [27]
The\ slope-point\ form:y-y_1=m(x-x_1)\\The\ slope-intercept\ form:y=mx+b\\--------------------\\m=-3;\ (4;\ 4)\to x_1=4\ and\ y_1=4\\\\\boxed{y-4=-3(x-4)}\\\\y-4=-3x+12\ \ \ |add\ 4\ to\ both\ sides\\\boxed{y=-3x+16}
7 0
3 years ago
Amir says the graph of<br> y= x2 + 16 has -4 as a zero. Is Amir correct?<br> Explain.
sukhopar [10]

Answer:

  • Amir is wrong

Step-by-step explanation:

The function y = x² + 16 is positive for any value of x, hence the graph has no intersection with the x-axis and therefore has no real zero's.

If no zero's, -4 also can't be its zero.

8 0
2 years ago
Read 2 more answers
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