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Ulleksa [173]
3 years ago
10

Find sin theta if theta is an angle in standard position and the point with coordinates (3, -4) lies on the terminal side of an

angle.

Mathematics
1 answer:
alisha [4.7K]3 years ago
4 0

Answer:

\sin \theta=-\frac{4}{5}

Step-by-step explanation:

Given:

Angle is in standard position which means the starting ray is at the origin. The terminal side has coordinates (3, -4).

So, the 'x' value is 3 and 'y' value id -4.

Using Pythagoras Theorem, we find the hypotenuse.

Hypotenuse = \sqrt{3^2+(-4)^2}=\sqrt{9+16}=\sqrt{25}=5

Now, using the sine ratio for the angle, we have

\sin \theta=\frac{Opposite}{Hypotenuse}\\\sin \theta=\frac{-4}{5}\\\sin \theta=-\frac{4}{5}

Therefore, the value of \sin \theta is -\frac{4}{5}.

The value is negative as the point (3, -4) lies in the fourth quadrant and sine ratio is negative in the fourth quadrant,

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About how much difference in average leaf width would you find in two forests whose average annual rainfalls are near 1500 mm bu
goblinko [34]

Answer:

4.80 mm to 2 d.p

Step-by-step explanation:

w = f(r)

f'(r) = (dw/dr) = (0.0218 mm/mm)

So, for a small difference in rainfall of 220 mm, what is the corresponding small difference in width of leaves in the two forests given.

One definition of a derivative or a rate of change is that it is the ratio of very small differences in the dependent variable to very small differences in the independent variable.

Mathematically,

(dw/dr) = (Δw/Δr) for very small Δw and Δr.

0.0218 = (Δw/220)

Δw = 0.0218 × 220 = 4.796 mm = 4.80 mm to 2 d.p

Hope this Helps!!!

5 0
3 years ago
What is the answer please.... need help
mina [271]

Answer:

f\left(x\right)=x^3-6x^2+3x+10 is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Step-by-step explanation:

Given the function

f\left(x\right)=x^3-6x^2+3x+10

As the highest power of the x-variable is 3 with the leading coefficients of 1.

  • So, it is clear that the polynomial function of the least degree has the real coefficients and the leading coefficients of 1.

solving to get the zeros

f\left(x\right)=x^3-6x^2+3x+10

0=x^3-6x^2+3x+10              ∵  f(x)=0

as

Factor\:x^3-6x^2+3x+10\::\:\left(x+1\right)\left(x-2\right)\left(x-5\right)=0

so

\left(x+1\right)\left(x-2\right)\left(x-5\right)=0    

Using the zero factor principle

if  ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)

x+1=0\quad \mathrm{or}\quad \:x-2=0\quad \mathrm{or}\quad \:x-5=0

x=-1,\:x=2,\:x=5

Therefore, the zeros of the function are:

x=-1,\:x=2,\:x=5

f\left(x\right)=x^3-6x^2+3x+10 is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Therefore, the last option is true.    

8 0
3 years ago
True of false? the slope of thia equation is zero <br>y=x+5<br>​
Ugo [173]

Answer:

false

Step-by-step explanation:

no solution

5 0
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Answer:

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Step-by-step explanation:

polynomial degree:224

leading term:7x^223

leading coefficient:7

8 0
3 years ago
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7 0
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