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3 years ago

If sinθ= 1/2, and 0°<θ<180°, the smaller value of θ is ° and the larger value of θ is °.

Mathematics

2 answers:

Mrrafil [7]3 years ago

8 0

Answer:

Smaller value - 30; Larger value - 150

Step-by-step explanation:

This is shown on a Radian Circle.

To prove that this is true sin(30) = 1/2 & sin(150) = 1/2

Zielflug [23.3K]3 years ago

4 0

Answer: the smaller value of θ is 30° and the larger value of θ is 150°.

Step-by-step explanation:

We know, for the notable angles and values of the trigonometric functions (you can find a lot of tables of it) that sin(30°) = 1/2 and also that sin(150°) = 1/2

Then if 0°<θ<180°, the smallest value of θ is 30°, and the biggest value is 150°

You also can use the inverse function

if sinθ= 1/2

then θ= asin(1/2)

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if the measure of the third angle of the triangle is 20 more than three times the measure of either of the other two angles, fin

valentina_108 [34]

You should multiply 20 time 8 to get your answer

5 0

3 years ago

The area of a billboard sign is represented by (x+12)^2 square feet. Find the product

QveST [7]

(x+12)^2=(x+12)(x+12) = x^2 +12x + 12x + 144
= x^2+24x+144

6 0

3 years ago

Given that rectangle LMNO with coordinates L(0,0), M(3,0), N(3,7), O(0,7), P is the midpoint of LM⎯⎯⎯, and Q is the midpoint of

Elina [12.6K]

The midpoint of a line divides the line into equal segments.

The option that proves PQ = LO is (a)

The given parameters are:

$\mathbf{L = (0,0)}$

$\mathbf{M = (3,0)}$

$\mathbf{N = (3,7)}$

$\mathbf{O = (0,7)}$

P is the midpoint of LM.

So, we have:

$\mathbf{P = \frac{LM}{2}}$

$\mathbf{P = (\frac{(0 +3}{2},\frac{0+0}{2})}$

$\mathbf{P = (\frac{3}{2},0)}$

Q is the midpoint of NO.

So, we have:

$\mathbf{Q = \frac{NO}{2}}$

$\mathbf{Q = (\frac{(3 +0}{2},\frac{7+7}{2})}$

$\mathbf{Q = (\frac{3}{2},7)}$

Distance PQ is calculated as follows:

$\mathbf{d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}}$

This gives:

$\mathbf{PQ = \sqrt{(3/2 - 3/2)^2 + (0 - 7)^2}}$

$\mathbf{PQ = \sqrt{ 7^2}}$

$\mathbf{PQ = 7}$

Distance LO is calculated as follows:

$\mathbf{LO = \sqrt{(0 - 0)^2 + (0 - 7)^2}}$

$\mathbf{LO = \sqrt{ 7^2}}$

$\mathbf{LO=7}$

So, we have:

$\mathbf{PQ = 7}$

$\mathbf{LO=7}$

Thus:

$\mathbf{PQ = LO}$

Hence, the correct option is (a)

Read more about distance and midpoints at:

brainly.com/question/11231122

8 0

2 years ago

Ten less than twice a number is the same as 7 times the number. find the number

Dafna11 [192]

The number is -2 // Math [n is equivalent to number]: 2n -10 = 7n
2n = 7n + 10
-5n = 10
-n = 2
n = -2
// Hope this helped, please give me brainliest, thanks!! //

6 0

3 years ago

Use Lagrange multipliers to find the maximum and minimum values of (i) f(x,y)-81x^2+y^2 subject to the constraint 4x^2+y^2=9. (i

sp2606 [1]

i. The Lagrangian is

$L(x,y,\lambda)=81x^2+y^2+\lambda(4x^2+y^2-9)$

with critical points whenever

$L_x=162x+8\lambda x=0\implies2x(81+4\lambda)=0\implies x=0\text{ or }\lambda=-\dfrac{81}4$

$L_y=2y+2\lambda y=0\implies2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_\lambda=4x^2+y^2-9=0$

If $x=0$ , then $L_\lambda=0\implies y=\pm3$ .
If $y=0$ , then $L_\lambda=0\implies x=\pm\dfrac32$ .
Either value of $\lambda$ found above requires that either $x=0$ or $y=0$ , so we get the same critical points as in the previous two cases.

We have $f(0,-3)=9$ , $f(0,3)=9$ , $f\left(-\dfrac32,0\right)=\dfrac{729}4=182.25$ , and $f\left(\dfrac32,0\right)=\dfrac{729}4$ , so $f$ has a minimum value of 9 and a maximum value of 182.25.

ii. The Lagrangian is

$L(x,y,z,\lambda)=y^2-10z+\lambda(x^2+y^2+z^2-36)$

with critical points whenever

$L_x=2\lambda x=0\implies x=0$ (because we assume $\lambda\neq0$ )

$L_y=2y+2\lambda y=0\implies 2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_z=-10+2\lambda z=0\implies z=\dfrac5\lambda$

$L_\lambda=x^2+y^2+z^2-36=0$

If $x=y=0$ , then $L_\lambda=0\implies z=\pm6$ .
If $\lambda=-1$ , then $z=-5$ , and with $x=0$ we have $L_\lambda=0\implies y=\pm\sqrt{11}$ .

We have $f(0,0,-6)=60$ , $f(0,0,6)=-60$ , $f(0,-\sqrt{11},-5)=61$ , and $f(0,\sqrt{11},-5)=61$ . So $f$ has a maximum value of 61 and a minimum value of -60.

5 0

3 years ago

If sinθ= 1/2, and 0°<θ<180°, the smaller value of θ is __° and the larger value of θ is __°.

If sinθ= 1/2, and 0°<θ<180°, the smaller value of θ is ° and the larger value of θ is °.