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4 years ago

If sin(theta°) = 0.5, and we know that cos (theta°) < 0, then what is the smallest possible positive value of theta?

Mathematics

1 answer:

9966 [12]4 years ago

4 0

Answer:

150°

Step-by-step explanation:

It is given that CosФ < 0 i.e CosФ is negative.

Therefore, the minimum the value of Ф for which CosФ <0 will be in the second quadrant i.e 90° < Ф < 180°.

Now it is also given that, Sin Ф =0.5 {the value of SinФ is positive because Sin value is positive in second quadrant.}

⇒ Ф =180° - Sin⁻¹ (0.5) = 180°-30° =150° (Answer)

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For g (2) = 3x – 2 find the value of x<br> -<br> for which g(x) = 10

hram777 [196]

Answer:

For g (2) = 3x -2 find the value of x

The value of x is 2

for which g(x) = 10?

10 = 3x -2

10 +2 = 3x

12 = 3x

x = 12/3

x = 4

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3 years ago

Lao needs to be at school at 8:25 am. It takes her 15 minutes to walk to school, 3/12 hour to eat breakfast, 0.75 hours to get d

Sati [7]

Answer:

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3 years ago

State the slope and y intercept of Y=5x=7

lys-0071 [83]

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3 years ago

Read 2 more answers

Any help someone....<br> (solve triangles using the law of sines)

gogolik [260]

Answer:

$B=51$ °

Step-by-step explanation:

Use the Law of Sines as follows:

$\frac{sinC}{c} =\frac{sinB}{b}$

Insert the appropriate values:

$\frac{sin100}{14} =\frac{sinB}{11}$

Isolate B. Multiply both sides by 11:

$11*(\frac{sin100}{14} )=11*(\frac{sinB}{11})\\\\\frac{11*sin100}{14} =sinB\\\\sinB=\frac{11*sin100}{14}$

Use the inverse:

$B=sin^{-1}(\frac{11*sin100}{14})$

Insert the equation into a calculator and round to the nearest degree:

$B=50.7\\\\B=51$

Angle B is 51°.

:Done

7 0

3 years ago

The length and width of a rectangle are measured as 30cm and 24cm, respectively, with an error in measured of

larisa [96]

Answer:

The maximum error in the calculated area of rectangle is 5.4 cm².

Step-by-step explanation:

Given : The length and width of a rectangle are measured as 30 cm and 24 cm, respectively, with an error in measured of at most 0.1 cm in each.

To find : Use differentials to estimate the maximum error in the calculated area of rectangle ?

Solution :

The area of the rectangle is $A=l\times w$

The derivative of the area is equal to the partial derivative of area w.r.t. length times the change in length plus the partial derivative of area w.r.t. width times the change in width.

i.e. $dA=\frac{\partial A}{\partial L}\Delta L+\frac{\partial A}{\partial W}\Delta W$

Here, $\frac{\partial A}{\partial L}=30,\ \frac{\partial A}{\partial W}=24 ,\ \Delta L=0.1,\ \Delta W=0.1$

Substitute the values,

$dA=(30)(0.1)+(24)(0.1)$

$dA=3+2.4$

$dA=5.4$

Therefore, the maximum error in the calculated area of rectangle is 5.4 cm².

7 0

3 years ago