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anastassius [24]
2 years ago
13

Help? Use the law of sines to find the length of side c

Mathematics
2 answers:
____ [38]2 years ago
8 0

Answer:

C

Step-by-step explanation:

Using the law of Sines in ΔABC

\frac{a}{sinA} = \frac{c}{sinC}, that is

\frac{37}{sin42} = \frac{c}{sin41.5} ( cross-  multiply )

c × sin42° = 37 × sin41.5° ( divide both sides by sin42° )

c = \frac{37(sin41.5)}{sin42} ≈ 36.64 ( to 2 dec. places )

Ahat [919]2 years ago
5 0
<h3>✽ - - - - - - - - - - - - - - - ~<u>Hello There</u>!~ - - - - - - - - - - - - - - - ✽</h3>

➷ a/sinA = c/sinC

Substitute in the values:

37/sin(42) = c/sin(41.5)

Multiply both sides by sin(41.5)

37/sin(42) x sin(41.5) = c

Solve:

c = 36.63999457

The correct answer would be C. 36.64

<h3><u>✽</u></h3>

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

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Rama09 [41]

Answer:

Option B. 3/4 > 2/3 is the correct answer.

Step-by-step explanation:

Lets look at the options one by one

A. 1/2 < 1/3

This inequality converts into 0.5<0.33 which is false.

B. 3/4 > 2/3

This simplifies to 0.75>0.666 which is true.

C. -1/4 < - 2/3

This converts to -0.25<-0.666 which makes the inequality false.

D. -1 > 3/4​

This converts to -1>0.75 which makes the inequality false.

Hence,

Option B. 3/4 > 2/3 is the correct answer.

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If f(x) = -3x+1 then f^-1(x) =
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Answer:

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f(x) = -3x+1

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x = -3y +1

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or, y = 1 - x /3

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The students in Mr. Wilson's Physics class are making golf ball catapults. The
Mnenie [13.5K]

Answer:

Part a) About 48.6 feet

Part b) About 8.3 feet

Part c) The domain is 0 \leq x \leq 48.6\ ft and the range is 0 \leq y \leq 8.3\ ft

Step-by-step explanation:

we have

y=-0.014x^{2} +0.68x

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

where

x is the ball's  distance from the catapult in feet

y is the flight of the balls in feet

Part a)  How far did the ball fly?

Find the x-intercepts or the roots of the quadratic equation

Remember that

The x-intercept is the value of x when the value of y is equal to zero

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

in this problem we have

-0.014x^{2} +0.68x=0

so

a=-0.014\\b=0.68\\c=0

substitute in the formula

x=\frac{-0.68(+/-)\sqrt{0.68^{2}-4(-0.014)(0)}} {2(-0.014)}

x=\frac{-0.68(+/-)0.68} {(-0.028)}

x=\frac{-0.68(+)0.68} {(-0.028)}=0

x=\frac{-0.68(-)0.68} {(-0.028)}=48.6\ ft

therefore

The ball flew about 48.6 feet

Part b) How high above the ground did the ball fly?

Find the maximum (vertex)

y=-0.014x^{2} +0.68x

Find out the derivative and equate to zero

0=-0.028x +0.68

Solve for x

0.028x=0.68

x=24.3

<em>Alternative method</em>

To determine the x-coordinate of the vertex, find out the midpoint  between the x-intercepts

x=(0+48.6)/2=24.3\ ft

To determine the y-coordinate of the vertex substitute the value of x in the quadratic equation and solve for y

y=-0.014(24.3)^{2} +0.68(24.3)

y=8.3\ ft

the vertex is the point (24.3,8.3)

therefore

The ball flew above the ground about 8.3 feet

Part c) What is a reasonable domain and range for this function?

we know that

A  reasonable domain is the distance between the two x-intercepts

so

0 \leq x \leq 48.6\ ft

All real numbers greater than or equal to 0 feet and less than or equal to 48.6 feet

A  reasonable range is all real numbers greater than or equal to zero and less than or equal to the y-coordinate of the vertex

so

we have the interval -----> [0,8.3]

0 \leq y \leq 8.3\ ft

All real numbers greater than or equal to 0 feet and less than or equal to 8.3 feet

8 0
2 years ago
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