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2 years ago

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

Mathematics

2 answers:

____ [38]2 years ago

8 0

Answer:

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

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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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3 years ago

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Answer:

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3 years ago

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

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

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

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

$0 \leq x \leq 48.6\ ft$

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

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

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2 years ago