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4 years ago

For the right triangle shown, which expression represents the length of CA?

Mathematics

1 answer:

strojnjashka [21]4 years ago

6 0

Step-by-step explanation:

The attached figure shows a right angled triangle right angle at C. It is required to find the length of CA.

Here, CA = x = perpendicular

BC = base

AB = hypotenuse =7

Angle B is 40 degrees

We know that the ratio of perpendicular to the hypotenuse is equal to sine of angle.

$\sin\theta=\dfrac{P}{H}\\\\\sin(40)=\dfrac{x}{7}$

x is length of CA

$x=7\sin(40)$

So, the correct option is (A).

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Can anyone explain how to do this? I looked at the book and now I'm even more confused. 50 points

katrin [286]

The piecewise function is basically the result of two different functions combined together. If x is 0 or larger, then h(x) = x+4. Otherwise, if x < 0, then h(x) = -x-4

No matter what number you pick for x, the h(x) function will be used in some way. So the domain is the set of all real numbers. To write this in interval notation, we write $(-\infty, \infty)$ which means we start off at negative infinity and go to positive infinity. This is basically saying "the entire real number line". Since we can't actually reach either infinity, we always use parenthesis with them. Never use square brackets with either infinity

From the graph (see attached image below), we see that (0,-4) is the lowest point. This means y = -4 is the smallest y output possible, though we can't actually reach it because of the open circle at (0,-4). We can get any other larger y value. So the range is therefore: $(-4, \infty)$ meaning we start at -4 and head off to positive infinity. The curve parenthesis next to -4 the reader "exclude -4 as part of the range". There is an open hole or gap here. Another way to state the range is to write y > -4

8 0

3 years ago

Write a linear equation given the point and slope or two points.

attashe74 [19]

Answer:

1) y = -2x - 1

2) y = -3/4 + 3

3) y = 4x + 9

4) y = - 5/3x - 2

Step-by-step explanation:

b = y - m*x

1) (-7,13) and slope: -2

b = 13 - (-2)*(-7)

b = 13 - 14

b = -1

2) (4,6) lope = -3/4

b = 6 - (-3/4)*(-4)

b = 6 - 3

b = 3

y = -3/4x + 3

3) (-5,-11) and (3,-7)

Slope: (1 - - 11)/(-2 - - 5) = 12/3

= 4

b = -11 - (4) (-5) = 9

b = 9

4) slope = (-7 - 8)/(3- - 6)

= -15/9 = - 5/3

b = 8 - (-5/3)*(-6)

b = 8 - 10

b = -2

5 0

3 years ago

Order of fractions from least to greatest 0.4,5/8,38%

Angelina_Jolie [31]

4/10,5/8,38/100
2 10,8,100
2 5, 4, 50
2 5, 2, 25
5 5, 1, 25
5 1, 1, 5
1, 1, 1
2*2*2*5*5=200
4/10=4*200110
=80
5/8=5*200/8
=125
38/100=38*200/100
=76
76,80,125

The answer is 38%,0.4,5/8

8 0

3 years ago

Solve the system by substitution. 3x-6 = y -6x + y = 9

fomenos

Answer:

https://www.tiger-algebra.com/drill/y=-3x_6;y=9/

Step-by-step explanation:

5 0

3 years ago

Please answer the question below (ABOUT VECTORS AND MAGNITUDE)

coldgirl [10]

(a) v appears to have a fixed direction along the positive x-axis. If ||u|| = 150 N, ||v|| = 220 N, then when θ = 30°, you have

u = (150 N) (cos(30°) i + sin(30°) j ) ≈ (129.904 i + 75 j ) N

v = (220 N) (cos(0°) i + sin(0°) j ) = (220 i ) N

(i and j are the unit vectors in the positive x and y directions)

and their sum is

u + v ≈ (349.904 i + 75 j ) N

with magnitude

||u + v|| ≈ √((349.904)² + (75)²) N ≈ 357.851 N ≈ 357.9 N

and at angle φ made with the positive x-axis such that

tan(φ) ≈ (75 N) / (349.904 N) → φ ≈ 12.098° ≈ 12.1°

(b) Letting θ vary from 0° to 180° would make v a function of θ :

u = (150 N) (cos(θ) i + sin(θ) j ) = (150 cos(θ) i + 150 sin(θ) j ) N

Then

u + v = ((220 + 150 cos(θ)) i + (150 sin(θ)) j ) N

→ M = ||u + v|| = √((220 + 150 cos(θ))² + (150 sin(θ))²) N

M = √(48,400 + 66,000 cos(θ) + 22,500 cos²(θ) + 22,500 sin²(θ)) N

M = 10 √(709 + 660 cos(θ)) N

(c) As a function of θ, u + v makes an angle α with the positive x-axis such that

tan(α) = (150 sin(θ) / (220 + 150 cos(θ))

→ α = tan⁻¹((15 sin(θ) / (22 + 15 cos(θ)))

(d) Filling in the table is just a matter of evaluating M and α for each of the given angles θ. For example, when θ = 0°,

M = 10 √(709 + 660 cos(0°)) N = 370 N

α = tan⁻¹((15 sin(0°) / (22 + 15 cos(0°))) = 0°

When θ = 30°, you get the same result as in part (a).

When θ = 60°,

M = 10 √(709 + 660 cos(60°)) N ≈ 323.3 N

α = tan⁻¹((15 sin(60°) / (22 + 15 cos(60°))) = 23.8°

and so on.

4 0

3 years ago