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pantera1 [17]
3 years ago
5

A ladder is propped against the side of a building making a 30 degree

Mathematics
1 answer:
Dahasolnce [82]3 years ago
3 0
Using trigonometry. This picture shows a ‘not to scale’ sketch of the following question. Use Sin 30. Sin 30= opposite/hypo tenuse. That is simply dividing 6 ft over Sin 30 that is 12 feet.

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What is the slope and y-intercept of the equation 6x - 1 = 3y - 10?*
Ad libitum [116K]

Answer:

m = 2, b = 3

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + b ( m is the slope and b the y- intercept )

Given

6x - 1 = 3y - 10 ( add 10 to both sides )

6x + 9 = 3y ( divide all terms by 3 )

2x + 3 = y , that is

y = 2x + 3 ← in slope- intercept form

with slope m = 2 and y- intercept b = 3

6 0
3 years ago
If you could help me on all of them you a live saver
PtichkaEL [24]

Answer:

(1.37) AUB = { 1,2,3,4,5,6}

(1.38) AUC = { 1,2,3,4,5 }

(1.39)BUC = { 1,2,3,4,5,6}

(1.40) { 2,4 }

(1.41) { 1,3,5 }

(1.42) { phi }

(1.43) AU(BUC) = { 1,2,3,4,5,6 }

(1.44) { phi }

(1.45) {1,2,3,4,5}

(1.46) { 1,2,3,4,5 }

\huge\underline\red{Hope\:that\:helps}

8 0
3 years ago
1-Suppose the path of a baseball follows the path graphed by the quadratic function ƒ(d) = –0.6d2 + 5.4d + 0.8 where d is the ho
jasenka [17]

Answer:

1. The correct option B.

2.The correct option A.

Step-by-step explanation:

The given function is

f(d)=-0.6d^2+5.4d+0.8

Where f(d) is the height of the ball at horizontal distance d.

Put f(d)=0, to find the distance where the ball touch the ground.

0=-0.6d^2+5.4d+0.8

Quadratic formula:

d=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

Using the quadratic formula we get

d=\frac{-5.4\pm \sqrt{(5.4)^2-4(-0.6)(0.8)}}{2(-0.6)}

d=-0.146,9.146

Therefore the ball is in air between d=-0.146 to d=9.146.

The distance can not be negative, therefore the ball remains in the air between d=0 to d=9.146.

9.146\approx 9.15

Therefore the correct option is B.

2.

The given equation is

y=(x-1)^2+1                 .... (1)

The standard form of parabola is

y=a(x-h)^2+h             .... (2)

Where, a is constant and (h,k) is vertex.

On comparing (1) and (2), we get

a=1

h=1

k=1

Since the value of a is positive, therefore it is an upward parabola. The vertex of the parabola is (1,1).

Put x=0 in the given equation.

y=(0-1)^2+1

y=1+1=2

Therefore the y-intercept is (0,2) and option A is correct.

8 0
3 years ago
The area of a rectangle is 21 yd.² and the length of a rectangle is 1 yard or less than twice the width find the dimensions of t
emmasim [6.3K]

Answer:

length = 6 yards

width = 3.5 yards

Step-by-step explanation:

the length of a rectangle is 1 yard less than twice the width

LEt 'w' be the width of the rectangle

and 'l' be the length of the rectangle

length l = 2 times the width -1

so l = 2w -1

Area of the rectangle = length * width

Area = l * w

we know length is 2w-1  and given area = 21

21 = (2w-1) * w

21= 2w^2 - w

subtract 21 on both sides

0= 2w^2 - w - 21

now solve for 'w'

0= 2w^2 +6w -7w- 21

0 = 2w(w+3)-7(w+3)

0= (2w-7)(w+3)

2w-7 = 0                    and w+3 =0

w = 7/2                            w=-3

width cannot be negative so we ignore w=-3

width w= 3.5

length = 2w-1= 2(3.5) - 1= 7-1=6

length = 6 yards

5 0
4 years ago
1. What is the reason for Statement 3 of the two-column proof?
OLEGan [10]

1. Angle addition postulate (this could be wrong)

2. Definition of bisect

3. Definition and postulate

4. Screenshot at the bottom

Last one I can't help, my bad.

3 0
3 years ago
Read 2 more answers
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