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Neporo4naja [7]
2 years ago
6

A pair of parallel lines is cut by a transversal:

Mathematics
2 answers:
notsponge [240]2 years ago
7 0
60 degress add it up
Vladimir79 [104]2 years ago
3 0

Answer:

60

Step-by-step explanation:

1. parallel angle = 85

2. 180 - 85 = 95

3. 180 - (95+25) = 60

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<img src="https://tex.z-dn.net/?f=2%20%20%5Csqrt%7B2%7D%20" id="TexFormula1" title="2 \sqrt{2} " alt="2 \sqrt{2} " align="absm
True [87]

Answer:

2

Step-by-step explanation:

The square root of 2 is 1 and 1 multpied by 2 = 1

2\sqrt{2} \\2(1)=2

7 0
3 years ago
Which description can be written as the expression StartFraction 8 Over 21 n EndFraction?
frutty [35]

Answer: It is the quotient of 21 times and number and 8.

5 0
3 years ago
How many times does 9 go into 12?
tamaranim1 [39]

Answer:

one time.

Step-by-step explanation:

5 0
3 years ago
The graph h = −16t^2 + 25t + 5 models the height and time of a ball that was thrown off of a building where h is the height in f
Thepotemich [5.8K]

Answer:

part 1) 0.78 seconds

part 2) 1.74 seconds

Step-by-step explanation:

step 1

At about what time did the ball reach the maximum?

Let

h ----> the height of a ball in feet

t ---> the time in seconds

we have

h(t)=-16t^{2}+25t+5

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

The vertex represent a maximum

so

The x-coordinate of the vertex represent the time when the ball reach the maximum

Find the vertex

Convert the equation in vertex form

Factor -16

h(t)=-16(t^{2}-\frac{25}{16}t)+5

Complete the square

h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+5+\frac{625}{64}

h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}\\h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}

Rewrite as perfect squares

h(t)=-16(t-\frac{25}{32})^{2}+\frac{945}{64}

The vertex is the point (\frac{25}{32},\frac{945}{64})

therefore

The time when the ball reach the maximum is 25/32 sec or 0.78 sec

step 2

At about what time did the ball reach the minimum?

we know that

The ball reach the minimum when the the ball reach the ground (h=0)

For h=0

0=-16(t-\frac{25}{32})^{2}+\frac{945}{64}

16(t-\frac{25}{32})^{2}=\frac{945}{64}

(t-\frac{25}{32})^{2}=\frac{945}{1,024}

square root both sides

(t-\frac{25}{32})=\pm\frac{\sqrt{945}}{32}

t=\pm\frac{\sqrt{945}}{32}+\frac{25}{32}

the positive value is

t=\frac{\sqrt{945}}{32}+\frac{25}{32}=1.74\ sec

8 0
3 years ago
Algebra 1 polynomials hellllp​
defon
A. 40x^2 + 240x + 200
Shape: Rectangle
Formula: A = LW
A = (10x + 10)(4x + 20)
= (10x)(4x) + (10x)(20) + (10)(4x) + (10)(20)
= 40x^2 + 200x + 40x + 200
= 40x^2 + 240x + 200
————————————————————-
2. 57,600 ft^2
Width:
4x + 20 = 160
4x = 140
x = 35
Plug-In:
10x + 10
10(35) + 10
350 + 10
360 = Length
————————————-
A = LW
A = 360 • 160
A = 57,600 ft^2
8 0
2 years ago
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