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IRINA_888 [86]
3 years ago
15

I need help please and thank you

Mathematics
1 answer:
Sphinxa [80]3 years ago
8 0
The answer is I believe: 21
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Please help me solve
vlabodo [156]
(2x+5)=Inscribed Angle
50 = Arc Length
Inscribed Angle=1/2Arc Length
2x+5=1/2(50)
2x+5=25
2x=20
X=10
4 0
3 years ago
M – 2.3 – 0.7m = 1.6
posledela

Answer: m = 13 ;)

Step-by-step explanation:

5 0
3 years ago
A bag contains 3 yellow, 1 red, and 2 green beads. Charlene draws a bead, keeps it out, then draws another bead. What’s the prob
Archy [21]

The probability would be 1/15.

3 0
3 years ago
Use the Distributive Property​ to simplify the expression<br> 4(​c​ - 2)
xeze [42]

Answer

Step-by-step explanation:

4xc=4c

4x-2= -8

4c-8

3 0
3 years ago
I need help with 1-3 i don’t understand that problems and its already over due
Nina [5.8K]

Step-by-step explanation:

For a quadratic equation y = ax² + bx + c, the vertex (the maximum or minimum point) is at x = -b/(2a).

1) y = -0.5t² + 2t + 38

The maximum is at:

t = -2 / (2 × -0.5)

t = 2

The maximum height is:

y = -0.5(2)² + 2(2) + 38

y = 40

The coordinates of the vertex are (2, 40).  That means the missile reaches a maximum height of 40 km after 2 minutes.

2) y = -4.9t² + 12t + 1.6

The maximum is at:

t = -12 / (2 × -4.9)

t = 1.22

The maximum height is:

y = -4.9(1.22)² + 12(1.22) + 1.6

y = 8.95

The coordinates of the vertex are (1.22, 8.95).  That means the missile reaches a maximum height of 8.95 m after 1.22 seconds.

3) y = -0.04x² + 0.88x

The maximum is at:

x = -0.88 / (2 × -0.04)

x = 11

The maximum height is:

y = -0.04(11)² + 0.88(11)

y = 4.84

The maximum height of the tunnel is 4.84 meters.

The maximum width is when y = 0.

0 = -0.04x² + 0.88x

0 = -0.04x (x − 22)

x = 22

The maximum width is 22 feet.

7 0
4 years ago
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