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Zarrin [17]
2 years ago
11

I need help !! Can someone answer this mathematics question for algebra 1. Question number five pls.

Mathematics
1 answer:
Ymorist [56]2 years ago
7 0

Answer:

3x cubed- 3x squared- 11x -22

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The sky ride at an amusement park spans 4 x 10 feet. Over the course of the day, Anna rode the sky ride 7 times. How many feet d
Furkat [3]
This question tells that the sky ride is a rectangle.
This means the perimeter or the total distance round the figure = 2L + 2B
= 2(4)+ 2(10)
=8+20
=28 feet
Anna rose 7 times
Distance rode= 7(28)
=196 feet
8 0
2 years ago
Helpppppppppppppppppppppppppppppppp
Sonbull [250]
The equation is y=(x-7)^2+7.

We are looking for a function with a vertex above the x-axis and a function that opens upward (has coefficient a > 0).

The first function opens downward and intersects the x-axis. The second function has a vertex below the x-axis. The third function satisfies our requirements. The fourth function has a vertex on the x-axis.

We can solve this algebraically with the knowledge that the real solutions of a quadratic are its x-intercepts. If there are no x-intercepts (because it lies entirely above or below the x-axis), then there are no real solutions. This is true when the discriminant b^2-4ac \ \textless \  0. You can see that from the quadratic formula. This holds true for both answers A and C, so to find the correct one, we remember that when the coefficient a of the x^2 term is positive, the graph opens upwards, so we choose C.
5 0
3 years ago
Read 2 more answers
9 centimeters y'all it grows .7 centimeters each day what is the function
emmasim [6.3K]

9 + .7d

Should be your answer!

8 0
2 years ago
(will mark as brainliest)
Helga [31]

Find the x and y-intercepts of the equation. Then calculate the points at constant intervals and plot and connect those points on a makeshift graph. A quadratic equation can have 2 solutions, 1 solution, or none. The corresponding graphs(if two solutions) would most probably cross over the x-axis twice.

Hope this helps.

4 0
2 years ago
A plane is descending into the airport. After 5 minutes it is at a height of 6500 feet. After 7 minutes it is at a height of 590
Jet001 [13]

Let's assume

height of plane  in feet =h

time in minutes =t

we are given

A plane is descending into the airport. After 5 minutes it is at a height of 6500 feet

so, we get one point as (5,6500)

After 7 minutes it is at a height of 5900 feet

so, we get another point as (7,5900)

we can use point slope form of line

y-y_1=m(x-x_1)

points as

(5,6500)

x1=5, y1=6500

(7,5900)

x2=7 , y2=5900

Calculation of slope(m):

m=\frac{y_2-y_1}{x_2-x_1}

now, we can plug values

m=\frac{5900-6500}{7-5}

m=-300

Equation of line:

we can use formula

y-y_1=m(x-x_1)

we can plug values

h-6500=-300(t-5)

h=-300t+8000

Time of landing:

we can set h=0

and then we can solve for t

h=-300t+8000=0

t=\frac{80}{3}min..............Answer

5 0
3 years ago
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