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Illusion [34]
2 years ago
14

The archway of the main entrance of a university is modeled by the quadratic equation y = -x2 + 6x. The university is hanging a

banner at the main entrance at an angle defined by the equation 4y = 21 − x. At what points should the banner be attached to the archway?
A(1, 5.5) and (5.25, 6.56)
B(1, 5) and (5.25, 3.94)
C(1.5, 4.87) and (3.5, 4.37)
D(1.5, 5.62) and (3.5, 6.12)
EThere is no real solution.
Mathematics
2 answers:
katrin [286]2 years ago
5 0

Basically look for the intersection points of these two functions by setting them equal to each other.

-x^2 + 6x = 21/4 - x/4

-x^2 + (25/4)x - 21/4 = 0

x = 1 and x = 21/4


Plug these points back to any of the two original equations, you will get

(1,5) and (21/4, 63/16)


denpristay [2]2 years ago
4 0

1.5 ........ 5.25..... 3.94

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Point R is at (3, 1.3) and Point T is at (3, 2.4) on a coordinate grid. The distance between the two points is ____. (Input numb
snow_lady [41]
Point R and point T have same x coordinate so their distance is by y axis.
Ty-Ry=2.4-1.3=1.1
So the distance between two points is 1.1
3 0
2 years ago
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What number comes next 16, 06, 68, 88,
inessss [21]

So the given series is "16, 06, 68, 88, __"

Count all the cyclical opening in each of these numbers. For example in 16, there is a one cyclical loop present in it(the one in 6), similarly in 06 it is two(one in zero and one in 6), going ahead, in 68 it is 3(one in 6 and two in 8).

From here on things become simple: hence, the cyclical figures in these equations written down becomes 1,2,3,4,_,3.

Let's now try solving the above sequence, going by the logical reasoning the only number that can fill in the gap should be 4.

4 0
2 years ago
The area of a second soup can’s base is 5 pi inches squared. The can has a height of 10 inches. How would you use that informati
Ilia_Sergeevich [38]

Answer:

The amount of soup the can will hold is;

50π inches cubed = 157.08 inches cubed

Step-by-step explanation:

The amount of soup the can will hold is equal to the volume of the can.

Volume of the can = base area × height

Given;

Base Area = 5π inches squared

height = 10 inches

Volume of can = 5π × 10 = 50π inches cubed

The amount of soup the can will hold is;

50π inches cubed = 157.08 inches cubed

8 0
2 years ago
Read 2 more answers
PLEASE HELP ME!
algol13

Step-by-step explanation:

1.\sum_{i=1}^{5}3i

The simplest method is "brute force".  Calculate each term and add them up.

∑ = 3(1) + 3(2) + 3(3) + 3(4) + 3(5)

∑ = 3 + 6 + 9 + 12 + 15

∑ = 45

2.\sum_{k=1}^{4}(2k)^{2}

∑ = (2×1)² + (2×2)² + (2×3)² + (2×4)²

∑ = 4 + 16 + 36 + 64

∑ = 120

3.\sum_{k=3}^{6}(2k-10)

∑ = (2×3−10) + (2×4−10) + (2×5−10) + (2×6−10)

∑ = -4 + -2 + 0 + 2

∑ = -4

4. 1 + 1/4 + 1/16 + 1/64 + 1/256

This is a geometric sequence where the first term is 1 and the common ratio is 1/4.  The nth term is:

a = 1 (1/4)ⁿ⁻¹

So the series is:

\sum_{j=1}^{7}(\frac{1}{4})^{j-1}

5. -5 + -1 + 3 + 7 + 11

This is an arithmetic sequence where the first term is -5 and the common difference is 4.  The nth term is:

a = -5 + 4(n−1)

a = -5 + 4n − 4

a = 4n − 9

So the series is:

\sum_{j=1}^{5}(4j-9)

5 0
2 years ago
The owner of an office building is expanding the length and width of a parking lot by the same amount. The lot currently measure
Gnoma [55]

The best and most correct answer among the choices provided by your question is the second choice or letter B "20".

The current area is 120(80)=9600 and he want to expand it by 4400 so that the new area will be 9600+4400=14000 

14000=(120+x)(80+x) 

14000=9600+200x+x^2 

x^2+200x-4400=0 

x^2-20x+220x-4400=0 

x(x-20)+220(x-20)=0 

(x+220)(x-20)=0, since x is an increase it must be greater than zero so 

x=20ft 

(120+20)(80+20)=14000ft^2


I hope my answer has come to your help. Thank you for posting your question here in Brainly.

4 0
1 year ago
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