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sasho [114]
3 years ago
7

Please help me on this thank u

Mathematics
1 answer:
Artyom0805 [142]3 years ago
5 0
If h=0 represents the base of the model, then ...
  The bridge touches the base of the model at 2 feet and 6 feet from the leftmost edge of the model.

_____
When in doubt, use your calculator to graph it.

Here, the solution is not so hard. We're looking for h(x) = 0.
Multiply by -2
  (x -4)² -4 = 0
  (x -4) = ±√4
  x = 4 ±2 = {2, 6}

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Y=x²-4x+6<br> y=x+2<br><br><br> Please help
Ira Lisetskai [31]

Answer:

(1, 3), (2, 4).

Step-by-step explanation:

y= x² - 4x + 6

y = x + 2

Substitute for y in the first equation:

x + 2 = x²- 4x + 6

x^2 - 5x + 4 = 0

(x - 1)(x - 4) = 0

x = 1, 4

When x = 1 , y = 1 + 2 = 3.

When x = 4, y = 4 + 2 = 6.

7 0
3 years ago
an office space in new-york city measures 48 by 56. if it sells for 565 per square foot, what is the total cost of the office sp
inessss [21]
I added 48+56 and got 104
3 0
3 years ago
HELP! I put the formula but I don’t know the radius
ahrayia [7]

Answer:

C. 8,624

Step-by-step explanation:

recall that the formula for circumference is

Circumference = 2πr  ( = given as 85 meters)

hence,

85 = 2πr

r = 85/(2π)

given h = 15m

volume of cylinder

= πr²h

= π (85/2π)² 15  (using calculator and assuming π = 3.14)

= 8628.58

Comparing with the choices, the closest choice (within rounding error) is C. 8,624

8 0
3 years ago
The area of a right triangle with one leg measuring 5 ft. and hypotenuse measuring 13 ft. is_______ a0. Type a numerical answer
Snowcat [4.5K]

Area of the triangle = (1/2)*base*height

For right triangle base and height can be legs.

We have one leg = 5 ft. (Lets think it is a base.)


We need to find the other leg.

We are going to use Pythagorean theorem.

5² + b²=13²

b²=144

b=12 (It is going to be our height.)


Area of the triangle = (1/2)*5*12= 30 ft²

Area of the triangle = 30 ft²


5 0
3 years ago
Mathematics
Vladimir [108]

Answer:

there no data

Step-by-step explanation:

Find the median of the first and third quartiles81, 69 ,90, 79, 59, 125, 95, 116, 183, 97, 92, 50, 122, 55, 9250, 55, 59, 69, 79, 81, 90,   92,     92, 95, 97, 116, 122, 125, 183.The first quartile: 50, 55, 59, 69, 79, 81, 90: median of the first quartile is 69The third quartile: 92, 95, 97, 116, 122, 125, 183: median of the third quartile is 116.The interquartile range is the difference between the median of both the first and third quartile:  which is 116 - 69 = 47.There are no outliers in this set of data

4 0
2 years ago
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