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Andre45 [30]
2 years ago
6

If x is directly proportional to y, and x = 3.4 when y = 2, find y when x = 5.1.

Mathematics
2 answers:
Marina CMI [18]2 years ago
7 0
The answer is 3 ..........................
Wewaii [24]2 years ago
4 0
The answer is D. 2.3 because it has to be between 2 and 3.4, so the answer isn´t C. The answer also cannot be 3>x. Therefore, the answer is within the 2´s.
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The function A(b) relates the area of a trapezoid with a given height of 10 and
Natali5045456 [20]

The function of the trapezoid area is:

A(x)=(B+b)*h/2

Where B and b are the bases and h is the height.

With the given data: h=10 B and b =7 and x (it may vary which one is bigger)

-----------

So that function becomes:

A(x)=(7+x)*10/2

A(x)=(7+x)*5

----------

So if you want the inverse function, you have to operate to find x:

A(x)/5=7+x

A(x)/5-7=x

----------

So the new function is:

x(A)=A/5-7

6 0
2 years ago
Which geometric solids would model the tent?
aivan3 [116]

a cylinder and a cone. the cone would go on the top, and the cylinder on the bottom.

8 0
2 years ago
Read 2 more answers
What is the cricket’s hang time (amount of time the cricket is airborne)? Round to the nearest thousandth h(t) = –16t2 + 14t
Stels [109]
We have that
h(t) = –16t²<span> + 14t
using a graph tool
see the attached figure

</span><span>the cricket’s hang time is the difference  between point (0,0) and (0.875,0)
</span>(0.875-0)=0.875 sec

the answer is 0.875 sec

8 0
2 years ago
Read 2 more answers
Barbara drives between Miami, Florida, and West Palm Beach, Florida. She drives 50 mi in clear weather and then encounters a thu
NeX [460]

Distance traveled in clear weather = 50 miles

Distance traveled in thunderstorm = 15 miles

Let speed in clear weather = x

⇒ Speed in thunderstorm = x-20

Total time taken for trip = 1.5 hours

We need to determine average speed in clear weather (i.e. x) and average speed in the thunderstorm (i.e. x-20 ).

Total time taken for trip = Time taken in clear weather + Time taken in thunderstorm

⇒ Total time taken for trip = \frac{Distance covered in clear weather}{Speed in clear weather} + \frac{Distance covered in thunderstorm}{Speed in thunderstorm}

⇒ 1.5 = \frac{50}{x} + \frac{15}{x-20}

⇒ 1.5 = \frac{50(x-20)+15(x)}{(x)(x-20)}

⇒ 15*x*(x-20) = 10*[50*(x-20)+15*x]

⇒ 15x² - 300x = 500x - 10,000 + 150x

⇒ 15x² - 300x = 650x - 10,000

⇒ 15x² - 950x + 10,000 = 0

⇒ 3x² - 190x + 2,000 = 0

The above equation is in the format of ax² + bx + c = 0

To determine the roots of the equation, we will first determine 'D'

D = b² - 4ac

⇒ D = (-190)² - 4*3*2,000

⇒ D = 36,100 - 24,000

⇒ D = 12,100

Now using the D to determine the two roots of the equation

Roots are: x₁ = \frac{-b+\sqrt{D}}{2a} ; x₂ = \frac{-b-\sqrt{D}}{2a}

⇒ x₁ = \frac{-(-190)+\sqrt{12,100}}{2*3} and x₂ = \frac{-(-190)-\sqrt{12,100}}{2*3}

⇒ x₁ = \frac{190+110}{6} and x₂ = \frac{190-110}{6}

⇒ x₁ = \frac{300}{6} and x₂ = \frac{80}{6}

⇒ x₁ = 50 and x₂ = 13.33

So speed in clear weather can be 50 mph or 13.33 mph. However, we know that in thunderstorm was 20 mph less than speed in clear weather.

If speed in clear weather is 13.33 mph then speed in thunderstorm would be negative, which is not possible since speed can't be negative.

Hence, the speed in clear weather would be 50 mph, and in thunderstorm would be 20 mph less, i.e. 30 mph.

7 0
2 years ago
Line jk passes through points j(-4,-5) and k(-6,3 if the equation of the lines is written in slope-intercept form,y=my+b what is
lidiya [134]
Written in 2-point form, the equation of the line is
  y = (y2-y1)/(x2-x1)·(x-x1) +y1
  y = (3-(-5))/(-6-(-4))·(x-(-4)) + (-5)
  y = 8/-2·(x +4) - 5
  y = -4x -21

The value of b is -21.

6 0
2 years ago
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