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3 years ago

10

Find the slope of the line that passes through (10, 9) and (1,4).

2 answers:

pantera1 [17]3 years ago

8 0

Answer:

From the values given

x₁= 10 y₁=9

x₂=1 y₂=4

m= y₂ - y₁/x₂ - x₁

m = (4 - 9)/(1-10)

m= -5/-9

m=5/9.

Rom4ik [11]3 years ago

6 0

Answer:

1/2

Step-by-step explanation:

Slope = rise / run (delta y / delta x)

delta y = y1 - y2 = 9 - 4 = 5

delta x = x1 - x2 = 10 - 1 = 10

slope = 5 / 10 = 1/2

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Lf f(x) = 5x, what is f^-1(x)?

Nonamiya [84]

Answer: $f^{-1}(x)=\frac{x}{5}$

Step-by-step explanation:

By definition the domain of an inverse function $f^-1(x)$ is the range of f(x) and the range of the inverse function is equal to the domain of the principal function f(x).

If you have a function $f(x)=5x$ , then to find the inverse function, follow these steps:

1. Make $y=f(x)$

$f(x)=y=5x$

$y=5x$

2. Solve for the variable "x":

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3 years ago

Please solve i will give brainiest 100 point question ****** do the whole page please need to pass or i will fail its my final t

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Answer:

1. Find the difference between the areas.

<u>Area of the small rectangle</u>: $(x+2)(x+7)=x^2+7x+2x+14=x^2+9x+14$

<u>Area of the big rectangle</u>: $(x+9)(x+11)=x^2+11x+9x+99=x^2+20x+99$

The difference is: $11x+85$

$( x^2+20x+99)- (x^2+9x+14)=x^2+20x+99-x^2-9x-14=11x+85$

2.

You can solve this question just by looking at the graph.

a) The height is 4 meters.

$f(d)=h=-2d^2+7d+4$

To find the height of the bleachers, we should consider the moment before the shoot, when the distance is equal to 0.

$f(0)=h=-2(0)^2+7(0)+4$

$h=4$

The height is 4 meters.

b) 9 meters.

For $d=1$

$f(1)=h=-2(1)^2+7(1)+4$

$f(1)=h=-2+7+4$

$h=9$

b) The ball travels 4 meters.

But to calculate it, it is when $h=0$

$0=-2d^2+7d+4$

Using the quadratic formula:

$d=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$d=\frac{-7 \pm \sqrt{7^2-4\left(-2\right)4}}{2\left(-2\right)}$

$d=\frac{-7\pm\sqrt{81}}{-4}$

$d=\frac{-7\pm9}{-4}$

It will give us to solutions, once it is a quadratic equation, but we are talking about a positive distance.

$d=-\frac{1}{2} \text{ or }d=4$

3.

In this question, we have to find the area of the cylinder and the sphere.

From the information given, we have

a = 5mm and d = 5mm, therefore the radius is 2.5 mm.

The volume of a cylinder:

$V=\pi r^2h$

$V=\pi (2.5)^2 \cdot 5$

$V=31.25 \pi$

$V_{c} \approx 98.17 \text{ m}^3$

The volume of the sphere:

$V=\frac{4}{3} \pi r^2$

$V_{s} \approx 65.4 \text{ m}^3$

The volume of the capsule is approximately $163.57 \text{ m}^3$

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4 years ago