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4 years ago

A town doubles its size every 30 years. If the population is currently 10,000, what will the population be in 60 years?

Mathematics

1 answer:

Effectus [21]4 years ago

8 0

Answer: 20,000

Step-by-step explanation:

every 30 years 10,000 adds to the population

30 years=10,000 people

60 years=20,000 people

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Yoo can someone do this for me

o-na [289]

Answer:

Angle 5

Step-by-step explanation:

Adjacent angles share a common side and a common vertex.

6 0

3 years ago

Calculate the size of angle DBE.6. Calculate the size of angle DBE.

Lina20 [59]

Angle D = 47’

Angle B = 116’
32 + 32 = 64
180 - 64 = 116’

Angle E = 17’
116 + 47 = 163
180 - 163 = 17

Idk what to do after wards

5 0

4 years ago

A ball is thrown upward from ground level. Its height h, in feet, above the ground after t seconds is h-48t -16t^2.

Marat540 [252]

Answer:

36 feet.

Step-by-step explanation:

We have been given that a ball is thrown upward from ground level. Its height h, in feet, above the ground after t seconds is $h(t)=-48t -16t^2$ . We are asked to find the maximum height of the ball.

We can see that our given equation is a downward opening parabola, so its maximum height will be the vertex of the parabola.

To find the maximum height of the ball, we need to find y-coordinate of vertex of parabola.

Let us find x-coordinate of parabola using formula $x=-\frac{b}{2a}$ .

$x=-\frac{-48}{2(-16)}$

$x=-\frac{48}{32}$

$x=-\frac{3}{2}$

So, the x-coordinate of the parabola is $-\frac{3}{2}$ . Now, we will substitute $x=-\frac{3}{2}$ in our given equation to find y-coordinate of parabola.

$h(t)=-48t -16t^2$

$h(-\frac{3}{2})=-48(-\frac{3}{2})-16(-\frac{3}{2})^2$

$h(-\frac{3}{2})=-24(-3)-16(\frac{9}{4})$

$h(-\frac{3}{2})=72-4*9$

$h(-\frac{3}{2})=72-36$

$h(-\frac{3}{2})=36$

Therefore, the maximum height of the ball is 36 feet.

3 0

3 years ago

Simplify and determine the coefficient of (- x)(3y)(-2x).

Valentin [98]

Answer:

6yx2

Step-by-step explanation:

brackets means multiply hence -x×3y×-2x

6 0

3 years ago

Compare the process of solving |x – 1| + 1 < 15 to that of solving |x – 1| + 1 > 15. Check all of the following you includ

zepelin [54]

The required Comparison of the inequalities are

The |x – 1| + 1 > 15 represents the value of x lies between 13<x<15.

The range of values encompassing the region's junction is (-13, 15).

If x is more than or equal to 15, then x-11+1>15 indicates the value of x is greater than or equal to 13. None of the regions in the intersection are empty.

<h3>What is inequality?</h3>

When comparing two numbers, an inequality indicates whether one is less than, larger than, or not equal to the other.

We take into account the various variables of the inequality

|x – 1| + 1 > 15

Therefore

|-x-1|+1-1<15-1

|-x-1|-1 <14

13<x<15

The required region lies between the inequality -13 <x< 15.

Simplify the inequality Ix-11+1 > 15 we get,

|x-1|+1 > 15

|x+1| +1-1 >15-1

|x-1| > 14

x> 15

x<-13

If x has a value between -13 and x + 15, then the expression "|x-1|+1+115" is true. The range "(-13, 15)" contains the intersection of the region.

If "|x-1|+1>15" then either "x >15" or "x-13" applies to the value of x. This region's intersection is unoccupied.