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2 years ago

Which of the following is an example of natural selection?

Mathematics

1 answer:

SVETLANKA909090 [29]2 years ago

4 0

Answer:

A) Hummingbirds drinking nectar from a plant.

Step-by-step explanation:

Here are some examples of natural selection: In a habitat there are red bugs and green bugs. The birds prefer the taste of the red bugs, so soon there are many green bugs and few red bugs. The green bugs reproduce and make more green bugs and eventually there are no more red bugs.

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A home has a triangular backyard. The second angle of the triangle is 4degrees more than the first angle. The third angle is 8de

amid [387]

Answer:

1st angle = X

2nd angle = X + 7

3rd angle = X + 18

Then (sum. them): 3X + 25 = 180 ==> X = 51.67 degree

Therefore,

1st angle = 51.67 degree

2nd angle = 58.67 degree

3rd angle = 69.67 degree

5 0

3 years ago

Read 2 more answers

What is the value of 125 ⅔ ?

horrorfan [7]

Answer:125.666

Step-by-step explanation:

8 0

2 years ago

PLEASE HELP!

____ [38]

<h3>Answer:</h3>

A. x = -2
B. (-2, -3), (-3, -1)
C. x = 0

<h3>Step-by-step explanation:</h3>

Part A. The solution is represented by the point at which the graphs intersect: (-2, -3). The x-value that makes p(x) = f(x) is x = -2.

___

Part B. The point found in Part A is one solution to f(x). The graph shows the line has a slope of -2, so another point will be 1 to the left and 2 up: (-3, -1). So, two solutions are ...

... (-2, -3) and (-3, -1)

___

Part C. The graphs of p(x) and g(x) intersect at the point (0, 2). This means

... p(0) = g(0) = 2

So, x = 0 is the solution to the equation p(x) = g(x).

7 0

3 years ago

Use polar coordinates to find the volume of the given solid. Inside both the cylinder x2 y2 = 1 and the ellipsoid 4x2 4y2 z2 = 6

Anton [14]

The Volume of the given solid using polar coordinate is: $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta$

V= $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta$

<h3>What is Volume of Solid in polar coordinates?</h3>

To find the volume in polar coordinates bounded above by a surface z=f(r,θ) over a region on the xy-plane, use a double integral in polar coordinates.

Consider the cylinder, $x^{2}+y^{2} =1$ and the ellipsoid, $4x^{2}+ 4y^{2} + z^{2} =64$

In polar coordinates, we know that

$x^{2}+y^{2} =r^{2}$

So, the ellipsoid gives

$4{(x^{2}+ y^{2)} + z^{2} =64$

4( $r^{2}$ ) + $z^{2}$ = 64

$z^{2}$ = 64- 4( $r^{2}$ )

z=± $\sqrt{64-4r^{2} }$

So, the volume of the solid is given by:

V= $\int\limits^{2\pi}_ 0 \int\limits^1_0{} \, [\sqrt{64-4r^{2} }- (-\sqrt{64-4r^{2} })] r dr d\theta$

= $2\int\limits^{2\pi}_ 0 \int\limits^1_0 \, r\sqrt{64-4r^{2} } r dr d\theta$

To solve the integral take, $64-4r^{2}$ = t

dt= -8rdr

rdr = $\frac{-1}{8} dt$

So, the integral $\int\ r\sqrt{64-4r^{2} } rdr$ become

= $\int\ \sqrt{t } \frac{-1}{8} dt$

= $\frac{-1}{12} t^{3/2}$

= $\frac{-1}{12} (64-4r^{2}) ^{3/2}$

so on applying the limit, the volume becomes

V= $2\int\limits^{2\pi}_ {0} \int\limits^1_0{} \, \frac{-1}{12} (64-4r^{2}) ^{3/2} d\theta$

= $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(64-4(1)^{2}) ^{3/2} \; -(64-4(2)^{0}) ^{3/2} ] d\theta$

V = $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta$

Since, further the integral isn't having any term of $\theta$ .

we will end here.

The Volume of the given solid using polar coordinate is: $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta$

Learn more about Volume in polar coordinate here:

brainly.com/question/25172004

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3 0

1 year ago

I need the answers plz

garik1379 [7]

Answer:

These problems are an example of equations with two unknowns. The way these equations are solved is that we write these equations one under the another.

If both equations have, such is the case here, same parts, we can simply cancel the same parts out and subtract the rest of equatuons. That way, we are left with only one unknown (the other one was eliminated), which makes it easy to solve.

After we have found the value of an unknown, we just plug it back into any of the starting equations and solve for the second unknown.

2. Adult ticket costs $12 and child ticket costs $14.

3. Adult ticket costs $10 and child ticket costs $5.

4. One daylily costs $9 and one bush of ornamental grass costs $2.

5. A van can carry 15 and a bus can carry 56 students.

Step-by-step explanation:

2. If we mark the price of one adult ticket with x and the price of one child ticket with y, we get that:

- first day: 7x + 12y = $252

- second day: 7x + 10y = $224

Now, we can make a system:

7x + 12y = 252

7x + 10y = 224

We can now subtract these two equations and 7x will cancel out, so we get:

12y - 10y = 252 - 224

2y = 28

y = 14

Now, we can plug the value of y into any of the two equations:

7x + 10y = 224

7x + 140 = 224

7x = 84

x = 12

3. Similarly, if we mark the price of one adult ticket with x and the price of one child tickey with y, we'll get a system:

x + 12y = 70

x + 9y = 55

Again, if we subtract these two, x will cancel out, so we have:

12y - 9y = 70 - 55

3y = 15

y = 5

Now, we plug the value of y into any of the two equations, and we get:

x + 9y = 55

x + 45 = 55

x = 10

4. Using the same principle, we can mark the price of one daylily with x and the price of one bunch of ornamental grass with y, we'll get a system:

12x + 11y = 130

12x + 12y = 132

Again, we subtract so that 12x cancel out and we get:

11y - 12y = 130 - 132

-y = -2

If we get minuses on both sides, we can simply multiply both sides with -1 and we get:

y = 2

Again, we plug y:

12x + 12y = 132

12x + 24 = 132

12x = 108

x = 9

5. If we mark number of students in a van with x and the number of students in a bus with y, we get a system:

2x + 12y = 702

2x + y = 86

As you've probably already noticed the pattern, we subtract equations and cancel 2x out to get:

12y - y = 702 - 86

11y = 616

y = 56

Once again, we plug the value of y into any equation:

2x + y = 86

2x + 56 = 86

2x = 30

x = 15

4 0

3 years ago