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mamaluj [8]
3 years ago
15

Show how many ways to make 20 in a web

Mathematics
1 answer:
Hitman42 [59]3 years ago
4 0

Answer:

get 20 different shaped using 20 sides and 20 corners and 20 20 degree angles duh

Step-by-step explanation:

You might be interested in
Stanley makes paper constellations. The following data points represent how many stars were in each
____ [38]

Answer:

Check Explanation and the attached image.

Step-by-step explanation:

The histogram of the set of data presented is presented in the attached image to.this solution

The histogram represents data by indicating the frequency of distribution on the y-axis and the sets of variables indicated on the x-axis.

With the constellations named numbers 0 to 6, the frequency of each constellation, that is, the number of stars in each constellation corresponds to the height of the bar representing each constellation.

Hope this Helps!!!

5 0
3 years ago
Pls only answer if yk it ill give correct one BRAINLIEST!!
Dmitry [639]

Answer:

X=10, x=4

Step-by-step explanation:

first, divide 14 by 2, and square it; giving you

x^2-14x+(7)^2=-40

Because 7^2 is 49, you have to add 49 to the other side; giving you

x^2-14x+(7)^2=9

Then factor the left side; giving you (x-7)^2=9

Take the square root of both sides, giving you

(x-7)^2= plus or minus 3

Then solve to get two solutions, 10 and 4

8 0
3 years ago
NEED HELP
solmaris [256]

The solutions to the systems of equations are:

1. (2, 3) (see attachment below). [one solution]

2. no solution

3. (3, 13) [one solution]

<h3>Solution to a System of Equations?</h3>

The solution to a system of equations is the x-value and y-value that will make both equations true. It can be found either using a graph, by elimination method, or substitution method as explained below.

1. Using graph to solve y = 2x - 1 and y = 4x - 5:

The solution is the point where both lines intersect which is: (2, 3) (see attachment below). [one solution]

2. Solving using substitution method:

x = -5y + 4 ---> eqn. 1

3x + 15y = -1 ---> eqn. 2

Substitute x = -5y + 4 into eqn. 2

3(-5y + 4) + 15y = -1

-15y + 12 + 15y = -1

-15y + 15y = -1 - 12

0 = -13 (this shows that there is no solution)

3. Using elimination method:

14x = 2y + 16 ---> eqn. 1

5x = y + 2 ---> eqn. 2

1(14x = 2y + 16)

2(5x = y + 2)

14x = 2y + 16 ----> eqn. 3

10x = 2y + 4 -----> eqn. 4

Subtract

4x = 12

x = 12/4

x = 3

Substitute x = 3 into eqn. 2

5(3) = y + 2

15 = y + 2

15 - 2 = y

13 = y

y = 13

The solution is: (3, 13).

Learn more about the solution of a system of equations on:

brainly.com/question/13729904

#SPJ1

5 0
2 years ago
Which ordered pairs lie on the graph of the exponential function f (x) = 3(0.2)^x
photoshop1234 [79]

The ordered pairs which lie on the graph of the exponential function, f(x) = 3(0.2)^x is; Choice A; (0, 3).

<h3>Ordered pair Solution;</h3>

The given exponential function is;

  • f (x) = 3(0.2)^x

By testing all values; Only the pair; (0, 3) is a solution to the graph of the exponential given.

By testing;

  • f (0) = 3(0.2)⁰.

  • Where, (0.2)⁰ = 1 (from indices).

  • f(0) = 3 × 1 = 3.

Read more on ordered pairs;

brainly.com/question/5754926

4 0
2 years ago
B) Let g(x) =x/2sqrt(36-x^2)+18sin^-1(x/6)<br><br> Find g'(x) =
jolli1 [7]

I suppose you mean

g(x) = \dfrac x{2\sqrt{36-x^2}} + 18\sin^{-1}\left(\dfrac x6\right)

Differentiate one term at a time.

Rewrite the first term as

\dfrac x{2\sqrt{36-x^2}} = \dfrac12 x(36-x^2)^{-1/2}

Then the product rule says

\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 x' (36-x^2)^{-1/2} + \dfrac12 x \left((36-x^2)^{-1/2}\right)'

Then with the power and chain rules,

\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12\left(-\dfrac12\right) x (36-x^2)^{-3/2}(36-x^2)' \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} - \dfrac14 x (36-x^2)^{-3/2} (-2x) \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12 x^2 (36-x^2)^{-3/2}

Simplify this a bit by factoring out \frac12 (36-x^2)^{-3/2} :

\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-3/2} \left((36-x^2) + x^2\right) = 18 (36-x^2)^{-3/2}

For the second term, recall that

\left(\sin^{-1}(x)\right)' = \dfrac1{\sqrt{1-x^2}}

Then by the chain rule,

\left(18\sin^{-1}\left(\dfrac x6\right)\right)' = 18 \left(\sin^{-1}\left(\dfrac x6\right)\right)' \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac x6\right)'}{\sqrt{1 - \left(\frac x6\right)^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac16\right)}{\sqrt{1 - \frac{x^2}{36}}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{3}{\frac16\sqrt{36 - x^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18}{\sqrt{36 - x^2}} = 18 (36-x^2)^{-1/2}

So we have

g'(x) = 18 (36-x^2)^{-3/2} + 18 (36-x^2)^{-1/2}

and we can simplify this by factoring out 18(36-x^2)^{-3/2} to end up with

g'(x) = 18(36-x^2)^{-3/2} \left(1 + (36-x^2)\right) = \boxed{18 (36 - x^2)^{-3/2} (37-x^2)}

5 0
2 years ago
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