If the expression (x + 5 )to the 6th is expanded what is the coefficient of the linear X term?

There are 12 red pens, 20 blue pens and 5 black pens in a bag. What is the probability that you will pick a black pen? Round

Vlada [557]

Answer:

The probability of getting a black pen from the bag is 13.5%.

Step-by-step explanation:

To determine the percentage probability that a black pen will be obtained from the bag, the following calculation must be performed:

12 + 20 + 5 = 37

37 = 100%

5 = X

(5 x 100) / 37 = X

500/37 = X

13.5 = X

Thus, it is determined that the totality of pens is 100%, and through a cross multiplication the percentage of the 5 black pens with respect to the total of pens is obtained.

Therefore, the probability of getting a black pen from the bag is 13.5%.

5 0

3 years ago

PLEASE HELP THANK YOU SO MUCH IN ADVANCE. IF POSSIBLE PLEASE EXPLAIN HOW YOU GOT YOUR ANSWER

natka813 [3]

Answer:

Step-by-step explanation:

To be right triangle has to verify the Pythagorean Theorem

hypothenuse² = (side 1 )² + (side 2)²

Since the hypothenuse is the largest of the sides we consider the largest number the hypothenuse and check the relation.

16, 30, 34

34² = 16²+30²; 1156 = 1156 ✅

16, 30, 34 are measurements of a right triangle

8, 14, 16

16² = 8² +14²; 256 = 260 ❌

8, 14, 16 is not a right triangle

10, 12, 16

16² = 10² +12²; 256 = 244 ❌

10, 12, 16 is not a right triangle

5, 12, 13

13² = 12²+5²; 169 = 169 ✅

5, 12, 13 are measurements of a right triangle

8 0

2 years ago

Use series to verify that<br><br> <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{

SVETLANKA909090 [29]

$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

-----------------------

Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.

6 0

3 years ago

What value of x will make the given triangles congruent?

Molodets [167]

Because you're trying to make them congruent, equal the two expressions to each other.

So, for #10 you would equal them so it would look like this.
2x+2=5x-19

Then you would just go ahead and simplify

2x+2=5x-19
-2x -2x
---------------
2= 3x -19
+19 +19
---------------
21=3x
--- ----
3 3

x=7

This means that x should be 7. You can check this just by plugging it in

2x+2
2(7)+2 = 16

5x-19
5(7)-19= 16

Same with #11.

x+8=3x-14
-x -x
--------------
8=2x-14
+14 +14
--------------
22=2x
--- ----
2 2

x=11

Check.

x+8
11+8= 19

3x-14
3(11) - 14 = 19

6 0

3 years ago

Which of the following transformations does not

svlad2 [7]

Answer:

(9x, 9y) dilation

Step-by-step explanation:

(9, 9y) is the answer because F is 180 degrees rotations, G is not real, H is translations, and J is dilations and dilations is multiplying which makes the transformation non congruent.

Your welcome

7 0

3 years ago

If the expression (x + 5 )to the 6th is expanded what is the coefficient of the linear X term?​

If the expression (x + 5 )to the 6th is expanded what is the coefficient of the linear X term?