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olga_2 [115]
4 years ago
8

What two numbers add up to 12 but multiply to 20

Mathematics
2 answers:
marta [7]4 years ago
7 0

Answer:

I think it’s 10plus2 then u multiply

Step-by-step explanation:

Elan Coil [88]4 years ago
5 0

Answer:

10 and 2

Step-by-step explanation:

10 + 2 adds up to twelve. 10 x 2 is equal to 20

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Steve picks 55 pounds of pears. He packs an equal amount of pears into 6bags. He then has4 pounds of pears left. What is the wei
user100 [1]
9 pounds in each bag with 4 remaining.
6 0
3 years ago
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Suppose that g (x) varies inversely with x and g (x)=0.2 when x = 0.1.
CaHeK987 [17]

Answer : 0.0125

Suppose that g (x) varies inversely with x and g (x)=0.2 when x = 0.1.

If x varies inversely with y then y=k/x

Where k is constant of proportionality

g (x) varies inversely with x , so g(x) = \frac{k}{x}

g (x)=0.2 when x = 0.1

Plug in the values and solve for k

g(x) = \frac{k}{x}

0.2 = \frac{k}{0.1}

Multiply 0.1 on both sides

0.02 = k

so g(x) = \frac{0.02}{x}

Now we need to find g(x) when x= 1.6

Plug in 1.6 for x  and find out g(x) in the above equation

g(x) = \frac{0.02}{1.6}

g(x)= 0.0125



4 0
3 years ago
Bob has a standard deck of playing cards. If he randomly draws a J, K, 2, and 2, what is the probability that the next card he d
nlexa [21]

Answer:

The correct answer has already been given (twice). I'd like to present two solutions that expand on (and explain more completely) the reasoning of the ones already given.


One is using the hypergeometric distribution, which is meant exactly for the type of problem you describe (sampling without replacement):


P(X=k)=(Kk)(N−Kn−k)(Nn)


where N is the total number of cards in the deck, K is the total number of ace cards in the deck, k is the number of ace cards you intend to select, and n is the number of cards overall that you intend to select.


P(X=2)=(42)(480)(522)


P(X=2)=61326=1221


In essence, this would give you the number of possible combinations of drawing two of the four ace cards in the deck (6, already enumerated by Ravish) over the number of possible combinations of drawing any two cards out of the 52 in the deck (1326). This is the way Ravish chose to solve the problem.


Another way is using simple probabilities and combinations:


P(X=2)=(4C1∗152)∗(3C1∗151)


P(X=2)=452∗351=1221


The chance of picking an ace for the first time (same as the chance of picking any card for the first time) is 1/52, multiplied by the number of ways you can pick one of the four aces in the deck, 4C1. This probability is multiplied by the probability of picking a card for the second time (1/51) times the number of ways to get one of the three remaining aces (3C1). This is the way Larry chose to solve the this.

Step-by-step explanation:


6 0
3 years ago
Read 2 more answers
Find the indefinite integral. (Use C for the constant of integration.) <br> e2x 25 e4x dx.
Sladkaya [172]

Answer:

The solution is  \frac{1}{10} * tan^{-1}[\frac{e^{2x}}{5} ] +  C

Step-by-step explanation:

From the question

    The function given is  f(x) =  \frac{e^{2x}}{ 25 + e^{4x}} dx

The  indefinite integral is  mathematically represented as

          \int\limits  {\frac{e^{2x}}{ 25 + e^{4x}}} \, dx

Now  let  e^{2x} =  u

=>   \frac{du}{dx} 2e^{2x}

=>   2 e^{2x}dx =  du

So

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= \frac{1}{2} \int\limits  {\frac{1}{ 25 + u^2)} } \, du

=  \frac{1}{2} \int\limits  {\frac{1}{ 5^2 + u^2)} } \, du

= \frac{1}{2} \frac{tan^{-1} [\frac{u}{5} ]}{5}  +  C

Now substituting for  u

\frac{1}{10} * tan^{-1}[\frac{e^{2x}}{5} ] +  C

3 0
3 years ago
7)<br> 2 1<br> Please help giving Brainly
Mama L [17]

Answer:

You should use MathPapa.

Step-by-step explanation:

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3 years ago
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