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3 years ago

Help me pleaseeeeeeeeeeee

Mathematics

1 answer:

ycow [4]3 years ago

7 0

Answer:I would say the very first one, but I could be way off- ._.

Step-by-step explanation:

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Nine more than 3 times a number is the same as 6 less than twice the number

Ludmilka [50]

Answer:

Step-by-step explanation:

Let the number = x

3x + 9 = 2x - 6 Looks like you are going to get something negative here.

Subtract 2x from both sides.

3x - 2x + 9 = 2x - 2x - 6

x + 9 = -6

Subtract 9 from both sides.

x + 9 - 9 = - 6 - 9

Combine

x = - 15

7 0

3 years ago

(6X1)+(5X1/10) I need help on this

hram777 [196]

Answer:

13/2

Step-by-step explanation:

Hope this helps if you need steps let me know.

3 0

3 years ago

An urn contains 6 red marbles and 4 black marbles. Two marbles are randomly drawn (one by one) from the urn without replacement.

strojnjashka [21]

Answer: The required probability is $\dfrac{2}{15}.$

Step-by-step explanation: Given that an urn contains 6 red marbles and 4 black marbles. Two marbles are randomly drawn one by one from the urn without replacement.

We are to find the probability that both drawn marbles are black.

Let E and F denote the events of two marbles one by one without replacement and let S and S' denote the corresponding sample spaces.

Then, we have

$n(E)=^4C_1=4,\\\\n(F)=^3C_1=3,\\\\n(S)=^{10}C_1=10,\\\\n(S')=^9C_1=9.$

Therefore, the probability that both marbles are red is given by

$p=P(E)\times P(F)=\dfrac{n(E)}{n(S)}\times\dfrac{n(F)}{n(S')}=\dfrac{4}{10}\times\dfrac{3}{9}=\dfrac{2}{15}.$

Thus, the required probability is $\dfrac{2}{15}.$

4 0

3 years ago

Order form least to greatest -3/10, -4/15, -2/5

EleoNora [17]

Convert into decimal:

-3/10 = -0.3

-4/15 = -0.2667

-2/5 = -0.4

-0.4 < -0.3 < -0.267

-2/5 < -3/10 < -4/15

7 0

3 years ago

Read 2 more answers

Need some help with indefinite integrals pleasee

goldenfox [79]

Observe that $(x^2+2x+2)'=2x+2$ .

You have that $\int \frac{2x}{x^2+2x+2}\; dx=\int \frac{2x+2-2}{x^2+2x+2}\; dx=\int \frac{2x+2}{x^2+2x+2}\; dx+\int \frac{1}{x^2+2x+2}\; dx=I_1+2I_2$ .

To compute $I_1$ you set $y=x^2+2x+2$ , so $dy=2x+2\; dx$ . Therefore, $I_1=\int \frac{1}{y}\; dy=\ln |y|=\ln |x^2+2x+2|=\ln (x^2+2x+2)$ .

For $I_2$ observe that $x^2+2x+2=(x+1)^2+1$ . Let $y=x+1$ . Then $dy=dx$ and $I_2=\int \frac{1}{y^2+1}=\arctan y=\arctan (x+1)$ .

So, $\int \frac{2x}{x^2+2x+2}=\ln (x^2+2x+2)+2\arctan (x+1)+c$ .

8 0

4 years ago