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Phoenix [80]
3 years ago
6

-6x + 3 = 15 whats the solution to finding x

Mathematics
1 answer:
Anarel [89]3 years ago
5 0

Answer:

x=-2

Step-by-step explanation:

Subtract 3 from both sides

-6x=15-3

Simplify 15 -3 to 12

-6x=12

divide both sides by -6

x= -12/6

Simplify 12/6 to 2

x=-2

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Please help me I really need help
barxatty [35]

Answer:

im not sure but i think its 0.25

Step-by-step explanation:

6 0
3 years ago
Can someone answer this please
ollegr [7]
The answer is symmetric because it’s equal
7 0
3 years ago
A marble is randomly selected from a bag. The probability of selecting a marble with dots on it is 0.2. The probability of selec
marin [14]

Answer:

0.5

Step-by-step explanation:

Let D be the event of selecting a marble with dots.

Let P be the event of selecting a purple marble.

The probability of selecting a marble with dots, P(D)=0.2

The probability of selecting a marble that is both purple and has dots, P(D \cap P)=0.1

We want to determine the probability of selecting a purple marble given that the marble has dots on it, P(P|D)

By the definition of conditional probability:

P(P|D)= \dfrac{P(P \cap D)}{P(D)} \\= \dfrac{0.1}{0.2}\\ =0.5

The probability of selecting a purple marble given that the marble has dots on it is 0.5.

4 0
3 years ago
What expression is equivalent to (9x^2 + 2x -7)(x - 4)
Dima020 [189]

Answer: C

Step-by-step explanation:

We can use the expanding rule to get that one if the expressions is

(9x^2 + 2x - 7)x + (9x^2 + 2x - 7)(-4)

3 0
2 years ago
Please help me right away with these problems.
viktelen [127]

Q.34

\sum\limits_{k=1}^{\infty}420\left(1.002\right)^{k-1}

The infinite geometric series is converges if |r| < 1.

We have r =1.002 > 1, therefore our infinite geometric series is Diverges

Answer: c. Diverges, sum not exist.

Q.35

\sum\limits_{k=1}^{\infty}-5\left(\dfrac{4}{5}\right)^{k-1}

The infinite geometric series is converges if |r| < 1.

We have r = 4/5 < 1, therefore our infinite geometric series is converges.

The sum S of an infinite geometric series with |r| < 1 is given by the formula :

S=\dfrac{a_1}{1-r}

We have:

a_1=-5\left(\dfrac{4}{5}\right)^{1-1}=-5\left(\dfrac{4}{5}\right)^0=-5\\\\r=\dfrac{4}{5}

substitute:

S=\dfrac{-5}{1-\frac{4}{5}}=-\dfrac{5}{\frac{1}{5}}=-5\cdot\dfrac{5}{1}=-25

Answer: c. Converges, -25.

4 0
3 years ago
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