Ask question

Ask question

All categories

4 years ago

8

Solve the equation Solve 3=-(-4+6)

1 answer:

katovenus [111]4 years ago

4 0

3=-1(-4+6)
pemdas
inside parenthasees first
-4+6=2

3=-1(2)
3=-2
false

You might be interested in

4th grade math patterns 18250, 18500, 19000, 20000 what's the next number?

irinina [24]

22000

The number that's being added in the sequence increases by 2x.

250 * 2 = 500, which is added to 18500 to equal 19000.

500 * 2 = 1000, which is added to 19000 to equal 20000.

1000 * 2 = 2000, which added to 20000 equals 22000.

6 0

4 years ago

Solve the equation using the quadratic formula. x^2-4x+3=0

almond37 [142]

Let's solve your equation step-by-step.

x2−4x+3=0

Step 1: Factor left side of equation.

(x−1)(x−3)=0

Step 2: Set factors equal to 0.
x-1=0 or x-3=0

For this one add one on both sides
x−1+1=0+1 x=1
For this one add 3 on both sides
x−3+3=0+3. x=3

x=1 or x=3

5 0

3 years ago

PLEASE FOR THE LOVE OF GOD

Georgia [21]

Answer: (4)

Step-by-step explanation:

7 0

3 years ago

Select whether the equation has a solution or not.

Stolb23 [73]

This equation has no solution

7 0

4 years ago

Two distinct number cubes, one red and one blue, are rolled together. Each number cube has sides numbered 1 through 6.

cestrela7 [59]

Answer:

$P(E_1 or E_2) = \frac{7}{12}$

Step-by-step explanation:

Given

Two cubes of side 1 - 6

Required

Probability that the outcome of the roll is an odd sum or a sum that is a multiple of 5

First, the sample space needs to be listed;

Let $C_r$ represent the Red cube

$C_b$ represent the Blue cube

S represent the sample space

$C_r = (1,2,3,4,5,6)\\C_b = (1,2,3,4,5,6)\\S = (2,3,4,5,6,7,3,4,5,6,7,8,4,5,6,7,8,9,5,6,7,8,9,10,6,7,8,9,10,11,7,8,9,10,11,12)$ S is gotten by getting the sum of $C_r$ and $C_b$

$n(S) = 36$

<em>Calculating the Probability</em>

Let $E_1$ represent the event that an outcome is an odd sum

$E_1 = (3,5,7,3,5,7,5,7,9,5,7,9,7,9,11,7,9,11)$

$n(E_1) = 18$

$P(E_1) = \frac{n(E_1)}{n(S)}$

$P(E_1) = \frac{18}{36}$

Let $E_2$ represent the event that an outcome is a multiple of 5

$E_2 = (5,5,5,5,10,10,10)$

$n(E_2) = 7$

$P(E_2) = \frac{n(E_2)}{n(S)}$

$P(E_2) = \frac{7}{36}$

Let $E_3$ represent the event that an outcome is an odd sum and a multiple of 5

$E_3 = E_1 and E_2$

$E_3 = (5,5,5,5)$

$n(E_3) = 4$

$P(E_3) = \frac{n(E_3)}{n(S)}$

$P(E_3) = \frac{4}{36}$

Calculating $P(E_1 or E_2)$

$P(E_1 or E_2) = P(E_1) + P(E_2) - P(E_1 and E_2)$

$P(E_1 or E_2) = P(E_1) + P(E_2) - P(E_3)$

$P(E_1 or E_2) = \frac{18}{36} + \frac{7}{36} - \frac{4}{36}$

$P(E_1 or E_2) = \frac{18 + 7 - 4}{36}$

$P(E_1 or E_2) = \frac{21}{36}$

$P(E_1 or E_2) = \frac{7}{12}$

Hence, the probability that the outcome of the roll is an odd sum or a sum that is a multiple of 5 is $\frac{7}{12}$

6 0

3 years ago