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2 years ago

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You roll a number cube one time. What is the probability that you roll a 6 or 4? Enter your answer as a fraction in simplest for

m.

1 answer:

garri49 [273]2 years ago

3 0

Answer:

1/3 because there are 6 numbers on a number cube and the probability of rolling a 6 or 4 is 2/6 and simplified it would be 1/3

Step-by-step explanation:

if it is right, please mark as brainliest

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Find the value of x and explain how you found it. Make sure to cite the problem and share what you see before you explain your s

klemol [59]

Answer:

THE VALUE OF X = 3 .

Step-by-step explanation:

From the Δ RQP and Δ TSP ;

⇒ ∠ TPS = ∠ RPQ ( since both are common angle )

⇒ RQ ≈ TS ( parallel lines )

⇒ ∠ PRQ = ∠ PST ( Corresponding angle )

∴ Δ RQP ≅ Δ TSP ( BY ASA RULE )

Now ,

$\frac{RP}{TP} = \frac{QP}{PS}$

$\frac{9 + X}{9} = \frac{6 + 2}{6}$

⇒ X = 3

6 0

3 years ago

The following steps are used to rewrite the polynomial expression (x+2y)(x-5y)

neonofarm [45]

(x+2y)(x-5y)
do foil method:
x^2-5xy+2xy-10y^2
simplify x^2-3xy-10y^2,
so d is correct.

7 0

3 years ago

Read 2 more answers

Four times the some of a number plus 2. Eight times a number minus 20

8_murik_8 [283]

4 x N + 2

8 x N -20
n= number hope this helps

8 0

3 years ago

What is the solution to the compound inequality? –6 ≤ 2x + 6 ≤ 6

maw [93]

∈ ≤ ≥

-6 ≤ 2x + 6 ≤ 6 ( seperate the two inequalities)

2x + 6 ≥ - 6

2x + 6 ≤ 6 (solve the inequalities)

x ≥ -6

x ≤ 0 (find the intersection)

solution is : x ∈ { -6, 0 }

7 0

3 years ago

Suppose you solved a second-order equation by rewriting it as a system and found two scalar solutions: y = e^5x and z = e^2x. Th

xenn [34]

Answer:

The solutions are linearly independent because the Wronskian is not equal to 0 for all x.

The value of the Wronskian is $\bold{W=-3e^{7x}}$

Step-by-step explanation:

We can calculate the Wronskian using the fundamental solutions that we are provided and their corresponding the derivatives, since the Wroskian is defined as the following determinant.

$W = \left|\begin{array}{cc}y&z\\y'&z'\end{array}\right|$

Thus replacing the functions of the exercise we get:

$W = \left|\begin{array}{cc}e^{5x}&e^{2x}\\5e^{5x}&2e^{2x}\end{array}\right|$

Working with the determinant we get

$W = 2e^{7x}-5e^{7x}\\W=-3e^{7x}$

Thus we have found that the Wronskian is not 0, so the solutions are linearly independent.

3 0

3 years ago