Rewrite the expression using Distributive Property, then simplify: -2(-x+3y)-3(x-5y)

Express tan W as a fraction in simplest terms. U W 24 6

Deffense [45]

90

Step-by-step explanation:vkjl

7 0

2 years ago

Question is in the picture please help

LiRa [457]

When given the graph of a function, the domain would include all the points that there is a graph. The strategy is to find what is not included.

What we are looking for are points of discontinuity. Think of it as when you remove your pencil from the paper.

From left to right, the graph stops at x = -3. So anything less than -3 is in the domain. Next, the graph starts up again at x =-1 after an asymptote (the vertical dashed lines). This piece goes to x = 4. So our domain is from -1 to 4.

Lastly, there's a jump from 4 to 5 and the graph goes on again. After 5, we take all the stuff more than it. So x > 5 is in the domain.

So x < -3, - 1 < x < 4, and x > 5 appears to be our domain. However, end points needed to be checked to see if we include them or not. Again we go left to right.

At x = -3 there is a filled (or closed) circle and that means we include -3.

At x = -1 there is an asymptote. Asymptotes are things you get close to but don't get to. (Think of it as the "I'm Not Touching" game you play on car trips.) So we exclude -1.

At x = 4 there is an unfilled (or open) circle and that means we exclude 4.

At x = 5 there is a filled circle so we include 5.

Now we refine our domain for the endpoints.

x ≤ -3, -1 < x < 4, x ≥ 5 is our domain.

The problem gives us intervals, and we gave it in inequalities. When we include an endpoint we use brackets - [ and } and when we exclude and endpoint we use parentheses - ( and ). Let's go back to x ≤ -3. Anything less works, and -3 is included (closed circle). That interval is (-∞, -3]. Next is the piece between -1 and 4. Since both are excluded, (-1,4) is our interval. We include 5 to write x ≥5 as the interval [5,∞).

Put the bolded ones all together and use the union, ∪, symbol to connect them, since something on the graph could be in any piece.

Our domain is (-∞, -3] ∪(-1,4) ∪ [5,∞).

5 0

3 years ago

55. (a) If alpha and beta are the roots of the equation xsquare+ px+q=0 and beta>alpha find the square of the

densk [106]

Answer:

√(p²-4q)

Step-by-step explanation:

Using the Quadratic Formula, we can say that

x = ( -p ± √(p²-4(1)(q))) / 2(1) with the 1 representing the coefficient of x². Simplifying, we get

x = ( -p ± √(p²-4q)) / 2

The roots of the function are therefore at

x = ( -p + √(p²-4q)) / 2 and x = ( -p - √(p²-4q)) / 2. The difference of the roots is thus

( -p + √(p²-4q)) / 2 - ( ( -p - √(p²-4q)) / 2)

= 0 + 2 √(p²-4q)/2

= √(p²-4q)

7 0

2 years ago

A bag contains two six-sided dice: one red, one green. The red die has faces numbered 1, 2, 3, 4, 5, and 6. The green die has fa

gayaneshka [121]

Answer:

the probability the die chosen was green is 0.9

Step-by-step explanation:

Given that:

A bag contains two six-sided dice: one red, one green.

The red die has faces numbered 1, 2, 3, 4, 5, and 6.

The green die has faces numbered 1, 2, 3, 4, 4, and 4.

From above, the probability of obtaining 4 in a single throw of a fair die is:

P (4 | red dice) = $\dfrac{1}{6}$

P (4 | green dice) = $\dfrac{3}{6}$ = $\dfrac{1}{2}$

A die is selected at random and rolled four times.

As the die is selected randomly; the probability of the first die must be equal to the probability of the second die = $\dfrac{1}{2}$

The probability of two 1's and two 4's in the first dice can be calculated as:

= $\begin {pmatrix} \left \begin{array}{c}4\\2\\ \end{array} \right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^4$

= $\dfrac{4!}{2!(4-2)!} ( \dfrac{1}{6})^4$

= $\dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^4$

= $6 \times ( \dfrac{1}{6})^4$

= $(\dfrac{1}{6})^3$

= $\dfrac{1}{216}$

The probability of two 1's and two 4's in the second dice can be calculated as:

= $\begin {pmatrix} \left \begin{array}{c}4\\2\\ \end{array} \right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^2 \times \begin {pmatrix} \dfrac{3}{6} \end {pmatrix} ^2$

= $\dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^2 \times ( \dfrac{3}{6})^2$

= $6 \times ( \dfrac{1}{6})^2 \times ( \dfrac{3}{6})^2$

= $( \dfrac{1}{6}) \times ( \dfrac{3}{6})^2$

= $\dfrac{9}{216}$

∴

The probability of two 1's and two 4's in both dies = P( two 1s and two 4s | first dice ) P( first dice ) + P( two 1s and two 4s | second dice ) P( second dice )

The probability of two 1's and two 4's in both die = $\dfrac{1}{216} \times \dfrac{1}{2} + \dfrac{9}{216} \times \dfrac{1}{2}$

The probability of two 1's and two 4's in both die = $\dfrac{1}{432} + \dfrac{1}{48}$

The probability of two 1's and two 4's in both die = $\dfrac{5}{216}$

By applying Bayes Theorem; the probability that the die was green can be calculated as:

P(second die (green) | two 1's and two 4's ) = The probability of two 1's and two 4's | second dice)P (second die) ÷ P(two 1's and two 4's in both die)

P(second die (green) | two 1's and two 4's ) = $\dfrac{\dfrac{1}{2} \times \dfrac{9}{216}}{\dfrac{5}{216}}$

P(second die (green) | two 1's and two 4's ) = $\dfrac{0.5 \times 0.04166666667}{0.02314814815}$

P(second die (green) | two 1's and two 4's ) = 0.9

Thus; the probability the die chosen was green is 0.9

8 0

3 years ago

Subtract -2k3 + k2 - 9 from 5k3 - 3k + 7.

tankabanditka [31]

5k3-3k+7-(-2k3+k2-9)
basically rewrite with simple algebra principles
15k-3k+7-(-6k+2k-9)
split up the parenthesis
15k-3k+7+6k-2k+9
sort them
15k-3k+6k-3k+7+9
short down by adding up the similar factors
answer: 18k+18
factorise
18(k+1)
both forms are right

8 0

3 years ago

Read 2 more answers