What is 11x - 3 = 5x + 11

Please help me with this problem......

kupik [55]

The red graph is the result of a vertical translation, 4 units down, of the black graph.

This means that, whenever the black graph associates

$x\mapsto f(x)$

The red graph must associate 4 less to the same input:

$x\mapsto f(x)-4$

So, the equation for the red graph is

$y=f(x)-4$

or, equivalently,

$y+4=f(x)$

7 0

3 years ago

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Solve each proportion

mamaluj [8]

Hello.

In order to solve a proportion, you should cross-multiply:

$\displaystyle\frac{7}{x} =\frac{5}{9}$

Multiply 7 times 9 and 5 times x:

5x=63

Divide both sides by 5:

x= $\displaystyle\frac{63}{5}$

Now let's solve the second proportion.

Multiply 8p times 6 and 12 times 15:

48p=180

Divide both sides by 48:

p=3.75

I hope it helps.

Have an outstanding day.

$\boxed{imperturbability}$

5 0

2 years ago

Find the area of the shaded portion in the equilateral triangle with sides 6.

LUCKY_DIMON [66]

Triangle is equilateral with sides of 6.
Therefore all angles = 60°
Ht of 30-60-90 is 3sr3 = 5.196
Area of 30-60-90 = 1/2×b×h = 3×5.196
Area = 15.59
Pie slice from each corner = 60/360×pi×r^2, with r = 3
1/6×pi×9 = 4.71 × 3 pie slices = 14.13
So, shaded inner region = area triangle - 3 pie corners = 15.59-14.13
= 1.46

7 0

3 years ago

Which statement describes the sequence −9, −3, 3, 9, 15?

Arada [10]

C. An arithmetic sequence with common difference 6

6 0

3 years ago

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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