Select the common unit rate you would use to solve a problem involving the speed of a vehicle.

What is the solution set to the inequality 2²-x≥6

madreJ [45]

Answer:

A is for apple

Step-by-step explanation:

abc

8 0

3 years ago

Help. help. help. no link pls.

kondaur [170]

Answer:

x + 2, x ≠ -2

Step-by-step explanation:

Note that F(x) = x^2 + 4x + 4 factors into (x + 2)^2, and that we are dividing this by g(x) = x + 2. The quotient is x + 2, which in this problem has been named h(x).

The quotient (f/g)(x) is therefore x + 2 EXCEPT that this is not defined at x = -2, because the denominator, x + 2, will be zero there, undefined.

8 0

3 years ago

a cell phone company plans to market a new smartphone. they have already sold 612 units durning the first week of the campaign.

Vadim26 [7]

The first term is 612.

The common ratio is 1.08 and

The recursive rule is $a_{n} = a^{n-1} \times r$

Step-by-step explanation:

the question to the problem is to write the values of the first term, common ratio, and expression for the recursive rule.

The first term :

In geometric sequence, the first term is given as $a_{1}$ .

⇒ $a_{1} = 612$

Now, the geometric sequence follows as 612, 661, ........

The common ratio (r) :

It is the ratio between two consecutive numbers in the sequence.

Therefore, to determine the common ratio, you just divide the number from the number preceding it in the sequence.

⇒ r = 661 divided by 612

⇒ r = 1.08

To find the recursive rule :

A geometric series is of the form a,ar,ar2,ar3,ar4,ar5........

Here, first term $a_{1} = a$ and other terms are obtained by multiplying by r.

Observe that each term is r times the previous term.
Hence to get nth term we multiply (n−1)th term by r .

The recursive rule is of the form $a_{n} = a^{n-1} \times r$

This is called recursive formula for geometric sequence.

We know that r = 1.08 and $a_{1}$ = 612.

To find the second term $a_{2}$ , use the recursive rule $a_{n} = a^{n-1} \times r$

⇒ $a_{2} = a^{2-1}\times r$

⇒ $a_{2} = a^{1}\times r$

⇒ $a_{2} = 612\times 1.08$

⇒ $a_{2} = 661$

3 0

4 years ago

A (0, 2) and B (6,6) are points on the straight line ABCD.

elena-14-01-66 [18.8K]

Answer:

(18, 14)

Step-by-step explanation:

We know that C and D lie on the line AB and BC = CD = AB. Then we need to use the distance formula and equation of the line AB to find the other two coordinates.

The distance formula states that the distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ , the distance is denoted by: $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ . Let's find the distance between A and B:

d = $\sqrt{(6-0)^2+(6-2)^2}=\sqrt{6^2+4^2} =\sqrt{36+16} =\sqrt{52} =2\sqrt{13}$

Now say the coordinates of D are (a, b). Then the distance between D and B will be twice of 2√13, which is 4√13:

4√13 = $\sqrt{(6-a)^2+(6-b)^2}$

Square both sides:

208 = (6 - a)² + (6 - b)²

Let's also find the equation of the line AB. The y-intercept we know is 2, so in y = mx + b, b = 2. The slope is (6 - 2) / (6 - 0) = 4/6 = 2/3. So the equation of the line is: y = (2/3)x + 2. Since (a, b) lines on this line, we can put in a for x and b for y: b = (2/3)a + 2. Substitute this expression in for b in the previous equation:

208 = (6 - a)² + (6 - b)²

208 = (6 - a)² + (6 - (2/3a + 2))² = (6 - a)² + (-2/3a + 4)²

208 = a² - 12a + 36 + 4/9a² - 16/3a + 16 = 13/9a² - 52/3a + 52

0 = 13/9a² - 52/3a - 156

13a² - 156a - 1404 = 0

a² - 12a - 108 = 0

(a + 6)(a - 18) = 0

a = -6 or a = 18

We know a can't be negative so a = 18. Plug this back in to find b:

b = 2/3a + 2 = (2/3) * 18 + 2 = 12 + 2 = 14

So point D has coordinates (18, 14).

8 0

4 years ago

Read 2 more answers

Find the value of x in each case:

Yakvenalex [24]

Nice... you might have put the wrong picture lol

5 0

3 years ago

Read 2 more answers