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iragen [17]
3 years ago
7

A Write an explicit formula for an, the nth term of the sequence 17, 15, 13, ...

Mathematics
1 answer:
yawa3891 [41]3 years ago
3 0

Tn=-2n+19

Step-by-step explanation:

Well in south africa we use Tn

Tn=a+(n-1)d

a=is the first term in the sequence which is 17

d=is the the difference between the terms which is - 2

Tn=17+(n-1)(-2)

Tn=17-2n+2

Tn=-2n+19

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(-8)^-2 / 3^-4 simplify the expression. Write your answer using only positiv exponents.
spin [16.1K]

Answer:

3^4 / 8^2 or 81/64

Step-by-step explanation:

(-8)^-2 / 3^-4 =

= 3^4 / (-8)^2

= 3^4 / 8^2

= 81/64

4 0
3 years ago
Find an equation for the line with the given properties. Express your answer using the general form or the slope-intercept form
Damm [24]

Answer:

y=-3x+4

Step-by-step explanation:

To get the equation in the slope-intercept form first, put the equation in slope-point form using the information given. The slope-point form is y-(-2)=-3(x-2). Then solve for y.

Distribute -3 to (x-2),

y-(-2)=-3x+6

Then, add -2 to both sides,

y=-3x+4

This is your final answer; the slope is -3 and the y-intercept is 4. There are also a few other ways to solve but I find this the easiest.

3 0
3 years ago
The corners of a meadow are shown on a coordinate grid. Ethan wants to fence the meadow. What length of fencing is required?
Nuetrik [128]

Answer:

34.6 units

Step-by-step explanation:

The lenght of fencing required is the total distance between point A to B, B to C, C to D, and D to A. That is the distance between all 4 corners of the meadow.

The coordinates of the corners of the meadow is shown on a coordinate plane in the attachment. (See attachment below).

Let's use the distance formula to calculate the distance between the 4 corners of the meadow using their coordinates as follows:

Distance between point A(-6, 2) and point B(2, 6):

AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

A(-6, 2)) = (x_1, y_1)

B(2, 6) = (x_2, y_2)

AB = \sqrt{(2 - (-6))^2 + (6 - 2)^2}

AB = \sqrt{(8)^2 + (4)^2}

AB = \sqrt{64 + 16} = \sqrt{80}

AB = 8.9 (nearest tenth)

Distance between B(2, 6) and C(7, 1):

BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

B(2, 6) = (x_1, y_1)

C(7, 1) = (x_2, y_2)

BC = \sqrt{(7 - 2)^2 + (1 - 6)^2}

BC = \sqrt{(5)^2 + (-5)^2}

BC = \sqrt{25 + 25} = \sqrt{50}

BC = 7.1 (nearest tenth)

Distance between C(7, 1) and D(3, -5):

CD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

C(7, 1) = (x_1, y_1)

D(3, -5) = (x_2, y_2)

CD = \sqrt{(3 - 7)^2 + (-5 - 1)^2}

CD = \sqrt{(-4)^2 + (-6)^2}

CD = \sqrt{16 + 36} = \sqrt{52}

CD = 7.2 (nearest tenth)

Distance between D(3, -5) and A(-6, 2):

DA = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

D(3, -5) = (x_1, y_1)

A(-6, 2) = (x_2, y_2)

DA = \sqrt{(-6 - 3)^2 + (2 - (-5))^2}

DA = \sqrt{(-9)^2 + (7)^2}

DA = \sqrt{81 + 49} = \sqrt{130}

DA = 11.4 (nearest tenth)

Length of fencing required = 8.9 + 7.1 + 7.2 + 11.4 = 34.6 units

8 0
3 years ago
G(t) = 3t + 4; Find g(−3)
Bess [88]
G(-3)= 3(-3)+4

g(-3) = -5
7 0
3 years ago
A field test for a new exam was given to randomly selected seniors. The exams were graded, and the sample mean and sample standa
swat32
All our answers lie in the above statement.

Confidence Level:
The creator claims that 9 out 10 students will have the average score in the said range. Or in other words we can say that the creator is 90% confident about the result of the field test. So the confidence level is 90%. 

Margin of Error:
The average score lies within 4% of 70%. This means the margin of error is 4% i.e. the average scores can deviate from 70% by 4% .

Confidence Interval:
Lower Limit = 70% - 4% = 66%
Upper Limit = 70% + 4% = 74%

Interpretation:
The exam creator is 90% confident that the average scores of seniors will be between 66% and 74%. 
4 0
2 years ago
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