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GenaCL600 [577]
2 years ago
13

Write an equation for the nth term of the arithmetic sequence. Then find a40.

Mathematics
1 answer:
Olegator [25]2 years ago
8 0

Answer:

The nth term is: a_n = \frac{1}{4} + \frac{1}{4}(n-1)

a40 = 10

Step-by-step explanation:

Arithmetic sequence:

In an arithmetic sequence, the difference between consecutive terms is always the same, and this difference is called common difference.

The nth term of a sequence is given by:

a_n = a_1 + (n-1)d

In which a_1 is the first term and d is the common difference.

1/4,1/2

This means that:

d = \frac{1}{2} - \frac{1}{4} = \frac{2}{4} - \frac{1}{4} = \frac{1}{4}

1/4

This means that a_1 = \frac{1}{4}

The nth term is:

a_n = a_1 + (n-1)d

a_n = \frac{1}{4} + \frac{1}{4}(n-1)

Then find a40.

a_{40} = \frac{1}{4} + \frac{1}{4}(40-1) = \frac{1}{4} + \frac{39}{4} = \frac{40}{4} = 10

So

a40 = 10

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In the kite , AK=9 , JK=15 , and AM=16 .
Veronika [31]

Answer:

The Length of JM is 20.

Step-by-step explanation:

Given,

JKLM is a kite in which JL and KM are the diagonals that intersect at point A.

Length of AK = 9    

Length of JK = 15  

Length of AM = 16

Solution,

Since JKLM is a kite. And JL and KM are the diagonals.

And we know that the diagonals of a kite perpendicularly bisects each other.

So, JL ⊥ KM.

Therefore ΔJAK is aright angled triangle.

Now according to Pythagoras Theorem which states that;

"The square of the hypotenuse is equal to the sum of the square of base and square of perpendicular".

JK^2=KA^2+AJ^2

On putting the values, we get;

(15)^2=9^2+AJ^2\\\\225=81+AJ^2\\\\AJ^2=225-81=144

On taking square root onboth side, we get;

\sqrt{AJ^2} =\sqrt{144}\\\\AJ=12

Again By Pythagoras Theorem,

AJ^2+AM^2=JM^2

On putting the values, we get;

JM^2=(12)^2+(16)^2\\\\JM^2=144+256=400

On taking square root onboth side, we get;

\sqrt {JM^2}=\sqrt{400}\\\\JM=20

Hence The Length of JM is 20.

4 0
3 years ago
Problem: a circle is inscribed in a square. The squares perimeter is 16 times the square root of 6. What is the probability of s
Oksi-84 [34.3K]
I got 21% at selecting a point outside.
Explanation: Since the perimeter is 16sqrt (6), each side must be 4sqrt (6)
This also means that the diameter is 4sqrt (6).

To find the area of a circle, we need to find the radius (2sqrt (6)).
Area of circle is pi × (2sqrt (6))^2 = 24pi or approximately 75.4 units^2
Area of square is 4sqrt (6)^2 = 96

Thus, let's take the area of the circle and subtract that from the area of our square to yield approx. 20.6 units^2
and now divide through by 96 to yield 21%
8 0
3 years ago
-9q = 63 what would be q
Snezhnost [94]

Answer:

-7

Step-by-step explanation:

-9q = 63

q= 63/-9 = -7

7 0
3 years ago
What equation represents the line that passes through the points (-3 7 and (3 3?
dexar [7]


<span>Hello : let  A(-3,7)    B(3,3)
the slope is :   (YB - YA)/(XB -XA)
(3-7)/(3+3)  = -2/3
an equation is : y=ax+b     a is a slope</span>

y = (-2/3)x +b

 

the line through point B (3,3) :  3 = (-2/3)(3)+b<span>
<span>b =5 
the equation is : y =(-2/3)x+5</span></span>

3 0
3 years ago
The domain of a quadratic function is all real numbers and the range is y s 2. How many x-intercepts does the
mel-nik [20]

Answer:

There are two x-intercepts.

A quadratic function whose maximum degree two. Therefore, Total two x-intercept possible.

We are given the range of quadratic function y less than equal to 2. It means parabola is downward whose maximum value 2.

Here y is greater than 2 and parabola form downward. Here must be two x-intercept.

If a>0 then form open up parabola.

If a<0 then open down parabola.

Here open down parabola. So, we get total two x-intercept.

Please see the attachment for diagram.

5 0
3 years ago
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