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Alisiya [41]
3 years ago
13

Find the 64th term of the following arithmetic sequence. 6, 14, 22, 30,

Mathematics
1 answer:
olga55 [171]3 years ago
7 0

Answer:

<h2>510</h2><h2 />

Step-by-step explanation:

30-22=8

22-14=8

14-6=8

then

→ the common difference r = 8

→ the first term = 6

THerefore, (accoriding to the standard explicit formula)

the 64th term = term 1 + (64 - 1)×r

                       = 6 + 63×8

                       = 510

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Let f(x) = 9x3+ 21x2^ - 14 and g(x) = 3x + 1. Find "<br>f (x) over g (x)​
kondor19780726 [428]

Answer:

\frac{f(x)}{g(x)}=3x^2+6x-2-\frac{12}{3x+1}

Step-by-step explanation:

The first function is f(x)=9x^3+21x^2-14

The second function is g(x)=3x+1.

\frac{f(x)}{g(x)}=\frac{9x^3+21x^2-14}{3x+1}

We perform the long division as shown in the attachment to obtain the quotient as: Q(x)=3x^2+6x-2 and remainder R=-12.

Therefore:

\frac{f(x)}{g(x)}=3x^2+6x-2-\frac{12}{3x+1}

where x\ne -\frac{1}{3}

7 0
4 years ago
Write an inequality using absolute value if the answer is given double inequality:
gayaneshka [121]

Answer:

Answer: 67 ≤ x ≤ 73 Step-by-step explanation: Given inequality is: |x − 70| ≤ 3 We have to find the value of x. If |x| ≤ a then x ≤ a and x ≥ -a . Applying this rule to given inequality,we get x - 70 ≤ 3 and x - 70 ≥ -3 Adding 70 to both sides of above both inequality,we get x-70+70 ≤ 3+70 and x-70+70 ≥ -3+70 Adding like terms,we get x ≤ 73 and x ≥ 67 Combining above two inequalities ,we get 67 ≤ x ≤ 73 which is the answer.

Read more at Answer.Ya.Guru – https://answer.ya.guru/questions/4282353-what-are-the-solutions-to-the-absolute-value-inequality-x-70-3.html

8 0
3 years ago
What is the equation for each parabola in y=a(x-r)(x-s) form?<br><br>Please explain it step by step
ozzi
1. first parabola is up, so a are going to be positive
y = a(x-r)(x-s), 
roots -7,-3, so we can write y=a(x+7)(x+3)   [1]
we can take point (-5,-2), substitute into equation [1], and find value of a
-2=a(-5+7)(-5+3)
-2=a*2*(-2)
a=1/2
so first parabola has an equation 
y=1/2(x+7)(x+3) 

2. second parabola
y=a(x-r)(x-s)      roots are 2 and 4 so we ca write
y=a(x-2)(x-4) take point (3,4) and substitute into this equation
4 =a(3-2)(3-4)
4=-a
a=-4, and it will look down
y=-4(x-2)(x-4)

6 0
3 years ago
The net of a solid figure is shown below:
olganol [36]
Each square is 3 by 3, so the area is 3*3 = 9. We can leave it as 3*3 since that's what the answer choices show.

There are 6 of these squares, so the total area of this figure is 6*3*3
The total area of the net is the same as the total surface area when we fold along the lines to form a 3D cube

Final Answer: 6 x 3 x 3 square inches which is choice D
4 0
3 years ago
Suppose that you have three consecutive positive integers where the product of the smaller and larger number is equal to 15 more
EleoNora [17]

Answer:

The numbers are either {-3, -2, -1} or {7, 8, 9}.

Step-by-step explanation:

Let the smallest integer be represented by x. The middle number is x+1, and the largest number is x+2.

The product of the smaller and larger number is (x)(x+2).

15 more than 6 times the middle number is 6(x+1)+15.

(x)(x+2)=6(x+1)+15

x^2+2x=6x+6+15

x^2+2x=6x+21

x^2-4x-21=0

(x-7)(x+3)=0

x=-3 or x=7

The numbers are either {-3, -2, -1} or {7, 8, 9}.

3 0
3 years ago
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