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zaharov [31]
3 years ago
13

Resolve to me these exercises please?

Mathematics
1 answer:
Dima020 [189]3 years ago
7 0
A)4^(n+3)=8^14
2^(2×(n+3))=2^(3×14)
2^(2n+6)=2^42
2^2n=2^36
n=18

b) (assuming a : is divide)
3^(2n+1)=9^17/3^3
3^(2n+1)=3^(2×16)/3^3
3^(2n+1)=3^29
3^2n=3^28
2n=28
n=14

d) (6^n)^4×36=216^10
6^4n×6^2=6^(3×10)
6^(4n+2)=6^30
6^4n=6^28
4n=28
n=7

e)7^(n^2)÷7=49^24
7^(n^2-1)=7^(2×24)
7^(n^2)=7^49
n^2=49
n=7

g)15^(n+4)÷5^(n+4)=81^6
3^(n+4)×5^(n+4)÷5^(n+4)=3^(4×6)
3^(n+4)=3^24
n=20

h)81^n÷9^n+9^(n+2)÷9=90÷9^6
9^2n÷9^n+9^(n+2)÷9=9*10/9^6
9^n+9^(n+1)=10/9^5
I don't know where to go from here

I)what?
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2 to the power of 14
mrs_skeptik [129]
2 to the power of 14 means 2^14.
That would be 16384.
6 0
3 years ago
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Consider a population list x, with a CV= 10%. A second population list, y, has a CV= 20%. The linear regression of y on x is y =
ch4aika [34]

Answer:

The correlation coefficient 0.5 depicts that there is moderate positive correlation between y and x.

Step-by-step explanation:

The linear regression equation is

y= a+bx

The given linear equation is

y= 10x

Here, slope=b=10 and intercept=a=0.

From above equation,

ybar=10xbar

We are given that CVx=0.1 and CVy=0.2

where CV=(standard deviation/mean)*100

We know that

b=r(Sy/Sx)

Multiplying by xbar/ybar on both sides

(xbar/ybar)b=r(Sy/Sx)(xbar/ybar)

(xbar/ybar)b=r[(Sy/ybar)/(Sx/xbar))]

(xbar/ybar)b=r[CVy/CVx]

As CVy/CVx=(Sy/ybar)*100/(Sx/xbar)*100=(Sy/ybar)/(Sx/xbar).

By putting ybar=10xbar, b=10, CVx=0.1 and CVy=0.2.

(xbar/10xbar)10=r(0.2/0.1)

1=r(0.2/0.1)

1=r(2)

r=0.5

There is moderate positive correlation between y and x.

6 0
2 years ago
Given the arithmetic sequence an = 4 − 3(n − 1), what is the domain for n?
Oxana [17]

Answer:

The answer is All integers where n ≥ 1

Hope this helps!


8 0
3 years ago
Given the vertex of a quadratic function, find the axis of symmetry.
motikmotik

(i) The equation of the axis of symmetry is x = - 5.

(ii) The coordinates of the vertex of the parabola are (h, k) = (4, - 18). The x-value of the vertex is 4.

(iii) According to the <em>vertex</em> form of the <em>quadratic</em> equation, the parabola opens down due to <em>negative</em> lead coefficient and has a vertex at (2, 4), which is a <em>maximum</em>.  

<h3>How to analyze and interpret quadratic functions</h3>

In this question we must find and infer characteristics from three cases of <em>quadratic</em> equations. (i) In this case we must find a formula of a axis of symmetry based on information about the vertex of the parabola. Such axis passes through the vertex. Hence, the equation of the axis of symmetry is x = - 5.

(ii) We need to transform the <em>quadratic</em> equation into its <em>vertex</em> form to determine the coordinates of the vertex by algebraic handling:

y = x² - 8 · x - 2

y + 18 = x² - 8 · x + 16

y + 18 = (x - 4)²

In a nutshell, the coordinates of the vertex of the parabola are (h, k) = (4, - 18). The x-value of the vertex is 4.

(iii) Now here we must apply a procedure similar to what was in used in part (ii):

y = - 2 · (x² - 4 · x + 2)

y - 4 = - 2 · (x² - 4 · x + 2) - 4

y - 4 = - 2 · (x² - 4 · x + 4)

y - 4 = - 2 · (x - 2)²

According to the <em>vertex</em> form of the <em>quadratic</em> equation, the parabola opens down due to <em>negative</em> lead coefficient and has a vertex at (2, 4), which is a <em>maximum</em>.  

To learn more on quadratic equations: brainly.com/question/1863222

#SPJ1

5 0
1 year ago
Can you choose a starting point so that the first 5 numbers in your sequence are all positive? Explain your reasoning.
marin [14]

Answer:

Take the starting point as the first natural number that is 1.

Step-by-step explanation:

To find: a starting point so that the first 5 numbers in the sequence are all positive

Solution:

Natural numbers are numbers that are used for counting.

Natural numbers are basically positive integers.

Take the starting point as the first natural number that is 1.

The next point as 2.

The next point as 3.

The next point as 4.

The next point as 5.

So, the sequence becomes 1,2,3,4,5,... so that the first 5 numbers in the sequence are positive.

4 0
2 years ago
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