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mel-nik [20]
3 years ago
6

The following integers are listed in order from greatest to least. 10, 7, -3, -7

Mathematics
1 answer:
alexgriva [62]3 years ago
4 0

Answer:

true

Step-by-step explanation:

negative numbers are smaller than positives and bigger negative numbers like -7 are lower than smaller negative numbers like -3

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2.10 grater than 2.09<br> true or false
shepuryov [24]

Answer:

2.10 is greater than 2.09

Step-by-step explanation:

because 2.10 has more equation than 2.09

5 0
3 years ago
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The speed you are going determines the distance you travel before coming to a complete stop after pressing the brake.
notka56 [123]

Answer: A and D

Step-by-step explanation:

4 0
3 years ago
What angle do the minute and hour hands of a clock make at 8:00
svlad2 [7]
The angle that the minute and hour hands make when they are at 8:00 is obtuse
6 0
3 years ago
What is the missing factor in the equation? Assume x0 and y0.
denpristay [2]

Answer:

y^3/4x

Step-by-step explanation:

Given that the expression 6x^2/5y multiplied by some unknown gives answer as 3xy^2/10

We know that if ab = c, the b = c/a

Using the same principle, we have

unknown value = 3xy^2/10 divided by 6x^2/y

Use the fact that a/b divided by c/d = ad/bc

We get unknown = 5y(3xy^2)/6x^2 (10)

=15xy^3/60x^2

Simplify the constant part 15/60 as 1/4

xy^3 /x^2y

X term becomes x/x^2 = 1/x

so multiplying term =y^3/4x

5 0
3 years ago
How do you solve x+4=10^x
raketka [301]
Two ways to solve the problem:

A. By graphing f(x)=10^x and g(x)=x+4.
The intersection(s) will be the solution.
See graph below, approximate solutions: (-4,0),(0.66, 4.67)

B. Refine approximate solutions using Newton's method
Let h(x)=f(x)-g(x) = 10^x-x-4 ...............(1)
we calculate the derivative, h'(x) = log(10)*10^x-1   [ note:log(x) means ln(x) ]
and use Newton's iterative formula to find successive approximations to the root, basically refining the approximate solutions.
The iterative formula for nth approximation x_n is given by
x_n = x_{n-1} - h(x_n) / h'(x_n)....(2)

Using initial approximation (-4,0), we have x0=-4
x1
=x0-h(x0)/h'(x0)
=-4 - h(-4)/h'(-4)
=-4 - (10^(-4)-(-4)-4)/(log(10*10^(-4)-1)
=-4 - (1/10000)/(log(10)/10000-1)
=-3.9999000

Repeating the same for second approximation, x2, we get
x2=-3.99989997696619 which is accurate to 14 places after decimal

Now we can refine the other approximate solution, x0=0.66 to find
x1=0.669356
x2=0.6692468481102326
x3=0.669246832877748
x4=0.6692468328777476
So we will accept x=0.669246832877748

So the solutions are S={-3.99989997696619,0.6692468328777476}  both accurate to 14 places after the decimal

3 0
3 years ago
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