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Vlad1618 [11]
3 years ago
6

How many solutions does the following polynormial function in the number system (3x5-20x4-x3)3

Mathematics
1 answer:
icang [17]3 years ago
5 0
I think the answer is -60 i calcuate it :/
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The sum of measures of two supplementary angles is 90°
Vedmedyk [2.9K]

Answer:

Step-by-step explanation:

Never true Two angles are called supplementary when their measures add up to 180 degrees.

8 0
3 years ago
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Use the given graph. Determine the period of the function.
Ivahew [28]

Answer:

3

Step-by-step explanation:

Period of a function is the period after which the function attains the same value

in the graph attached with this problem we can see that

f(0)=1

the value of x for which function f(x) attains the value 1 again is at

x=3

f(3)=1

similarly , we see

f(6)=1 , f(9)=1

Hence we see that after every increased value of x by 3 units , we attain the same value of function . hence the period of the function is 3

6 0
3 years ago
Which linear equation has a slope of 3 and a y-intercept of -2?
natka813 [3]

Slope-intercept form:  y = mx + b

(m is the slope, b is the y-intercept or the y value when x = 0 --> (0, y) or the point where the line crosses through the y-axis)

You know:

m = 3

b = -2        So substitute/plug it into the equation

y = mx + b

y = 3x + (-2)

y = 3x - 2      Your answer is the 2nd option

7 0
3 years ago
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What is the solution of -8/2y-8=5/y+4-7y+8/y^2-16
Mnenie [13.5K]
<h2>Answer:</h2>

The solution of the given equation is:

                         y=6

<h2>Step-by-step explanation:</h2>

The expression is given by:

\dfrac{-8}{2y-8}=\dfrac{5}{y+4}-\dfrac{7y+8}{y^2-16}

Now on solving for the given equation

\dfrac{-8}{2(y-4)}=\dfrac{5}{y+4}-\dfrac{7y+8}{(y-4)(y+4)}

since,

a^2-b^2=(a-b)(a+b)\\\\so,\\\\y^2-16=y^2-4^2\\\\i.e.\\\\y^2-16=(y-4)(y+4)

Hence, we get:

\dfrac{-4}{y-4}=\dfrac{5\times (y-4)-(7y+8)}{(y+4)(y-4)}\\\\i.e.\\\\\dfrac{-4}{y-4}=\dfrac{5y-20-7y-8}{(y-4)(y+4)}

-4\times (y-4)(y+4)=(-2y-28)(y-4)\\\\i.e.\\\\(y-4)(-4\times (y+4))=(-2y-28)(y-4)\\\\i.e.\\\\(y-4)(-4y+16)=(-2y-28)(y-4)\\\\i.e.

(y-4)(-4y+16)-(-2y-28)(y-4)=0\\\\i.e.\\\\(y-4)(-4y-16+2y+28)=0\\\\i.e.\\\\(y-4)(-2y+12)=0\\\\i.e.

y-4=0\ or\ -2y+12=0\\\\i.e.\\\\y=4\ or\ 2y=12\\\\i.e.\\\\y=4\ or\ y=6

but y≠ 4

since, the denominator of the term in the left side of the given equality and the second term in the right side of the given equality will be zero and hence, the expression will be not defined.

Hence, the value of y is: 6

7 0
3 years ago
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Is the point on the line ?
Oxana [17]

Answer:

i think the answer is no i used tge desmo graphing calculator and tht point wasnt on it

3 0
3 years ago
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