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Elis [28]
3 years ago
6

Composite numbers are

Mathematics
2 answers:
oee [108]3 years ago
6 0
Composite numbers are numbers that are not prime.
Murljashka [212]3 years ago
6 0
A composite number is a positive integer that can be formed by multiplying two smaller positive integers. “
Equivalently, it is a positive integer that has at least one divisor other than 1 and itself” -wikipedia
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What is the equation of the following line written in general form?
Arturiano [62]

The line drops 4 units between the points (1, 6) and (6, 2) as it goes over 5 units. Thus the point-slope form of the equation can be written as

... y - k = m(x - h) . . . . . . line with slope m through point (h, k)

... y - 6 = (-4/5)(x - 1)

Multiplying by 5 and subtracting the right side gives ...

... 5y - 30 = -4x +4

... 4x + 5y - 34 = 0 . . . . . equation in general form

6 0
3 years ago
A segment is on a number line with endpoints at −5.3 and 8.7.<br> What is the length of the segment?
Kisachek [45]

Answer:

about 6.3✔

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Solve the given inequality :
tigry1 [53]

Answer:

x < -5  or  x = 1  or  2 < x < 3  or  x > 3

Step-by-step explanation:

Given <u>rational inequality</u>:

\dfrac{(x-1)^2(x-2)^3}{(x^2-5x+6)^2(x+5)}\geq 0

\textsf{Factor }(x^2-5x+6):

\implies x^2-2x-3x+6

\implies x(x-2)-3(x-2)

\implies (x-3)(x-2)

Therefore:

\dfrac{(x-1)^2(x-2)^3}{(x-3)^2(x-2)^2(x+5)}\geq 0

Find the roots by solving f(x) = 0  (set the numerator to zero):

\implies (x-1)^2(x-2)^3=0

\implies (x-1)^2=0\implies x=1

\implies (x-2)^3=0 \implies x=2

Find the restrictions by solving f(x) = <em>undefined  </em>(set the denominator to zero):

\implies (x-3)^2(x-2)^2(x+5)=0

\implies (x-3)^2=0 \implies x=3

\implies (x-2)^2=0 \implies x=2

\implies (x+5)=0 \implies x=-5

Create a sign chart, using closed dots for the <u>roots</u> and open dots for the <u>restrictions</u> (see attached).

Choose a test value for each region, including one to the left of all the critical values and one to the right of all the critical values.

Test values:  -6, 0, 1.5, 2.5, 4

For each test value, determine if the function is positive or negative:

f(-6)=\dfrac{(-6-1)^2(-6-2)^3}{(-6-3)^2(-6-2)^2(-6+5)}=\dfrac{(+)(-)}{(+)(+)(-)}=+

f(0)=\dfrac{(0-1)^2(0-2)^3}{(0-3)^2(0-2)^2(0+5)}=\dfrac{(+)(-)}{(+)(+)(+)}=-

f(1.5)=\dfrac{(1.5-1)^2(1.5-2)^3}{(1.5-3)^2(1.5-2)^2(1.5+5)}=\dfrac{(+)(-)}{(+)(+)(+)}=-

f(2.5)=\dfrac{(2.5-1)^2(2.5-2)^3}{(2.5-3)^2(2.5-2)^2(2.5+5)}=\dfrac{(+)(+)}{(+)(+)(+)}=+

f(4)=\dfrac{(4-1)^2(4-2)^3}{(4-3)^2(4-2)^2(4+5)}=\dfrac{(+)(+)}{(+)(+)(+)}=+

Record the results on the sign chart for each region (see attached).

As we need to find the values for which f(x) ≥ 0, shade the appropriate regions (zero or positive) on the sign chart (see attached).

Therefore, the solution set is:

x < -5  or  x = 1  or  2 < x < 3  or  x > 3

As interval notation:

(- \infty,-5) \cup x=1 \cup (2,3) \cup(3,\infty)

4 0
2 years ago
Which equation represents a linear function that has a slope of 4/5 and a y-intercept of –6?
erastovalidia [21]
Use the form y=mx+b where m is the slope and b is the y-intercept: y=4/5x-6
5 0
4 years ago
Need help on finding the equation for both graphs
Assoli18 [71]

for #1 it is y=-3/2x+4 the -3/2 is the slope and the 4 is the y intercept

for #2 it is y=6/+2 the 6/2 is the slope and the 2 is the y-intercept

the y-intercept is where the line passes through the y axis

5 0
3 years ago
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