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3 years ago

ON THIS YOU HAVE TO GRAHP EACH NUMBER INTO A NUMBERLINE

Mathematics

2 answers:

Stolb23 [73]3 years ago

8 0

Answer: the answer will be plotted on the number line close or farther to zero

Step-by-step explanation: the answers has to be behind the zero

Nikolay [14]3 years ago

8 0

Answer:

you put 0 in the middle of each line then you put positives on the right side or top and negatives on the left side or bottom. :)

Step-by-step explanation:

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In 5 years, Dad will be three times as old as his daughter Jill will be then. If the sum of their present ages is 50, how old ar

grandymaker [24]

First, we know (D is dad, J is Jill) that D+J=50.
Second, in 5 years D will be 3 times as old so
(D+5)= 3*(J+5)
Moving on, if we know D+J=50 then D=(50-J) is true.Taking that relationship and substituting for D in the other equation, it becomes
50 - J + 5 = 3 * (J + 5)
Then, it becomes 50 - J + 5 = (3 * J) + (3 * 5)
Then, 55 - J = (3 * J) + 15
Then, 55 - 15 = (3 * J) + J
which becomes 40 = 4*J

6 0

3 years ago

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A car travels 30 1/5 miles in 2/3 of an hour. Whwt is the average speed, in miles per hour, of the car

LiRa [457]

Answer:

45 3/10 mph

Step-by-step explanation:

30 1/5 /2 = 15 1/10

15 1/10 x 3 = 45 3/10 mph

8 0

3 years ago

Solve the given inequality :

tigry1 [53]

Answer:

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

Step-by-step explanation:

Given rational inequality:

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

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

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

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

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

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) = undefined (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 roots and open dots for the restrictions (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

PLEASE PLEASE PLEASE HELP!!!!

lesya [120]

The answer is the fourth one

5 0

3 years ago

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How would you multiply fractions to whole numbers?

kvv77 [185]

You could just multiply the whole number with the numerator and then divide the answer by the denominator. Or you could write the whole number over 1, and then simplify the numbers before multiplying straight across.

2/3 x 36
2/3 x 36/1
2/1 x 12/1 (simplified the 3 and the 36)
24/1=24 (just multiplied across)

3 0

3 years ago

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