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2 years ago

What is 6/4 in its simplst form

Mathematics

2 answers:

Readme [11.4K]2 years ago

5 0

Answer:

Fraction form: $\frac{3}{2}$
Decimal form: 1.5
Mixed number form: 1 $\frac{1}{2}$

Step-by-step explanation:

Find the greatest common factor between the two numbers in the fraction, in this case the largest number both can be divided by equally is 2, so,

Take $\frac{6}{4}$ and divide both parts by 2:

6 ÷ 2 = 3

4 ÷ 2 = 2

Which gives you $\frac{3}{2}$

arlik [135]2 years ago

3 0

It would be 3/2 i believe

Read more on nested functions;

brainly.com/question/15855840

#SPJ1

3 0

2 years ago

Please help asap 20 pts

ddd [48]

B is the answer. If you solve for x, the values you get are 2.7077 and -0.422.

6 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 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

Andre received $60 for his birthday and put it in his piggy bank. Each week, he puts 3 more dollars in his bank. The amount of m

Alina [70]

$D=3\cdot42+60\\
D=126+60\\
D=186$

3 0

3 years ago

Read 2 more answers

A pizza has a diameter of 16 inches.

Cloud [144]

I got 50.24
since 2*3.14*8=50.24
but if you want it rounded to the nearest tenth I guess it would be 50.2
Hope this helps :)

6 0

3 years ago

Read 2 more answers