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3 years ago

What Is The Quotient And Remainder Of 33÷9

Mathematics

1 answer:

Oduvanchick [21]3 years ago

4 0

Answer: 3.6 (repeating)

Step-by-step explanation:

9 only goes into 33 , 3 times considering 9 times 3 equals 27 and your leftovers or remainder is an irrational number which mean it repeats

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The sum of two even numbers is even. the sum of 6 and another number is even. what conjecture can you make about the other numbe

Basile [38]

We can say that the other number is even or it is 0.

6 0

2 years ago

Help Please What’s The Answer!!

JulsSmile [24]

Answer:

c) (6,0)

Step-by-step explanation:

The formula to find a slope is y=mx-b where m is the slope (to find the slope you find the change in y and the change in x) which in this case is -3. The change in y is -3 and the change in x is -1. If you put it on a graph the line is going down from left to right and it intersects the y-axis at 18 meaning if you keep sloping down in a straigt line it intersects the x-axis at (6,0).

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3 years ago

How many real-number solutions does the equation have?

Alexxx [7]

I believe the correct answer would be that it has one solution which is 5. Calculating x for the quadratic equation, you will only get one value so it should be one solution. Hope this answers the question. Have a nice day.

8 0

2 years ago

Read 2 more answers

Select the graph that shows the solution set for the following system.<br> x + y < 2<br> x>2

kompoz [17]

Answer:

Answer is in the attachment.

Step-by-step explanation:

To graph x>2 consider first x=2. x=2 is a vertical line and if you want to graph x>2 you need to shade to the right of the vertical line.

To graph x+y<2, I will solve for y first.

x+y<2

Subtract x on both sides:

y<-x+2

Consider the equation y=-x+2. This is an equation with y-intercept 2 and slope -1 or -1/1. So the line you have in that picture looks good for y=-x+2. Now going back to consider y<-x+2 means we want to shade below the line because we had y<.

Now where you see both shadings will be intersection of the shadings and will actually by your answer to system of inequalities you have. In my picture it is where you have both blue and pink.

I have a graph in the picture that shows the solution.

Also both of your lines will be solid because your question in the picture shows they both have equal signs along with those inequality signs.

Just in case my one graph was confusing, I put a second attachment with just the solution to the system.

5 0

2 years ago

1) Determine the discriminant of the 2nd degree equation below:

Aleksandr-060686 [28]

$\LARGE{ \boxed{ \mathbb{ \color{purple}{SOLUTION:}}}}$

We have, Discriminant formula for finding roots:

$\large{ \boxed{ \rm{x = \frac{ - b \pm \: \sqrt{ {b}^{2} - 4ac} }{2a} }}}$

Here,

x is the root of the equation.
a is the coefficient of x^2
b is the coefficient of x
c is the constant term

1) Given,

3x^2 - 2x - 1

Finding the discriminant,

➝ D = b^2 - 4ac

➝ D = (-2)^2 - 4 × 3 × (-1)

➝ D = 4 - (-12)

➝ D = 4 + 12

➝ D = 16

2) Solving by using Bhaskar formula,

❒ p(x) = x^2 + 5x + 6 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 5\pm \sqrt{( - 5) {}^{2} - 4 \times 1 \times 6 }} {2 \times 1}}}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 5 \pm \sqrt{25 - 24} }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 5 \pm 1}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = - 2 \: or - 3}}}$

❒ p(x) = x^2 + 2x + 1 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 2 \pm \sqrt{ {2}^{2} - 4 \times 1 \times 1} }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 2 \pm \sqrt{4 - 4} }{2} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 2 \pm 0}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = - 1 \: or \: - 1}}}$

❒ p(x) = x^2 - x - 20 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - ( - 1) \pm \sqrt{( - 1) {}^{2} - 4 \times 1 \times ( - 20) } }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ 1 \pm \sqrt{1 + 80} }{2} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{1 \pm 9}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = 5 \: or \: - 4}}}$

❒ p(x) = x^2 - 3x - 4 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - ( - 3) \pm \sqrt{( - 3) {}^{2} - 4 \times 1 \times ( - 4) } }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{3 \pm \sqrt{9 + 16} }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{3 \pm 5}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = 4 \: or \: - 1}}}$

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5 0

2 years ago

Read 2 more answers