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3 years ago

Draw 6 circles. Make 3 groups. Shade 1 group.

Mathematics

2 answers:

krek1111 [17]3 years ago

7 0

Answer:

ummm I think this is obvious

Step-by-step explanation:

2 circles in each group, shade in one group of 2.

Mrrafil [7]3 years ago

3 0

Answer:

you just gotta draw the shapes and put them together

Step-by-step explanation:

You might be interested in

How to add these mixed numbers

krek1111 [17]

Make them into improper fractions
(5*5)+3
(2*3)+2

Final fractions
28/5+8/3
Find a common denominator
Multiply 28/5 by 3
84/15
Multiply 8/3 by 5
40/15

Add - 84+40=124

8 0

3 years ago

Read 2 more answers

Whats the error?katie divided 4.25 by 0.25 and a got a quotient of 0.17.

m_a_m_a [10]

The answer is 17. The decimal point needed to be moved by 2 places to the right since the divisor is a decimal.

6 0

4 years ago

What are the ordered pairs of the

9966 [12]

Start by writing the system down, I will use $y$ to represent $f(x)$

$y=x^2-5x+3\wedge y=-3$

Substitute the fact that $y=-3$ into the first equation to get,

$-3=x^2-5x+3$

Simplify into a quadratic form ( $ax^2+bx+c=0$ ),

$x^2-5x+6=0$

Now you can use Vieta's rule which states that any quadratic equation can be written in the following form,

$x^2+x(m+n)+mn=0$

which then must factor into

$(x+m)(x+n)=0$

And the solutions will be $m,n$ .

Clearly for small coefficients like ours $5,6$ , this is very easy to figure out. To get 5 and 6 we simply say that $m=3, n=2$ .

This fits the definition as $5=3+2$ and $6=2\cdot3$ .

So as mentioned, solutions will equal to $m=3, n=2$ but these are just x-values in the solution pairs of a form $(x,y)$ .

To get y-values we must substitute 3 for x in the original equation and then also 2 for x in the original equation. Luckily we already know that substituting either of the two numbers yields a zero.

So the solution pairs are $(2,0)$ and $(3,0)$ .