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3 years ago

Wat 5x4 ???????? Help

Mathematics

2 answers:

antiseptic1488 [7]3 years ago

8 0

Please let this be a joke but it's 20

tino4ka555 [31]3 years ago

3 0

Ok 5×4 is 20... uhh hope this helps?

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PLEASE HELP ASAP 30 POINTS

Yuki888 [10]

Answer:

I don't know how to do please let me I will try solve the question

8 0

3 years ago

5+ 3x - 37 = 6x + 23 - 8x what is the value for x

iris [78.8K]

Answer:

The value of X is 11

X=11

Step-by-step explanation:

Isolate the variable by dividing each side by the factors that don't contain the cariable.

4 0

3 years ago

A circle has the order pairs (-1, 2) (0, 1) (-2, -1) what is the equation . Show your work.

olga55 [171]

We know that:

$(x-a)^2+(y-b)^2=r^2$

is an equation of a circle.

When we substitute x and y (from the pairs we have), we'll get a system of equations:

$\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}$

and all we have to do is solve it for a, b and r.

There will be:

$\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}\\\\\\
\begin{cases}1+2a+a^2+4-4b+b^2=r^2\\a^2+1-2b+b^2=r^2\\4+4a+a^2+1+2b+b^2=r^2\end{cases}\\\\\\
\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\\\\\
$

From equations (II) and (III) we have:

$\begin{cases}a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\--------------(-)\\\\a^2+b^2-2b+1-a^2-b^2-4a-2b-5=r^2-r^2\\\\-4a-4b-4=0\qquad|:(-4)\\\\\boxed{-a-b-1=0}$

and from (I) and (II):

$\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\end{cases}\\--------------(-)\\\\a^2+b^2+2a-4b+5-a^2-b^2+2b-1=r^2-r^2\\\\2a-2b+4=0\qquad|:2\\\\\boxed{a-b+2=0}$

Now we can easly calculate a and b:

$\begin{cases}-a-b-1=0\\a-b+2=0\end{cases}\\--------(+)\\\\-a-b-1+a-b+2=0+0\\\\-2b+1=0\\\\-2b=-1\qquad|:(-2)\\\\\boxed{b=\frac{1}{2}}\\\\\\\\a-b+2=0\\\\\\a-\dfrac{1}{2}+2=0\\\\\\a+\dfrac{3}{2}=0\\\\\\\boxed{a=-\frac{3}{2}}$

Finally we calculate $r^2$ :

$a^2+b^2-2b+1=r^2\\\\\\\left(-\dfrac{3}{2}\right)^2+\left(\dfrac{1}{2}\right)^2-2\cdot\dfrac{1}{2}+1=r^2\\\\\\\dfrac{9}{4}+\dfrac{1}{4}-1+1=r^2\\\\\\\dfrac{10}{4}=r^2\\\\\\\boxed{r^2=\frac{5}{2}}$

And the equation of the circle is:

$(x-a)^2+(y-b)^2=r^2\\\\\\\left(x-\left(-\dfrac{3}{2}\right)\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}\\\\\\\boxed{\left(x+\dfrac{3}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}}$

7 0

3 years ago

Worth 20 points

bezimeni [28]

Answer:

1. 1/8 ÷ 3/4 = 1/6

2. 3/5 ÷ 3/2 = 2/5

3. 4/1 ÷ 2/3 = 6

4.9/4 ÷ 6/5 = 15/8

5.22/4 ÷ 2/5 = 261/80

Step-by-step explanation:

Hope this Helped

5 0

3 years ago

Which of the sets of ordered pairs represents a function?

IgorC [24]

Answer:

Neither set A nor set B represents a function.

Step-by-step explanation:

According to the definition of a function, for each input there exists a unique output. In other word for each value of x there exists unique value of y.

In set A we have four ordered pair (-4,5),(1,-1),(2,-2) and (2,3).

It means for x=2, we have two values of y, i.e., -2 and 3.

Since for a single value of x there exist more than one value of y, therefore set A is not a function.

In set B we have four ordered pair (2,2),(3,-2),(9,3) and (9,-3).

It means for x=9, we have two values of y, i.e., -3 and 3.

Since for a single value of x there exist more than one value of y, therefore set B is not a function.

3 0

4 years ago

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