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4 years ago

How to write 4(x-8) in verbal expression

Mathematics

1 answer:

Trava [24]4 years ago

7 0

4 times x minus 8 is the answer

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(6x^2-5x+3)+(3x^2+7x-8) add

Mrac [35]

Answer:

$9 {x}^{2} + 2x - 5$

Step-by-step explanation:

I have answered ur question

8 0

3 years ago

The perimeter of a triangle is 65 meters. The second side is 5 meters less than the first side The third side is 7

tangare [24]

Answer:

The length of each side is 31.5m, 26.5m, 7m

Step-by-step explanation:

Let the length of the first side of the triangle be x meters.

Then, the second side is x-5 meters.

The third side is given as 7 meters.

The perimeter is 65 meters

This gives us the equation:

$65 = x + x - 5 + 7$

$65 - 7 + 5 = x + x$

$63 = 2x$

$x = \frac{63}{2} = 31.5$

The length of each side is 31.5m, 26.5m, 7m

4 0

4 years ago

Write the number 2 and a half million in figures

inysia [295]

The answer is 2 500 000

6 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

Prove that if a is a natural number, then there exist two unequal natural numbers k and l for which ak−al is divisible by 10.

EastWind [94]

Assume a is not divisible by 10. (otherwise the problem is trivial).
Define R(m) to be the remainder of a^m when divided by 10. 
R can take on one of 9 possible values, namely, 1,2,...,9. 
Now, consider R(1),R(2),......R(10). At least 2 of them must have the sames value (by the Pigeonhole Principle), say R(i) = R(j) ( j>i ) 
Then, a^j - a^i is divisible by 10.

8 0

3 years ago

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