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4 years ago

Can someone help me? photo attached

Mathematics

1 answer:

kykrilka [37]4 years ago

7 0

I don't think you can sorry

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Apply the laws of exponents, calculate the result and express the result in scientific notation, and as a decimal:

DochEvi [55]

We are given

$\frac{(4.1\times 10^{3})(2.8\times 10^{-7})}{(3.1\times 10^{-5})}$

We can move like terms altogether

$\frac{(4.1\times 10^{3})(2.8\times 10^{-7})}{(3.1\times 10^{-5})}=\frac{4.1\times 2.8}{3.1}\times \frac{10^{3}\times 10^{-7}}{10^{-5}}$

now, we can use property of exponent

$a^m\times a^n=a^{m+n}$

$=\frac{4.1\times 2.8}{3.1}\times \frac{10^{3-7}}{10^{-5}}$

We can use another exponent property

$\frac{a^m}{a^n} =a^{m-n}$

so, we get

$=\frac{4.1\times 2.8}{3.1}\times 10^{3-7+5}$

$=\frac{4.1\times 2.8}{3.1}\times 10^{1}$

$=3.70323\times 10^{1}$

So, the result in scientific notation is

$=3.70323\times 10^{1}$ ..........Answer

So, the result as a decimal is

$=37.0323$ .............Answer

8 0

4 years ago

What is the value of the expression!<br><br> Help

Sindrei [870]

The answer is 256 :)

3 0

4 years ago

Px^2-qx+r=0 solve by complete squaring method

Natasha2012 [34]

Step-by-step explanation:

px^2-qx+r

p(x^2-q/px)+r

p[(x^2-q/px +(q/2p)^2_(q/2p)^2)]+r

p[(x-q/2p)^2 -q^2/4p^2)]+r

p(x-q/2p)^2 -q^/4p+r

p(x-q/2p)^2-(q^2-4pr)/4p

3 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

Rounding to 1 significant figure, estimate the

Maslowich

Answer:

a) 7000

b) 20000

c) 300000

3 0

2 years ago

Read 2 more answers