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2 years ago

How do you graph 3 and 4??? I know you have to find the coordinates but how?? Please explain

Mathematics

1 answer:

Umnica [9.8K]2 years ago

8 0

So 3. put it on x. but the negative numbers . then 4 put. 5 on y

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53,806 in expanded form

mojhsa [17]

50,000 + 3,000 + 800 + 6

5 0

3 years ago

Help please, need to show work.

stiks02 [169]

9514 1404 393

Answer:

3. m∠A < 80°

Step-by-step explanation:

The exterior angle NGS, marked as 80°, has a measure that is the sum of the measures of the remote interior angles: GAN, GNA.

If we assume that angle measures cannot be zero, then we must have ...

m∠A +m∠N = 80°

m∠A = 80° -m∠N

Since m∠N > 0, the measure of angle A must be less than 80°.

This conclusion matches choice 3.

8 0

2 years ago

Solve 6^x+2 = 10 for x using the change of base formula

pashok25 [27]

Answer:

$x\approx-0.7149$

Step-by-step explanation:

$\displaystyle 6^{x+2}=10\\\\\log_6(6^{x+2})=\log_6(10)\\\\x+2=\frac{\log(10)}{\log(6)}\\\\x+2\approx1.2851\\\\x\approx-0.7149$

Recall that the change of base formula is $\displaystyle \log_a(b)=\frac{\log(b)}{\log(a)}$

3 0

1 year ago

Solve the equation<br> 5x = 20<br> X=. Pls helps fast test due soon! lol

hodyreva [135]

Answer:

x=4

Step-by-step explanation:

8 0

2 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