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3 years ago

13

PLS HELP ASAP ill give brainliest!! At a conference for 500 people, 30% of the participants are French, 15% are Americans, 5% ar

e Germans, and are of other nationalities. How many French participants are at the conference?
30
150
160
250

2 answers:

Arada [10]3 years ago

6 0

Answer:

150

Step-by-step explanation:

30/100 times 500/1 = 150

Shalnov [3]3 years ago

4 0

Your answer will be 150

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Determine whether the triangles are similar. or not.

mr Goodwill [35]

Answer:

A

Step-by-step explanation:

We know that ∠ $ABE$ ≅ ∠ $DBC$ because they are vertical angles, so that's one pair of congruent angles. However, we can't compare ∠ $BAE$ and ∠ $DCB$ as they have different measures, so we'll have to find the measure of either ∠ $AEB$ or ∠ $BDC$ . Let's go with the former. Using the fact that the sum of the measures of the interior angles of a triangle is $180$ °, we know that $m$ ∠ $AEB=180-(53+42)=180-95=85$ °. Now, we know that ∠ $AEB$ ≅ ∠ $DCB$ because their measures are equal. That's another pair of congruent angles. Therefore, the angles are similar by $AA$ . Hope this helps!

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2 years ago

Jon makes a map of his neighborhood for a presentation. The scale of his map is 1inch:250 feet, How many feet do 6.5 inches repr

anygoal [31]

B 1625

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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

Please solve.<br> -7 = z over 2 + 1

mafiozo [28]

-7=z/2+1

What you first need to do is to multiply both sides of the equation by 2

-14=z+2

Now you move the variable to the left side and change within it’s own sign

-z-14=2

Into

-z=2+14

Now you then add the numbers

-z=16

The last part you do is change the sign on both sides in this equation

z=-16

Therefore, z=-16 is your answer in number 16.

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2 years ago

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Lerok [7]

-24n - 30, i suggest using math.way before asking a math question aha

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2 years ago

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