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4 years ago

Find the missing side. Round to the nearest tenth.

Mathematics

1 answer:

ziro4ka [17]4 years ago

5 0

Answer:

$x=7.2\ units$

Step-by-step explanation:

we know that

In the right triangle of the figure

The tangent of angle of 67 degrees is equal to divide the opposite side to the angle of 67 degrees (17 units) by the adjacent side to angle of 67 degrees (x units)

$tan(67\°)=17/x$

Solve for x

$x=17/tan(67\°)$

$x=7.2\ units$

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Naddik [55]

Answer:

358

Step-by-step explanation:

In 1995, the boys were 50 %

Take the number of people at the convention in 1995 ( 716)

and multiply by 50%

716 * 50%

716 * .50

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6 0

3 years ago

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Subtract.<br><br> Eight-ninths minus one-third

Bingel [31]

8/9 - 1/3

To subtract these two fractions, first find a common denominator.

The common denominator will be the

least common multiple for the two denominators.

In this case, the least common multiple of 9 and 3 is 9.

So to subtract, a denominator of 9 is needed in each fraction.

8/9 already has 9 in the denominator so it stays the same.

To get a 9 in the denominator of 1/3, multiply top and bottom by 3.

This gives us 3/9.

So we have 8/9 - 3/9 which is 5/9.

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3 years ago

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Find the area of the given figure.

marissa [1.9K]

84.6 km2 is the answer

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3 years ago

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How many solutions does the following equation have: <br> 16y + 7 = -3y + 10 + 8y

joja [24]

Answer:

one solution only

Step-by-step explanation:

16y + 7 = -3y + 10 + 8y

16y + 3y - 8y = 10 - 7

11y = 3

y = 3/11

7 0

3 years ago

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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