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Tems11 [23]
2 years ago
9

A right triangle has a vertex at point M and a height of 5 units. The base of the triangle is on MN←→− and is 3 units long. One

of vertices of the triangle is at M.
Use the Polygon tool to draw the triangle.

Each segment on the grid represents 1 unit.

Mathematics
2 answers:
Fynjy0 [20]2 years ago
7 0

Answer:

Same as above.

Step-by-step explanation:

Julli [10]2 years ago
5 0

Answer:

We are given a triangle having vertex at point M.

The base of the triangle is MN which is 3 units long.

Now, the height of the triangle is given as 5 units.

So, plotting a line starting at M, which is 5 units long towards the left or right will give us the height of the triangle.

Moreover, as the base is 3 units long.

We will have that the point N is 3 units down or up from the point M.

Using Pythagoras theorem, get that the length of the third side is 5.8 units.

After joining these points and constructing straight lines, we will get the following right angles triangle having base 3 units and height 5 units.

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Step-by-step explanation:

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Which of the following is equivalent to sec2θ−1sec2θ for all values of θ for which sec2θ−1sec2θ is defined?
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According to the scenario, the following information is given:

= \frac{sec^2 \theta - 1 }{sec^2 \theta}

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olga55 [171]
We know that:

(x-a)^2+(y-b)^2=r^2

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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}\\\\\\&#10;\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}\\\\\\&#10;\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}\\\\\\&#10;

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}

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\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}}

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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}}

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ohaa [14]
Here’s your answer the A/= it means the answer is okay.

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3 years ago
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