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Liono4ka [1.6K]

4 years ago

7

In ΔABC, if the length of sides a, b, and c are 3.5 centimeters, 5.5 centimeters, and 6 centimeters respectively, what is m∠C to

two decimal places

2 answers:

Tamiku [17]4 years ago

8 0

I beleive it could be <span>(13/77)=80.28 degrees but I am not so sure >~<</span>

Luba_88 [7]4 years ago

6 0

<span>C=arccos(13/77)=80.28 degrees
</span><span>Look at attachment. </span>

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brilliants [131]

The perimeter of the triangle is 10x-1 cm.

Explanation:

Perimeter of the triangle:
(6x+2) + (x-4) + (3x+1)

= 6x + 2 + x - 4 + 3x + 1

= 10x - 1

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3 years ago

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Step-by-step explanation:

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3 years ago

On a coordinate plane, a triangle has points R (negative 1, 3), S (3, negative 2), and T (1, negative 4). Which reflection will

myrzilka [38]

Answer:

a reflection of ΔRST across the line y = –x

Step-by-step explanation:

A reflection across the line y = –x transforms point (x, y) into (-y, -x)

After reflecting ΔRST across the line y = –x we get:

R (-1, 3) -> (-3, 1)

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8 0

3 years ago

Read 2 more answers

Recall the equation for a circle with center (h, k) and radius r. At what point in the first quadrant does the line with equatio

lina2011 [118]

Answer:

The point of intersection is:

$\displaystyle \left(\frac{6\sqrt{5}}{5}, \frac{3\sqrt{5}}{5}+5\right)\approx \left(2.68, 6.34)$

Step-by-step explanation:

We want to find the point in QI at which the line with the equation:

$y=0.5x+5$

Intersect a circle with a radius of 3 and a center of (0, 5).

First, write the equation of a circle. The equation for a circle is given by:

$(x-h)^2+(y-k)^2=r^2$

Where (<em>h, k</em>) is the center and <em>r</em> is the radius.

Since our center is (0, 5), <em>h</em> = 0 and <em>k</em> = 5. The radius is 3. So, <em>r</em> = 3. Substitute:

$(x-0)^2+(y-5)^2=(3)^2$

Simplify:

$x^2+(y-5)^2=9$

At the point where the two equations intersect, its <em>x-</em>coordinate and <em>y-</em>coordinate will be the same. Therefore, we can substitute the equation of the line into the equation of the circle and solve for <em>x</em>. So:

$x^2+((0.5x+5)-5)^2=9$

Simplify:

$x^2+(0.5x)^2=9$

Square:

$x^2+0.25x^2=9$

Combine like terms:

$\displaystyle 1.25x^2=\frac{5}{4}x^2=9$

Solve for <em>x: </em>

<em /> $\displaystyle \begin{aligned} x^2&=\frac{36}{5} \\ x&=\pm\sqrt{\frac{36}{5}} \\ x&\Rightarrow \frac{6}{\sqrt{5}}=\frac{6\sqrt{5}}{5}\approx2.68\end{aligned}$ <em />

Note that since we are looking for the point of intersection in QI, <em>x</em> should be positive. So, we can ignore the negative answer.

To find the <em>y-</em>coordinate, substitute the <em>x-</em>value back into either equation. Using the linear equation:

$\displaystyle y=0.5\left(\frac{6\sqrt{5}}{5}\right)+5=\frac{3\sqrt{5}}{5}+5\approx 6.34$

So, the point of intersection in QI is:

$\displaystyle \left(\frac{6\sqrt{5}}{5}, \frac{3\sqrt{5}}{5}+5\right)\approx \left(2.68, 6.34)$

6 0

3 years ago

Find the perimeter of the following shape: <br> 10.8<br> 11<br> 11.4<br> 11.6

amm1812

Add all the sides up

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4 years ago