Here's another geometry thing I suck at...

Mathematics

1 answer:

solniwko [45]3 years ago

6 0

A = 133; b = 31; c = 82; d = 64.
Opposite angles in an inscribed quadrilateral are supplementary; this means that d + 116 = 180. Subtracting 116 from both sides, we have d = 64.

By the same theorem, c + 98 = 180; subtracting 98 from both sides, we have c = 82.

Inscribed angles are equal to 1/2 the measure of the intercepted arc. Using this, we have

116 = 1/2(a+99)

Multiplying both sides by 2, we have
232 = a+99

Subtract 99 from both sides:
232 - 99 = a + 99 - 99
133 = a

We also have that
82 = 1/2(133+b)

Multiplying both sides by 2, we have:
164 = 133 + b

Subtract 133 from both sides:
164 - 133 = 133 + b - 133
31 = b

You might be interested in

"Two less than the total of a number and five"

tino4ka555 [31]

Answer:

Step-by-step explanation:

6 0

2 years ago

Find the average rate of change of the function over the given interval

sattari [20]

$\bf slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ f(x_2)}}-{{ f(x_1)}}}{{{ x_2}}-{{ x_1}}}\impliedby 
\begin{array}{llll}
average\ rate\\
of\ change
\end{array}\\\\
-------------------------------\\\\$

$\bf h(t)=cot(t)\implies h(t)=\cfrac{cos(t)}{sin(t)}\quad 
\begin{cases}
t_1=\frac{\pi }{4}\\
t_2=\frac{3\pi }{4}
\end{cases}\implies \cfrac{h\left( \frac{3\pi }{4} \right)-h\left( \frac{\pi }{4} \right)}{\frac{3\pi }{4}-\frac{\pi }{4}}
\\\\\\$

$\bf \cfrac{\frac{cos\left( \frac{3\pi }{4} \right)}{sin\left( \frac{3\pi }{4} \right)}-\frac{cos\left( \frac{\pi }{4} \right)}{sin\left( \frac{\pi }{4} \right)}}{\frac{\pi }{2}}\implies \cfrac{-1-1}{\frac{\pi }{2}}\implies \cfrac{-2}{\frac{\pi }{2}}\implies -\cfrac{4}{\pi }\\\\\\
-------------------------------\\\\$

$\bf h(t)=cot(t)\implies h(t)=\cfrac{cos(t)}{sin(t)}\quad 
\begin{cases}
t_1=\frac{\pi }{3}\\
t_2=\frac{3\pi }{2}
\end{cases}\implies \cfrac{h\left( \frac{3\pi }{2} \right)-h\left( \frac{\pi }{3} \right)}{\frac{3\pi }{2}-\frac{\pi }{3}}
\\\\\\$

$\bf \cfrac{\frac{cos\left( \frac{3\pi }{2} \right)}{sin\left( \frac{3\pi }{2} \right)}-\frac{cos\left( \frac{\pi }{3} \right)}{sin\left( \frac{\pi }{3} \right)}}{\frac{9\pi -2\pi }{6}}\implies \cfrac{\frac{0}{-1}-\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}}{\frac{7\pi }{6}}\implies\cfrac{-\frac{1}{\sqrt{3}}}{\frac{7\pi }{6}}\implies -\cfrac{\sqrt{3}}{3}\cdot \cfrac{6}{7\pi }
\\\\\\
-\cfrac{2\sqrt{3}}{7\pi }$

8 0

3 years ago

What is 4b + 6 ≤ 18. I don't know how to solve it and don't know the answer.

wolverine [178]

Answer:

b≤3

Step-by-step explanation:

For this problem, you want to solve it similarly to linear equations.

We start with 4b+6≤18.

Subtracting 6 from both sides gives 4b≤12.

From there, in order to isolate the b, we must divide both sides by 4, leaving b≤3.

**This content involves solving linear inequalities, which you may wish to revise. I'm always happy to help!