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3 years ago

Find the circumference and area

Mathematics

2 answers:

Karo-lina-s [1.5K]3 years ago

8 0

I think the answer is 18.85:)

KATRIN_1 [288]3 years ago

5 0

18.85 I hope this helps you :)

You might be interested in

Performance Matters

spayn [35]

Answer:

the 1st, 2nd and 6th statements are true.

and probably the 3rd statement is true.

I am not sure about the 3rd statement, as I cannot read the original in your screenshot, and your transcribed description is probably not correct and contains typos.

but all you need is in the explanation below to decide, if the actual 3rd statement is true or not.

if "58 + 79 = 1268" actually means "5s + 7g = 1268" then it is true. otherwise it is false.

Step-by-step explanation:

g = number of general tickets sold

s = number of student tickets sold

so, in total

g + s = 234

tickets were sold.

and the revenue was

7g + 5s = $1,268

out of these 2 basic equating we get

g = 234 - s

and then

7(234 - s) + 5s = 1,268

1,638 - 7s + 5s = 1,268

-2s = -370

s = 185

g = 234 - s = 234 - 185 = 49

so, we know

185 student tickets were sold for 5×185 = $925

49 general tickets were sold for 7×49 = $343

7 0

2 years ago

Use the Law of Sines to find the measure of angle R to the nearest degree.

IrinaK [193]

$\frac{\sin(R)}{PQ} = \frac{\sin(Q)}{PR}$

$\frac{\sin(R)}{24.7} = \frac{\sin(43^{\circ})}{16.9}$

$\sin(R) = 24.7*\frac{\sin(43^{\circ})}{16.9}$

$\sin(R) \approx 0.9967668$

$R \approx \arcsin(0.9967668)$

$R \approx 85.39137896^{\circ}$

$R \approx 85^{\circ}$

The final answer is 85 degrees

4 0

3 years ago

I'm quite confused on how to solve 4 to the -2nd power. Could someone help?

leonid [27]

It's 1 over 4 to the 2nd or 1 over 16

6 0

3 years ago

Solve the equation 2 sec² x = 3 - tan x for the domain 0° ≤ x ≤ 360°

vodka [1.7K]

$2 \sec^2 x = 3- \tan x \\\\\implies 2(1 +\tan^2 x) = 3- \tan x \\\\\implies 2 + 2 \tan^2 x +\tan x -3 =0\\\\\implies 2 \tan^2 x + \tan x -1 =0\\\\\implies 2u^2 + u -1 =0~~~ ;[\text{set} \tan x = u]\\\\\implies u = \dfrac{-1\pm \sqrt{1-4\cdot 2 \cdot (-1)}}{2(2)}\\\\\implies u =\dfrac{-1 \pm\sqrt{9}}{4}\\\\\implies u = \dfrac{-1 \pm 3}{4}\\\\\implies u = \dfrac{2}{4}=\dfrac 12~~ \text{or} ~~u =\dfrac{-4}{4} =-1\\\\$

$\implies \tan x = \dfrac 12 ~~ \text{or} ~~ \tan x =-1~~~ ;[\text{Substitute back u =tan x}]\\\\\text{Now,}~ \\\\\tan x = -1,\\\\\implies x = n\pi - \dfrac{\pi}4\\\\\\\text{For interval,}~ [0,2\pi) \\\\x = \dfrac{3\pi}4, \dfrac{7\pi}4\\\\\text{In degrees,}~ x = 135^{\circ}, x =315^{\circ}\\\\\tan x = \dfrac 12\\\\\implies x = n\pi + \tan^{-1} \left(\dfrac 12 \right)\\\\\text{For interval} ~[0,2\pi),\\\\$

$x=\tan^{-1} \left(\dfrac 12 \right), \left[\pi +\tan^{-1} \left( \dfrac 12 \right)\right]\\\\\text{in degrees,} ~~x=\tan^{-1} \left(\dfrac 12 \right), \left[180^{\circ} +\tan^{-1} \left( \dfrac 12 \right)\right]\\\\\\\\\text{Combine all solutions:}\\\\x=\dfrac{3\pi}4, \dfrac{7\pi}4, \tan^{-1} \left(\dfrac 12 \right), \left[\pi +\tan^{-1} \left( \dfrac 12 \right)\right]\\\\\\$

$\text{In degrees,}~ x = 135^{\circ},~315^{\circ},~\tan^{-1} \left(\dfrac 12 \right), \left[180^{\circ} +\tan^{-1} \left( \dfrac 12 \right)\right]$

4 0

2 years ago

The function is defined by f(x)= x^2+3x-10

yaroslaw [1]

First you put (x+5) into the initial function wherever you see x so it becomes
(x+5)^2+3(x+5)-10=x^2+kx+30
(x^2+5x+25)+(3x+15)-10 simplified left side
x^2+8x+30 fully simplified left side
thus k=8
x^2+8x+30=0 to find 0s
-4 + 3.7416573867739i
<span>-4 - 3.7416573867739i
</span>these are the roots you find after using the quadratic formula
the second one is the smallest

8 0

3 years ago