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2 years ago

Find the requested angle supplement of 68.9

Mathematics

1 answer:

Minchanka [31]2 years ago

7 0

supplement of 68.9° = 180° - 68.9° = 111.1° = 111° 06'

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Someone please help!

svetoff [14.1K]

Answer:

a =6

b=4

c=2

step-by-step explanation:

$\sqrt{x^{12} y^{9}z^{5} }$ = $\sqrt{x^{12} y^{8}y z^{4} z}$ = $\sqrt{(x^{6}) ^{2} (y^{4}) ^{2} (z^{2})^{2} yz }$ = $x^{6}y^{4} z^{2} \sqrt{yz}$

7 0

2 years ago

A cylinder-shaped candle has a diameter of 5 inches and a height of 8 inches.

Arada [10]

V=hpir²
h=8
d=2r
d/2=r
5/2=2.5=r

V=8pi2.5²
V=8pi6.25
V=50pi
aprox pi=3.14

V=50 times 3.14
V=157 in³

157 cubic inches

5 0

3 years ago

Read 2 more answers

Find the length of side x to the nearest tenth. 30° 12 x х 60°

k0ka [10]

Answer:

13.9

Step-by-step explanation:

Use the Pythagorean Theorem

x=13.85 round to the nearest tenth 13.9

8 0

2 years ago

What is the solution to the equation

Dafna11 [192]

$\sqrt{4t+5}=3-\sqrt{t+5}\\\\D:4t+5\geq0\ \wedge\ t+5\geq0\ \wedge\ \sqrt{t+5}\leq3\\\\t\geq-\dfrac{5}{4}\ \wedge\ t\geq-5\ \wedge\ t\leq6$

therefore
$D:x\in\left< -\dfrac{5}{4};\ 6\right>$

$\sqrt{4t+5}=3-\sqrt{t+5}\ \ \ |^2\\\\(\sqrt{4t+5})^2=(3-\sqrt{t+5})^2\ \ \ |use:(a-b)^2=a^2-2ab+b^2\\\\4t+5=3^2-2\cdot3\cdot\sqrt{t+5}+(\sqrt{t+5})^2\\\\4t+5=9-6\sqrt{t+5}+t+5\ \ \ \ |-t\\\\3t+5=14-6\sqrt{t+5}\ \ \ \ |-14\\\\3t-9=-6\sqrt{t+5}\ \ \ \ |change\ signs$
$9-3t=6\sqrt{t+5}\ \ \ \ |:3\\\\3-t=2\sqrt{t+5}\ \ \ \ |^2\\\\(3-t)^2=(2\sqrt{t+5})^2\\\\3^2-2\cdot3\cdot t+t^2=4(t+5)\\\\9-6t+t^2=4t+20\ \ \ |-4t-20\\\\t^2-10t-11=0\\\\t^2+t-11t-11=0\\\\t(t+1)-11(t+1)=0\\\\(t+1)(t-11)=0\iff t+1=0\ \vee\ t-11=0\\\\t=-1\in D\ \vee\ t=11\notin D$
Answer: t = -1.

7 0

2 years ago

Read 2 more answers

1. X - 18 = 23

Thepotemich [5.8K]

Answer:

41
-5
42
21
4
7

Step-by-step explanation:

6 0

2 years ago

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