(-2,3) and slope m=9

Okay guys, I'm going to need your help today.As always, no false answers, and do it CORRECTLY, you get brainliest!Here it goes:

AleksAgata [21]

Answer:

B. 19 inches and 60 inches.

Step-by-step explanation:

Required

Select matching pairs

Represent diameter with D and Circumference with C

C is calculated as:

Where

A. D = 5m; C = 22m

Using:

15.71 does not approximates to 22m.

Hence, the pair do not match

B. D = 19 in; C = 60 in

Using:

59.698 approximates to 60.

Hence, the pair match

C. D = 33 cm; D = 80 cm.

Using:

103.686 does not approximates to 22m.

Hence, the pair do not match

From the analysis above, only option B matches the diameter with the circumference

5 0

3 years ago

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Law of sines: StartFraction sine (uppercase A) Over a EndFraction = StartFraction sine (uppercase B) Over b EndFraction = StartF

Vsevolod [243]

First of all, this problem is properly done with the Law of Cosines, which tells us

$a^2 = b^2 + c^2 - 2 b c \cos A$

giving us a quadratic equation for b we can solve. But let's do it with the Law of Sines as asked.

$\dfrac{\sin A}{a} = \dfrac{\sin B}{b} = \dfrac{\sin C}{c}$

We have c,a,A so the Law of Sines gives us sin C

$\sin C = \dfrac{c \sin A}{a} = \dfrac{5.4 \sin 20^\circ}{3.3} = 0.5597$

There are two possible triangle angles with this sine, supplementary angles, one acute, one obtuse:

$C_a = \arcsin(.5597) = 34.033^\circ$

$C_o = 180^\circ - C_a = 145.967^\circ$

Both of these make a valid triangle with A=20°. They give respective B's:

$B_a = 180^\circ - A - C_a = 125.967^\circ$

$B_o = 180^\circ - A - C_o = 14.033^\circ$

So we get two possibilities for b:

$b = \dfrac{a \sin B}{\sin A}$

$b_a = \dfrac{3.3 \sin 125.967^\circ}{\sin 20^\circ} = 7.8$

$b_o = \dfrac{3.3 \sin 14.033^\circ}{\sin 20^\circ} = 2.3$

Answer: 2.3 units and 7.8 units

Let's check it with the Law of Cosines:

$a^2 = b^2 + c^2 - 2 b c \cos A$

$0 = b^2 - (2 c \cos A)b + (c^2-a^2)$

There's a shortcut for the quadratic formula when the middle term is 'even.'

$b = c \cos A \pm \sqrt{c^2 \cos^2 A - (c^2-a^2)}$

$b = c \cos A \pm \sqrt{c^2( \cos^2 A - 1) + a^2}$

$b = 5.4 \cos 20 \pm \sqrt{5.4^2(\cos^2 20 -1) + 3.3^2}$

$b = 2.33958 \textrm{ or } 7.80910 \quad\checkmark$

Looks good.

6 0

3 years ago

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PLZ HELP ME. TYYYYYYYY

Rom4ik [11]

Answer:

89.1

Step-by-step explanation:

14/43 = 29/x, 14x = 1247, x = 89.1

7 0

3 years ago

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For $x$, $y$, and $z$ positive real numbers, what is the maximum possible value for\[\sqrt{\frac{3x 4y}{6x 5y 4z}} \sqrt{\frac{y

joja [24]

Answer:

Infinity

Step-by-step explanation:

Since x,y,z are positive rel numbers, we have that

$\dfrac{3x4y}{6x5y4z}=\dfrac{1}{10z}\\\\\\\dfrac{y2z}{6x5y4z}=\dfrac{1}{60x}\\\\\\\dfrac{2z3x}{6x5y4z}=\dfrac{1}{20y}$

Hence,

$\sqrt{\dfrac{3x4y}{6x5y4z}}\sqrt{\dfrac{y2z}{6x5y4z}}\sqrt{\dfrac{2z3x}{6x5y4z}}\\\\\\=\sqrt{\dfrac{1}{10z}\dfrac{1}{60x}\dfrac{1}{20y}}=\sqrt{\dfrac{1}{12000xyz}}$

Now let

$f(x,y,z)=\sqrt{\dfrac{1}{12000 xyz}}$

if we take x=y=1, we have

$f(1,1,z)=\sqrt{\dfrac{1}{12000z}}$

and so f(1,1,z) tends to infinity as z goes to 0.

7 0

4 years ago

Which graph represents y as a function of x? <br> A) A <br> B) B <br> C) C <br> D) D

Arlecino [84]

Answer:

where is the graph if you can put it i will give you the answer

Step-by-step explanation:

8 0

3 years ago