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Naily [24]
3 years ago
6

A certain quantity has a decade growth factor of 2.6. What is it’s yearly growth factor?

Mathematics
1 answer:
8090 [49]3 years ago
3 0

Answer:

31.2

Step-by-step explanation:

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You start at (8, 0). You move right 2 units. Where do you end?
frez [133]

Answer:

(10,0)

Step-by-step explanation:

Moving right 2 units will increase the x value by 2 since you are going further 'down the corridor'. Hope this helps.

8 0
2 years ago
At a pet store, there are 30 aquariums.
irakobra [83]

Answer:

f = (20 × 15) + (10 × 6)

Explanation:

You are just find how many fish are salt water and how many freshwater and adding them together. The total number of aquariums is irrelevant in this case.

8 0
3 years ago
Please help math asap 30 points
Bas_tet [7]

The answer is A

(5,3)

4 0
3 years ago
Read 2 more answers
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
Read 2 more answers
Solve each equation for the indicated variable abc = 1/2; b
ANEK [815]
Please rewrite your question so it makes sense. Thanks, and good luck! :D
6 0
3 years ago
Read 2 more answers
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