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3 years ago

Please Help ASAP! i always give brainliests and thx. My brain is fried help, its not hard just tedious PLEASE! ;)

Mathematics

2 answers:

Brrunno [24]3 years ago

3 0

Answer:

2) is the answer

Step-by-step explanation:

when your start to multiply the large candles by each number, you get a price and then you have to find the price for the small candles, so you multiply the remaining number of candles by the small candle price for each. If you try 10 and 12 for the big candles, the get remaining 10 small candles which is to much and 8 small candles is also to much. But when you try the second answer, you get a just right price for the large candles and then you get the remaining number of candles for the small candles and you'll get the price you are looking for

Maurinko [17]3 years ago

3 0

Answer:

i think the answer would be 8

Step-by-step explanation:

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Someone please help me!

Ivenika [448]

-1/4 i think, Y1-Y2/X1 -X2= the slope . -9-3= -12 and -6 - (-9)= -6 +9= 3. So simplify 3/-12 makes -1/4.

4 0

3 years ago

PLZ HELP QUICKLY, THIS IS IMPORTANT

Elza [17]

Answer: 626944

Step-by-step explanation:

6 0

3 years ago

–6 + 4k = 2(–2k + 5) + 6k

balu736 [363]

Answer:

k=8

Step-by-step explanation:

yes

5 0

3 years ago

What's 2sin(2theta)=1? Need it for webassign.

lutik1710 [3]

<em>Solve: </em> $2sin(2x) = 1$

Divide both sides by 2.
$sin(2x) = \frac{1}{2}$

Take the sine inverse of both sides:
$2x = sin^{-1}(\frac{1}{2})$
$2x = \frac{\pi}{6}, \frac{5\pi}{6}, 0 \leq x \leq 2\pi$
$x = \frac{\pi}{12}, \frac{5\pi}{12}$

But we know there are more solutions if we extend the domain. In fact, there are infinitely more solutions since the domain of sine is all real x values.
Thus, we can develop a general solution:

For every $\pi$ units, there is another solution for both $\frac{\pi}{12}$ and $\frac{5\pi}{12}$

General solutions:
$x = \pi n + \frac{\pi}{12}, n \in \mathbb{Z}$
$x = \pi n + \frac{5\pi}{12}, n \in \mathbb{Z}$

7 0

3 years ago

I need to solve question 3

AlexFokin [52]

3) 3.sec² Ф = 4.tan² Ф
but sec² Ф = 1/cos² Ф
3/cos² Ф = 4.tan² Ф
3/4 = (tan² Ф).(cos² Ф) ; tan² Ф = sin² Ф/cos² Ф

3/4 = (sin² Ф)x(cos² Ф)/(cos² Ф)

sin² Ф = 3/4

a) sin Ф = +(√3)/2 and b) sin Ф = - (√3)/2

sin Ф = + - (√3)/2 → Ф =π/3 + kπ/3

5 0

3 years ago