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zhenek [66]
3 years ago
7

Which expression is equivalent to -12(3x-)?

Mathematics
1 answer:
victus00 [196]3 years ago
7 0

-36×+9 is equivalent to this question.

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Need help.<br>I 2x +1 I &lt; -4
anastassius [24]


the absolute value of a number is always positive. Therefore you cannot have an absolute value equal to a negative number.

the answer would be "no solution"

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A bag contains 1 purple beads and 3 green beads. A bead is drawn and then replaced before drawing the second bead. Find the prob
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ggg

Step-by-step explanation:

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What is An equivalent fraction to 1/2 with a denominator as 10?
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Answer:5/10

Step-by-step explanation:

1/2 x 5 = 5/10

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Which symbol correctly compares the two measurementsone pound 16 oz ​
myrzilka [38]

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the = symbol

Step-by-step explanation:

there are 16 oz in a pound

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The sea ice area around the North Pole fluctuates between about 6 million square kilometers in September to 14 million square ki
Aleksandr [31]

Answer:

there are approximately 5.035 months when there is less than 9 million square meters of sea ice around the North Pole in a year.

Step-by-step explanation:

Given the data in the question;

Let S(t) represent the amount sea ice around the North Pole in millions of square meters at a given time t,

t is the number of months since January.

Now, we use a cosine curve to model this scenario

Vertical shift will be;

D = ( 6 + 14 ) / 2 = 20 / 2

D = 10

Next is the Amplitude;

|A| = ( 6 - 14 ) / 2

|A| = 4

Now, the horizontal stretch factor will be;

B = 2π / 12

B = π/6

Hence;

S(t) = 4cos( π/6 × t ) + 10 ----------- let this be equation 1

Now we find when there will be less than 9 million square meters of sea ice;

S(t) = 9

so we have

9 = 4cos( π/6 × (t-2) ) + 10

9 - 10 = 4cos( π/6 × (t-2) )

-1 = 4cos( π/6 × (t-2) )

-1/4 = cos( π/6 × (t-2) )

so we have;

cos⁻¹( -1/4 ) = π/6 × (t₁-2)  -------- let this be equation 2

2π - cos⁻¹( -1/4 ) = π/6 × (t₂-2)  -------- let this be equation 3

so we solve equation 2 and 3

we have'

t₁ - t₂ = 6/π × ( 2π - cos⁻¹( -1/4 ) - cos⁻¹( -1/4 ) )

t₁ - t₂ = 6/π × ( 2π - 2cos⁻¹( -1/4 )  

t₁ - t₂ = 6/π × ( π - cos⁻¹( -1/4 )  

t₁ - t₂ = 6/π × ( π - 104.4775 )

t₁ - t₂ = 6/π × ( π - 104.4775 )  

t₁ - t₂ = 5.035

therefore, there are approximately 5.035 months when there is less than 9 million square meters of sea ice around the North Pole in a year.

6 0
3 years ago
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