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Scilla [17]
3 years ago
15

Write an expression for the sequence of operations described below.

Mathematics
1 answer:
IgorC [24]3 years ago
7 0
U/(4^6-2)

would be the answer. Good luck!
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Find the greatest common factor of these three expressions. 5v2 , 14v4 , and 35
a_sh-v [17]

Question

Find the greatest common factor of these three expressions. 5v2 , 14v4 , and 35


Answer:

<h2>1</h2>

Step-by-step explanation:

assuming that 5v2 is 5², and 14v4 is 14^4

5² = 25

14^4 = 38416

35 = 35

The factors of 25 are: 1, 5, 25

The factors of 35 are: 1, 5, 7, 35

The factors of 38416 are: 1, 2, 4, 7, 8, 14, 16, 28, 49, 56, 98, 112, 196, 343, 392, 686, 784, 1372, 2401, 2744, 4802, 5488, 9604, 19208, 38416

--------------

the greatest common factor is 1



6 0
3 years ago
Rewrite the fraction in simplest form.<br> 18/24
Thepotemich [5.8K]
The greatest common fact is 6.

18 ÷ 6 = 3
24 ÷ 6 = 4

In simplest form, 18/24 = 3/4
4 0
3 years ago
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
What is the inverse of the given relation? Y=7x^2+4
EleoNora [17]

follow these steps:

1) y-4=7x^2

2) (y-4)/7 = x^2

x = \sqrt{ \frac{y - 4}{7} }

so you must inverse x and y names:

{f}^{ - 1} (x) = \sqrt{ \frac{x - 4}{7} }

7 0
3 years ago
A wheel in this scale drawing is 4 cm wide. The ratio of the drawing is 1:14. How wide is the actual wheel in cm?
bagirrra123 [75]

1:14 means that for every 1 cm it is drawn at it is actually 14 cm

 so multiply 4 x 14 = 56

 it is actually 56 cm wide

7 0
3 years ago
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