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3 years ago

Solve for x. Show all work. Then, plug in the x value you got to check your work 3(x+2)-4=4(x-3)

Mathematics

2 answers:

Amiraneli [1.4K]3 years ago

7 0

Answer:

-x= -14

Step-by-step explanation:

3(x+2)-4=4(x-3)

3(x+2)=3x+6

3x+6-4=3x+2

3x+2=4(x-3)

4(x-3)=4x-12

3x+2=4x-12

3x-4x=-1x

-12-2=-14

-x=-14

sergiy2304 [10]3 years ago

4 0

Answer:

Step-by-step explanation:

3(x+2)-4=4(x-3)

3x+6-4=4x-12

3x+2=4x-12

-3x. -3x

2=1x-12

+12. +12

14=1x

/1. /1

14=x

Then we put it back in the problem.

3(x+2)-4=4(x-3)

3(14+2)-4=4(14-3)

3(16)-4=4(11)

48-4=44

44=44

Can I please have brainliest :D

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If a=12 and b=23 what is the area of pencil decoration

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3 years ago

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The graphs below are based on government data on unemployment for 2017.

lord [1]

You already got the first two questions right: the group with the highest number of people unemployed / unemployment rate is the one with the highest bar in the respective histogram.

The two graph are related in this sense: take the first age group as reference. We know that 700 people from 16 to 19 are unemployed, and that those people represent 14% of the population, then we know that the 14% of the population between 16 and 19 is 700.

We can deduce that, if x is the number of people between 16 and 19, we have

$\dfrac{14}{100}x=700 \iff \dfrac{7}{50}x=700 \iff \dfrac{x}{50}=100 \iff x=5000$

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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 = 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

2 years ago

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2 years ago

Which function is increasingnon the interval (negative infinity, positive infinity)

BlackZzzverrR [31]

Answer:

h(x) = 2^(x) - 1.

Step-by-step explanation:

Let's look at each equation:

f(x) = -3x +7, Well as x increases, since it's multiplication, there are "going to be more" -3's, so it's going to be decreasing.

g(x) = -4(2^x). While 2^x is increasing, because "there are going to be more 2's multiplied by each other" as x increases, it's being multiplied by a negative number, so it's actually going to be decrasing

h(x) = 2^(x) - 1. Here's it's going to be increases as x goes towards infinity because "there are going to be more 2's multiplied by each other", and there isn't any negative sign, while there is a negative 1, it's constant, so the overall value will be increasing

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1 year ago