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3 years ago

Linear functions :( please help!

Mathematics

2 answers:

givi [52]3 years ago

7 0

Answer:

J y=4x+1

Step-by-step explanation:

when looking at the x column i will pick a couple of sets and plug them in to see if they fit and -1×4+1= -3

liraira [26]3 years ago

4 0

The answer to this question is “J”, y=4x+1

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Function g can be thought of as a translation (shifted) version of f(x)=x^2

dimaraw [331]

Answer:

(x+5)²

Step-by-step explanation:

To solve this we just need to shift the graph 5 spots to the left

to do this we need to add 5 to the x

(x+5)²

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3 years ago

You're studying how well students remember vocabulary after they stop studying a new language. The first student you study remem

SpyIntel [72]

Answer:

Exponential decay model

Step-by-step explanation:

A quick graph shows the remembrance pattern is not linear because the slope is not the same. Therefore a very likely remembrance pattern will be exponentially related.

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3 years ago

Solve the triangle A = 2 B = 9 C =8

VARVARA [1.3K]

Answer:

$\begin{gathered} A=\text{ 12}\degree \\ B=\text{ 114}\degree \\ C=54\degree \end{gathered}$

Step-by-step explanation:

To calculate the angles of the given triangle, we can use the law of cosines:

$\begin{gathered} \cos (C)=\frac{a^2+b^2-c^2}{2ab} \\ \cos (A)=\frac{b^2+c^2-a^2}{2bc} \\ \cos (B)=\frac{c^2+a^2-b^2}{2ca} \end{gathered}$

Then, given the sides a=2, b=9, and c=8.

$\begin{gathered} \cos (A)=\frac{9^2+8^2-2^2}{2\cdot9\cdot8} \\ \cos (A)=\frac{141}{144} \\ A=\cos ^{-1}(\frac{141}{144}) \\ A=11.7 \\ \text{ Rounding to the nearest degree:} \\ A=12º \end{gathered}$

For B:

$\begin{gathered} \cos (B)=\frac{8^2+2^2-9^2}{2\cdot8\cdot2} \\ \cos (B)=\frac{13}{32} \\ B=\cos ^{-1}(\frac{13}{32}) \\ B=113.9\degree \\ \text{Rounding:} \\ B=114\degree \end{gathered}$ $\begin{gathered} \cos (C)=\frac{2^2+9^2-8^2}{2\cdot2\cdot9} \\ \cos (C)=\frac{21}{36} \\ C=\cos ^{-1}(\frac{21}{36}) \\ C=54.3 \\ \text{Rounding:} \\ C=\text{ 54}\degree \end{gathered}$

3 0

1 year ago

Find the area of the shaded regions below. Give your answer as a completely simplified exact value in terms of π (no approximati

Svetach [21]

Answer:

Answer:18+4.5 $\pi \\$

Step-by-step explanation:

This shape is a triangle with a semicircle connected to it

and this triangle is a right triangle so side A=B

that means the other leg is 6.Knowing that we can solve

The formula for the area of a triangle is (base×height)÷2

so that means 6×6 equals 36 and if you divide that by 2 you get the 18.

Now we will deal with the semicircle. We know that both of the legs are 6 so that means the diameter is 6 and now we solve 6 divided by 2 equals 3 and we will have to square that and we will 9 and since it is a semicircle we have to divide it by 2 and that will give us 4.5 and since we have to express it in terms of pi it will be 4.5(pi) and then we add both of the areas

giving us 18+4.5(pi)

and do you go to RSM

8 0

3 years ago

Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br

Lina20 [59]

Answer:

The differential equation for the amount of salt A(t) in the tank at a time t > 0 is $\frac{dA}{dt}=12 - \frac{2A(t)}{500+t}$ .

Step-by-step explanation:

We are given that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another brine solution is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

The concentration of the solution entering is 4 lb/gal.

Firstly, as we know that the rate of change in the amount of salt with respect to time is given by;

$\frac{dA}{dt}= \text{R}_i_n - \text{R}_o_u_t$

where, $\text{R}_i_n$ = concentration of salt in the inflow $\times$ input rate of brine solution

and $\text{R}_o_u_t$ = concentration of salt in the outflow $\times$ outflow rate of brine solution

So, $\text{R}_i_n$ = 4 lb/gal $\times$ 3 gal/min = 12 lb/gal

Now, the rate of accumulation = Rate of input of solution - Rate of output of solution

= 3 gal/min - 2 gal/min

= 1 gal/min.

It is stated that a large mixing tank initially holds 500 gallons of water, so after t minutes it will hold (500 + t) gallons in the tank.

So, $\text{R}_o_u_t$ = concentration of salt in the outflow $\times$ outflow rate of brine solution

= $\frac{A(t)}{500+t} \text{ lb/gal } \times 2 \text{ gal/min}$ = $\frac{2A(t)}{500+t} \text{ lb/min }$

Now, the differential equation for the amount of salt A(t) in the tank at a time t > 0 is given by;

= $\frac{dA}{dt}=12\text{ lb/min } - \frac{2A(t)}{500+t} \text{ lb/min }$

or $\frac{dA}{dt}=12 - \frac{2A(t)}{500+t}$ .

4 0

3 years ago