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Oduvanchick [21]
3 years ago
9

What function equation is represented by the graph?

Mathematics
1 answer:
Ahat [919]3 years ago
5 0

In the given graph, lets mark two points:

Let A = (0,-3) and B = (4,-2)

The slope of the line is given by:

m = \frac{(-2)-(-3)}{(4)-(0)}

m = \frac{-2+3}{4}

m = \frac{1}{4}

The equation of the line is given by:

(y-y_1)=m(x-x_1)

(y-(-3))=(\frac{1}{4})(x-0)

4(y+3)=x

4y+12=x

4y=x-12

y= \frac{1}{4}x-3

Therefore, the function equation represented by the graph is: f(x)= \frac{1}{4}x-3

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B) The following patterns are made using small squares.
kotegsom [21]

Answer:

(N+2)^2+6

Step-by-step explanation:

8 0
2 years ago
7. John's seafood restaurant is trying to estimate its profits. John has found that on average, each meal served costs the resta
laiz [17]

Answer:

f(c) =17.12c -14.56c + 5.40(.5c) - 1.20(.5c)

Step-by-step explanation:

Profit is the difference between revenue (17.12c +5.40(.5c)) and cost(14.56c+1.20(0.5c)). That difference is expressed by the function shown above.

4 0
3 years ago
Read 2 more answers
Since at t=0, n(t)=n0, and at t=∞, n(t)=0, there must be some time between zero and infinity at which exactly half of the origin
Airida [17]
Answer: t-half = ln(2) / λ ≈ 0.693 / λ

Explanation:

The question is incomplete, so I did some research and found the complete question in internet.

The complete question is:

Suppose a radioactive sample initially contains N0unstable nuclei. These nuclei will decay into stable nuclei, and as they do, the number of unstable nuclei that remain, N(t), will decrease with time. Although there is no way for us to predict exactly when any one nucleus will decay, we can write down an expression for the total number of unstable nuclei that remain after a time t:

N(t)=No e−λt,

where λ is known as the decay constant. Note that at t=0, N(t)=No, the original number of unstable nuclei. N(t) decreases exponentially with time, and as t approaches infinity, the number of unstable nuclei that remain approaches zero.

Part (A) Since at t=0, N(t)=No, and at t=∞, N(t)=0, there must be some time between zero and infinity at which exactly half of the original number of nuclei remain. Find an expression for this time, t half.

Express your answer in terms of N0 and/or λ.

Answer:

1) Equation given:

N(t)=N _{0} e^{-  \alpha  t} ← I used α instead of λ just for editing facility..

Where No is the initial number of nuclei.

2) Half of the initial number of nuclei: N (t-half) =  No / 2

So, replace in the given equation:

N_{t-half} =  N_{0} /2 =  N_{0}  e^{- \alpha t}

3) Solving for α (remember α is λ)

\frac{1}{2} =  e^{- \alpha t} 

2 =   e^{ \alpha t} 

 \alpha t = ln(2)

αt ≈ 0.693

⇒ t = ln (2) / α ≈ 0.693 / α ← final answer when you change α for λ




4 0
3 years ago
The base of a triangle exceeds the height by 9 centimeters. If the area is 56 square centimeters, find the length of the base an
shepuryov [24]

Answer:

The height of the triangle is 7cm and the base is 16cm

Step-by-step explanation:

First of all we have to know the formula to calculate area of ​​a triangle

a = area = 56

b = base

h = heigth

a = (b * h)/2

we replace the known values and we make 2 equations

56cm² = (b * h)/2

b = h + 9cm

we replace b by (h + 9cm) in the first equation

56cm² = (h + 9cm * h)/2

56cm² * 2 = h² + 9h

0 = h² + 9h - 112cm²

we use bhaskara formula:

(-b (±) √ (b² - 4ac) ) / 2a

we replace with the known values

h = (-9 (±) √ (9² - 4*1*(-112) ) ) / 2*1

h = (-9 (±) √ (81 + 448) ) ) / 2

h = (-9 (±) √529) /2

h = (-9 (±) 23)/2

h1 = (-9 + 23) / 2

h1 = 14 / 2

h1 = 7

h2 = (-9 - 23) / 2

h2 = -32 / 2

h2 = -16

The height of the triangle is 7cm and the base is 16cm

8 0
3 years ago
You begin solving the equation 3+4x=513+4x=51 by subtracting 33 from both sides. Which is the best choice for Step 22?
bonufazy [111]
33 +4x = 513
4x = 480 (x = 480/4: x = 120)


Answer
Step 2 Divide each side by 4
7 0
3 years ago
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