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abruzzese [7]
3 years ago
15

The graph shows the functions f(x), p(x), and g(x): Graph of function g of x is y is equal to 2 multiplied by 1.5 to the power o

f x. The straight line f of x joins ordered pairs 3, 1 and 2, minus 1 and is extended on both sides. The straight line p of x joins the ordered pairs 1, 3 and 3, 1 and is extended on both sides. Part A: What is the solution to the pair of equations represented by p(x) and f(x)? (3 points) Part B: Write any two solutions for f(x). (3 points) Part C: What is the solution to the equation p(x) = g(x)? Justify your answer. (4 points)

Mathematics
1 answer:
Nataly_w [17]3 years ago
8 0
<h3>Answer:</h3>

A) (3, 1)

B) (3, 1), (2, -1)

C) (1, 3)

<h3>Step-by-step explanation:</h3>

A) The problem statement tells you the point (3, 1) is on both lines.

B) The problem statement tells you the line goes through points ...

... (3, 1) and (2, -1).

C) The problem statement tells you point (1, 3) is a solution to p(x). It also happens that g(1) = 2·1.5¹ = 3, so (1, 3) is also a solution to g(x).

(1, 3) is a solution to p(x) = g(x) because p(1) = 3 = g(1).

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Evalute 27⋅((3^3)^−1)
shtirl [24]

Answer: 0.81

Step-by-step explanation:

=27⋅((3^3)^−1)

Do the brackets first:

3^3 is the same as 3³ so it is <u>27</u>

then:

27^-1 equals <u> 0.03</u>

And lastly:

multiply 27 and 0.03

27(0.03)=<em><u>0.81</u></em>

8 0
3 years ago
Plz help I'm really bad at these
GREYUIT [131]
Y = mx + b, where m = slope = rise/run and b, the y intercept value

m = 1/1 = 1, then y = x + b.
We notice that y intercept is = - 1
And the equation is:

y = x - 1
6 0
3 years ago
Y = 60x + 20<br> y = 65x
dalvyx [7]

Answer:

4=x y=325

Step-by-step explanation:

60x + 20 = 65x

group the variables

20=5x

because you subtracted 60x from both

4=x

because you divided 5 from both

now substitute 5 for x

65×5 is 325

y=325

8 0
2 years ago
What is 5 minus 16 tenths
antoniya [11.8K]
The answer would be 3.4

Hope that helped : )
5 0
3 years ago
Read 2 more answers
Problem 4: Solve the initial value problem
pishuonlain [190]

Separate the variables:

y' = \dfrac{dy}{dx} = (y+1)(y-2) \implies \dfrac1{(y+1)(y-2)} \, dy = dx

Separate the left side into partial fractions. We want coefficients a and b such that

\dfrac1{(y+1)(y-2)} = \dfrac a{y+1} + \dfrac b{y-2}

\implies \dfrac1{(y+1)(y-2)} = \dfrac{a(y-2)+b(y+1)}{(y+1)(y-2)}

\implies 1 = a(y-2)+b(y+1)

\implies 1 = (a+b)y - 2a+b

\implies \begin{cases}a+b=0\\-2a+b=1\end{cases} \implies a = -\dfrac13 \text{ and } b = \dfrac13

So we have

\dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = dx

Integrating both sides yields

\displaystyle \int \dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = \int dx

\dfrac13 \left(\ln|y-2| - \ln|y+1|\right) = x + C

\dfrac13 \ln\left|\dfrac{y-2}{y+1}\right| = x + C

\ln\left|\dfrac{y-2}{y+1}\right| = 3x + C

\dfrac{y-2}{y+1} = e^{3x + C}

\dfrac{y-2}{y+1} = Ce^{3x}

With the initial condition y(0) = 1, we find

\dfrac{1-2}{1+1} = Ce^{0} \implies C = -\dfrac12

so that the particular solution is

\boxed{\dfrac{y-2}{y+1} = -\dfrac12 e^{3x}}

It's not too hard to solve explicitly for y; notice that

\dfrac{y-2}{y+1} = \dfrac{(y+1)-3}{y+1} = 1-\dfrac3{y+1}

Then

1 - \dfrac3{y+1} = -\dfrac12 e^{3x}

\dfrac3{y+1} = 1 + \dfrac12 e^{3x}

\dfrac{y+1}3 = \dfrac1{1+\frac12 e^{3x}} = \dfrac2{2+e^{3x}}

y+1 = \dfrac6{2+e^{3x}}

y = \dfrac6{2+e^{3x}} - 1

\boxed{y = \dfrac{4-e^{3x}}{2+e^{3x}}}

7 0
2 years ago
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