7. E(5, 4), F(8, 1) what is the slope of the line

Please someone help me with maths

Nata [24]

Answer:

a)= 5x+10

b)= 7y-6

c)= 6x-6

d)= 2x^2+8x-5

e)= 6x+9

f)= 4xy-4y^2+8y

g)= -12y^2+10y

h)= 6x+154

i)= 6x+21

h)= 2x^2+16

6 0

3 years ago

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Can someone please show me the steps in finding the answers to these two problems ?

ratelena [41]

Answer:

y = 8^(x-1)
x = 24

Step-by-step explanation:

The first step is always to look at the question to see what you are given and what you are asked for. The next step is to identify the relevant relations between what you have and what you need to find.

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1) You are given a sequence of (x, y) points where x increases by 1 from one point to the next, and y increases by a factor of 8. This tells you the value of y form a geometric sequence that is described by an exponential function.

The general form of the n-th term of a geometric sequence is ...

an = a1·r^(n-1)

where a1 is the value for n=1, and r is the common ratio.

If this were a geometric sequence, we could write the general term using a1=1 and r=8:

an = 1·8^(n-1)

Instead, this is a relation between x and y. It is the same relation as for the geometric sequence, but with different variable names. We have y instead of an, and we have x instead of n. So, the equation is ...

y = 8^(x -1) . . . . . . . or . . y = (1/8)8^x

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2) This equation is of the form ...

k·∛x = 6

Since k is also a cube root, it is easiest to cube both sides of the equation:

9x = 6³ = 216

Now, you can divide by 9 to find x:

x = 216/9

x = 24

5 0

3 years ago

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You earn $28 for washing 4 cars. How much do you earn for washing 3 cars?

a_sh-v [17]

Answer:

21 dollars

Step-by-step explanation:

8 0

3 years ago

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(5) Find the Laplace transform of the following time functions: (a) f(t) = 20.5 + 10t + t 2 + δ(t), where δ(t) is the unit impul

Aloiza [94]

Answer

(a) $F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}$

(b) $F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}$

Step-by-step explanation:

(a) $f(t) = 20.5 + 10t + t^2 +$ δ(t)

where δ(t) = unit impulse function

The Laplace transform of function f(t) is given as:

$F(s) = \int\limits^a_0 f(s)e^{-st} \, dt$

where a = ∞

=> $F(s) = \int\limits^a_0 {(20.5 + 10t + t^2 + d(t))e^{-st} \, dt$

where d(t) = δ(t)

=> $F(s) = \int\limits^a_0 {(20.5e^{-st} + 10te^{-st} + t^2e^{-st} + d(t)e^{-st}) \, dt$

Integrating, we have:

=> $F(s) = (20.5\frac{e^{-st}}{s} - 10\frac{(t + 1)e^{-st}}{s^2} - \frac{(st(st + 2) + 2)e^{-st}}{s^3} )\left \{ {{a} \atop {0}} \right.$

Inputting the boundary conditions t = a = ∞, t = 0:

$F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}$

(b) $f(t) = e^{-t} + 4e^{-4t} + te^{-3t}$

The Laplace transform of function f(t) is given as:

$F(s) = \int\limits^a_0 (e^{-t} + 4e^{-4t} + te^{-3t} )e^{-st} \, dt$

$F(s) = \int\limits^a_0 (e^{-t}e^{-st} + 4e^{-4t}e^{-st} + te^{-3t}e^{-st} ) \, dt$

$F(s) = \int\limits^a_0 (e^{-t(1 + s)} + 4e^{-t(4 + s)} + te^{-t(3 + s)} ) \, dt$

Integrating, we have:

$F(s) = [\frac{-e^{-(s + 1)t}} {s + 1} - \frac{4e^{-(s + 4)}}{s + 4} - \frac{(3(s + 1)t + 1)e^{-3(s + 1)t})}{9(s + 1)^2}] \left \{ {{a} \atop {0}} \right.$

Inputting the boundary condition, t = a = ∞, t = 0:

$F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}$

3 0

3 years ago

Which expression represents the distance between 4.7 and −11.2 on the number line?

Cerrena [4.2K]

Answer:

4.7 - ( -11.2)

Step-by-step explanation:

4.7 - ( - 11.2) = 15.9

15.9 is the distance between 4.7 and -11.2

Hopefully this helps you

pls mark brainlest

3 0

3 years ago