Can you check my work

What is 48+60 using distributive property

Sedaia [141]

108 you don't need the distributive property.

3 0

3 years ago

Read 2 more answers

How do you solve this

fiasKO [112]

(((x + y) / 3) + (1 / x)) / (5 + (15 / x))
The best way is to make it one fraction.
Multiply by ((3x/3x) / (3x/3x)) to remove the other fractions.
((x(x + y)) + 3(1)) / (5(3x) + 3(15))
(x^2 + xy + 3) / (15x + 45)
Then factor to simplify
(x^2 + xy + 3) / (3(x + 15))

4 0

3 years ago

PLS HELP I DONT REMEMBER HOW TO DO THIS

Gala2k [10]

Answer:

oh lord...ill try my best

Step-by-step explanation:

Let's start b writing down coordinates of all points:

A(0,0,0)

B(0,5,0)

C(3,5,0)

D(3,0,0)

E(3,0,4)

F(0,0,4)

G(0,5,4)

H(3,5,4)

a.) When we reflect over xz plane x and z coordinates stay same, y coordinate changes to same numerical value but opposite sign. Moving front-back is moving over x-axis, moving left-right is moving over y-axis, moving up-down is moving over z-axis.

A(0,0,0)

Reflecting

A(0,0,0)

B(0,5,0)

Reflecting

B(0,-5,0)

C(3,5,0)

Reflecting

C(3,-5,0)

D(3,0,0)

Reflecting

D(3,0,0)

b.)

A(0,0,0)

Moving

A(-2,-3,1)

B(0,-5,0)

Moving

B(-2,-8,1)

C(3,-5,0)

Moving

C(1,-8,1)

D(3,0,0)

Moving

im sorry think this is wrong I TRIED MY BESTTTT!!!!!!!! give me credit this is tiring

6 0

3 years ago

The dye dilution method is used to measure cardiac output with 3 mg of dye. The dye concentrations, in mg/L, are modeled by c(t)

Lemur [1.5K]

Answer:

Cardiac output: $F=0.055 L\s$

Step-by-step explanation:

Given : The dye dilution method is used to measure cardiac output with 3 mg of dye.

To Find : Find the cardiac output.

Solution:

Formula of cardiac output: $F=\frac{A}{\int\limits^T_0 {c(t)} \, dt}$ ---1

A = 3 mg

$\int\limits^T_0 {c(t)} \, dt =\int\limits^{10}_0 {20te^{-0.06t}} \, dt$

Do, integration by parts

$[\int{20te^{-0.6t}} \, dt]^{10}_0=[20t\int{e^{-0.6t} \,dt}-\int[\frac{d[20t]}{dt}\int {e^{-0.6t} \, dt]dt]^{10}_0$

$[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20}{0.6}\int {e^{-0.6t} \,dt]^{10}_0$

$[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20e^{-0.6t}}{(0.6)^2}]^{10}_{0}$

$[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-200e^{-6}}{0.6}+\frac{20e^{-6}}{(0.6)^2}]+\frac{20}{(0.60^2}$

$[\int{20te^{-0.6t}} \, dt]^{10}_0=\frac{20(1-e^{-6}}{(0.6)^2}-\frac{200e^{-6}}{0.6}$

$[\int{20te^{-0.6t}} \, dt]^{10}_0\sim {54.49}$

Substitute the value in 1

Cardiac output: $F=\frac{3}{54.49}$

Cardiac output: $F=0.055 L\s$

Hence Cardiac output: $F=0.055 L\s$

4 0

3 years ago

Determine whether the following statements are true and give an explanation or a counter example. a. The Trapezoid Rule is exact

Norma-Jean [14]

Answer:

statement is TRUE

statement is FALSE

statement is TRUE

Step-by-step explanation:

(a)

By using the Trapezoidal Rule, the definite integral can be computed by applying linear interpolating formula on each sub interval, and then sum-up them, to get the value of the integral

So, in computing a definite integral of a linear function, the approximated value occurred by using Trapezoidal Rule is same as the area of the region. Thus, the value of the definite integral of a linear function is exact, by using the Trapezoidal Rule.

Therefore, the statement is TRUE

(b) Recollect that for each rule of both the midpoint and trapezoidal rules, the number of sub-internals, n increases by a factor of a. then the error decreases by a factor of a^2.

So, for the midpoint rule, the number of sub-intervals, n is increased by a factor of 3, then the error is decreased by a factor of 32 = 9, not 8. Therefore, the statement is FALSE

(c) Recollect that for each rule of both the midpoint and trapezoidal rules, the number of sub-internals, n increases by a factor of a. then the error decreases by a factor of a^2.

So, for the trapezoidal rule, the number of sub-internals, n is increased by a factor of 4. then the error is decreased by a factor of 42 = 16

Therefore, the statement is TRUE

3 0

3 years ago