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3 years ago

Witch number is a palindrome 6554 or 606

Mathematics

2 answers:

Andrei [34K]3 years ago

7 0

606 is your answer .

vodka [1.7K]3 years ago

3 0

606 is a palindrome.

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One third of the sum of a number and four

STatiana [176]

Answer:

$\frac{1}{3} (n+4)$

That is what the expression would be

4 0

3 years ago

C=59(F−32)

neonofarm [45]

I says every time you INCREASE by 1 degree Farenheit you go up 59 degrees C, so it is false. Same for III, if you go up or down 1 degree C that's not a change of 59 in degrees F.

That leaves B, which is right if you experiment with the math.

8 0

3 years ago

Noelle has set a goal to run more than 50 mi this month. So far, she has run 10 mi. Her average running speed is 5 mph.

Vladimir79 [104]

Answer:

To meet her goal, Noelle must run more than $8\ hours$ the rest of the month

Step-by-step explanation:

Let

h-------> the number of hours Noelle runs for the rest of the month

we know that

The inequality that represent the situation is

$10+5h>50$

Solve for h

Subtract 10 both sides

$5h> 50-10$

$5h> 40$

Divide by 5 both sides

$h> 8\ hours$

3 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

Is it A. B. C. or D.

podryga [215]

Your answer would be b i'm pretty sure

5 0

3 years ago