All categories

3 years ago

Which equation is NOT a linear function?

Mathematics

2 answers:

Anni [7]3 years ago

4 0

Y=x^2-2

i got this one right

Paul [167]3 years ago

3 0

Answer:

$y = {x}^{2} - 2$

Step-by-step explanation:

because it's has two zeroes

hence it is known as the quadratic function

You might be interested in

I need help! <br> f(x)=-3x+6/4x-8

tia_tia [17]

Answer:

same

Step-by-step explanation:

8 0

2 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

2 years ago

The school store sells Midville High T-shirts for $12 each and Midville High shorts for $9 a pair. One afternoon they sold a tot

Flura [38]

Answer:
There will be 22 pairs of shorts were sold at a price of $9 each.

5 0

2 years ago

Please help with algebra 2 logarithms. brainliest if you explain and worth 25 points

Ronch [10]

Answer: 3log5(2) - 1 or ~0,29203

Step-by-step explanation:

2log5(4) - log5(10)

log5(4^2) - log5(10)

log5(4^2/10)

log5(16/10)

log5(8/5)

Log5(8) - log5(5)

log5(2^3) - 1

3log5(2) - 1

5 0

1 year ago

The authors of a paper presented a correlation analysis to investigate the relationship between maximal lactate level x and musc

ArbitrLikvidat [17]

Answer:

a) Sample correlation coefficient, r = 0.7411

bi) test statistic, t = 4.102

bii) P-value = 0.000736

Step-by-step explanation:

a) The formula for the sample correlation coefficient is given by the formula:

$r = \frac{S_{xy} }{\sqrt{S_{xx} S_{yy} }} }$

$S_{xx} = 2,648,130.357\\S_{yy} = 36.7376,\\S_{xy} = 7408.225$

$r = \frac{7408.225}{\sqrt{2648130.357*36.7376} }$

r = 0.7511

i) formula for the test statistic is given by the formula:

$t = \frac{r\sqrt{n-1} }{\sqrt{1 - r^{2} } }$

sample size, n = 4

$t = \frac{0.7511\sqrt{14-1} }{\sqrt{1 - 0.7511^{2} } }$

t = 4.102

ii) Degree of freedom, df = n -2

df = 14 -2

df = 12

The P-value is calculate from the degree of freedom and the test statistic using excel

P-value =(=TDIST(t,df,tail))

P-value = (=TDIST(4.1,12,1)

P-value = 0.000736

4 0

2 years ago