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3 years ago

5

Determine whether the polynomial is a difference of squares and if it is, factor it.

1 answer:

GREYUIT [131]3 years ago

8 0

Remember these three formulas (really, learn these by heart):

${(a + b)}^{2} = a^{2} + 2ab + b^{2}$

${(a - b)}^{2} = a^{2} - 2ab + b^{2}$

$(a + b)(a - b) = a^{2} - b^{2}$

The stated formula is a manifestation of the last of these three. So indeed,

$(y + 15)(y - 15) = y^{2} - 225$

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The value of 34 + 5.5 = . (1 point) 교

zheka24 [161]

Answer:

39.5

Step-by-step explanation:

you have to add 34 and then add 5.5 to it and then you'll get 39.5

8 0

2 years ago

Which equation is the slope intercept form of this equation?

AnnZ [28]

Answer:

C

Step-by-step explanation:

Y - 4 = -2x + 10 Multiply -2 to the inside the brackets

Y = -2x + 14 Add the 4 over

7 0

2 years ago

Dice Rolls: If you roll a pair of fair dice, what is the probability of each of the following? (round all answers to 4 decimal p

maria [59]

Answer:

(A)= 0

(B)= 1/9

(C)= 1/36

4 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

Double time pay is twice the regular time pay rate. If Rita, a nurse, earns $24 per hour for her regular pay rate and she is pai

Kitty [74]

Answer:24r+48h= $1248

Step-by-step explanation:

24*36+48*8

864 + 384 = 1248

Not sure if it's correct I'm not brilliant with equations xx

8 0

3 years ago