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Irina-Kira [14]

3 years ago

9

The cost to mall a package is $5 for the first 2 pounds and 40 cents for each additional ounce. Which of the following functions

represents the cost to mall a package if x is the number of ounces over 2 pounds?
plz hurry

1 answer:

erica [24]3 years ago

5 0

Answer:

Step-by-step explanation:

5 times 2 equales 10 divided by 40 eqauls 4 times 2 equals 8

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Solve (X + 6)(x - 4) = -16<br> O {-4, 6)<br> O {-4, 2]<br> O (-12, 22)

masya89 [10]

Answer:

x=2 or x=−4

Step-by-step explanation:

Let's solve your equation step-by-step.

(x+6)(x−4)=−16

Step 1: Simplify both sides of the equation.

x2+2x−24=−16

Step 2: Subtract -16 from both sides.

x2+2x−24−(−16)=−16−(−16)

x2+2x−8=0

Step 3: Factor left side of equation.

(x−2)(x+4)=0

Step 4: Set factors equal to 0.

x−2=0 or x+4=0

x=2 or x=−4

6 0

3 years ago

What is 25%(percent) Of 42

andrezito [222]

25% of 42=25 divided by 100 x 42
25\100 x 42=10.5

3 0

3 years ago

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PLEASE HELP I MIGHT FAIL MATH. For a charity event, raffle tickets are printed such that they have a vowel (excluding y) as the

Radda [10]

Answer:

Step-by-step explanation:

Remark

You better be issuing a very small number of tickets.

Tickets vowels: a e i o u

Primes < 10: 2 3 5 7

Probability(e and 5) = 1/5 * 1/4 = 1/20

Answer: 1/20

Answer:0.05

You should probabily give 0.05 as your answer.

8 0

2 years ago

Dana’s parents add $2 to her savings account for every $20 she puts in the account.

a_sh-v [17]

Answer:

A) p = $\frac{1}{10}$ d

Step-by-step explanation:

20(which is d) x $\frac{1}{10}$ = 2 (which is p)

cross checking always helps!

7 0

3 years ago

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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