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scZoUnD [109]
3 years ago
15

Hewp..Pls

Mathematics
1 answer:
timofeeve [1]3 years ago
7 0

Answer:

2y+4x=12

2*1+4*8=12

2+32=12

35=12

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Given that h(x) = 3x −19, find the value of x that makes h(x) = 71.
rosijanka [135]

Keywords:

<em>Equation, variable, value, clear </em>

For this case we have an equation with a variable of the form y = h (x). Where h (x) = 3x-19. Given the value ofh (x) = 71, we want to find the value of the variable "x". So, we have:

h (x) = 3x-19\\71 = 3x-19

We must clear "x", for this, we add "19" to both sides of the equation:

71 + 19 = 3x-19 + 19\\90 = 3x

We divide between "3" on both sides of the equation:

\frac {90} {3} = \frac {3x} {3}\\30 = x

Thus, the value of the variable "x" is 30.

Answer:

x = 30

Option A

6 0
3 years ago
Read 2 more answers
Emie ordered 3 buckets of fried chicken at $11.95 each 5 pints and Cole slaw at $2.19 each. The sales tax was $3.04. About how m
Oliga [24]

Answer:

Around $49 the actual answer is $49.84

3 0
3 years ago
What is the solution to the equation 37 x = 9 x + 4 ?<br> -7<br> − 1/7<br> 1/7<br> 7
natali 33 [55]

▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

let's solve for x ~

  • 37x =  9x + 4

  • 37x - 9x = 4

  • 28x = 4

  • x =  \dfrac{4}{28}

  • x =  \dfrac{1}{7}

5 0
2 years ago
Read 2 more answers
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
Emily and Kate were at a cheerleading competition this fall.They cheered 12 rounds. Each round lasted 8.24 minutes. Calculate th
trapecia [35]

Answer:

98.88

Step-by-step explanation:

Given: They cheered 12 rounds. Each round = 8.24

To find the total number, multiply how much one round is with the total number of rounds.

  8.24

×     12

-----------

  1648

+ 8240

----------

98.88

Each girl cheered 98.88 minutes. If you need this estimated, the answer would be 99 minutes.

4 0
3 years ago
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