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3 years ago

Correct the error a student made when solving the equation 4=-2(x-3). What is the correct solution?

Mathematics

2 answers:

Murrr4er [49]3 years ago

8 0

In the first step when the student did -2x-3 they didn’t make the sign positive. It should be 4=-2+6, not 4=-2-6

tankabanditka [31]3 years ago

3 0

Answer:

It should be positive 6

Step-by-step explanation:

The student multiplied a negative and a negative forgetting that it would become a positive. So in this case x would equal 1

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« 13. Given that R= {a,b,c}, S = {b, c, d}

Sphinxa [80]

Answer:

R= {a,b,c}

S = {b, c, d}

T = {c, d, e},

Step-by-step explanation:

R U S U T. ={a,b,c}U{b, c, d}U{c, d, e},={a,b,c,d,e}

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3 years ago

A circle has a sector with area 1/2pi and central angle of 1/9pi radians what is the area of the circle?

Mazyrski [523]

Answer:

Step-by-step explanation:

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3 years ago

Solve the system of equations.<br> y= 3x +5<br> y= 8x +3

Lady bird [3.3K]

Answer:

y=6.2 and x=0.4 or (0.4,6.2)

Step-by-step explanation:

Since both expressions are equal to y, we can set them equal to each other and solve for x:

3x+5=8x+3

5=5x+3

2=5x

2/5=x

0.4=x

Now to solve y:

y=8(2/5)+3

y=16/5+3

y=3.2+3

y=6.2

Check answers:

6.2=3(0.4)+5

6.2=1.2+5

6.2=6.2

6.2=8(0.4)+3

6.2=3.2+3

6.2=6.2

Answers work, so y=6.2 and x=0.4 or (0.4,6.2) as an ordered pair

5 0

3 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

100-99+98-97+96-95...2-1=

Simora [160]

100-99=1+98=99-97=2+96=98-95=3
2-1=1

5 0

3 years ago

Correct the error a student made when solving the equation 4=-2(x-3). What is the correct solution?​

Correct the error a student made when solving the equation 4=-2(x-3). What is the correct solution?