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il63 [147K]
3 years ago
15

Solving Quadratic Equations A B C D

Mathematics
1 answer:
mixas84 [53]3 years ago
8 0
I believe that it’s B.
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An equation is shown below: 6(2x - 11) 15 = 21 Write the steps you will use to solve the equation and explain each step.
kolbaska11 [484]

Answer:

x=6

Step-by-step explanation:

6(2x -11) 15 =21

6 × 2x = 12x

6 × -11 = -66

12x -66 + 15 = 21

-66 + 15 = -51

12x -51 = 21

-51 + 51 = 0

21 + 51 = 72

12x = 72

12x ÷ 2 = 0

72 ÷ 12 = 6

x = 6

3 0
2 years ago
3) Graph y=-x2+3 <br> State the vertex and axis of symmetry
klasskru [66]
Vertex (0,3)
Axis of symmetry x=0
8 0
3 years ago
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What is 3+x if x = 2. Give answer but do not give detailed explanation. Thank you.
goblinko [34]

Answer: 3 + x = 5.

Step-by-step explanation:

2 + 3 = 5, if 2 = x. Hope this helps :)

4 0
3 years ago
Read 2 more answers
5 is what percentage of 25? I’ll give brainliest:)
lord [1]

Hi there! :)

<h2>20%</h2><h2></h2>

To solve, we can simply divide and multiply by 100 to solve for the percentage:

5 / 25     = 5 ÷ 25 = 1/5 or 0.20

Multiply the decimal by 100 to solve for the percentage:

0.20 × 100 = 20%

3 0
3 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
3 years ago
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