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REY [17]
3 years ago
11

Need help with this plzzzz

Mathematics
2 answers:
sweet [91]3 years ago
8 0
I would say D my dear friend from what i see from the patterns hope it helps :) and sorry if its wrong i tried lol
jasenka [17]3 years ago
6 0
C Because the Ratio is 6:10 or 12:20 or 18:30 or 24:40 or 30:50

In this example you double both sides of the ratio, 
and then you keep doing it until you find the right ratio that works as an answer choice.

Tell me if i dont make any sense.

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Please help, this makes no sense to me
aev [14]

Answer:

7/11

Step-by-step explanation:

Let X equal the decimal number

Equation 1:

X=0.63¯¯¯¯¯

With 2 digits in the repeating decimal group,

create a second equation by multiplying

both sides by 102 = 100

Equation 2:

100X=63.63¯¯¯¯¯

Subtract equation (1) from equation (2)

100XX99X===63.63...0.63...63

We get

99X=63

Solve for X

X=6399

Find the Greatest Common Factor (GCF) of 63 and 99, if it exists, and reduce the fraction by dividing both numerator and denominator by GCF = 9,

63÷999÷9=711

Therefore

X=711

In conclusion,

0.63¯¯¯¯¯=711

used

https://www.calculatorsoup.com/calculators/math/decimal-to-fraction-calculator.php

6 0
3 years ago
On Monday, 365 students went on a
Tamiku [17]
45 on each bus. 365 minus 5 in cars is 360. Divided by 8 to get 45.
6 0
2 years ago
Read 2 more answers
3x(2-x)+2x(x-1)=5x(x+3)
mylen [45]

Step-by-step explanation:

6x - 3x^2 + 2x^2 -2x = 5x^2 + 15x

4x -x^2 = 5x^2 + 15x

-11x - 6x^2 = 0

-6x^2 - 11x = 0

use the quadratic formula

(-b±√(b²-4ac))/(2a)

3 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
2x2-3 evaluate each expression. Show your work
nirvana33 [79]

Answer:

1

Step-by-step explanation:

2x2 = 4

4-3=1

5 0
3 years ago
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