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Firlakuza [10]
3 years ago
13

PLEASE HELP THIS IS DUE IN 10 MINUTES!!!!!

Mathematics
2 answers:
andreev551 [17]3 years ago
4 0
<h3>Answer:</h3>

b1-h3, b2-h2, b3-h3.

Step-by-step explanation:

See the image.

Hope that helps! :)

(Can you please mark me as brainliest?)

Fynjy0 [20]3 years ago
3 0

Answer:

bottom B go with bottom h

and the other ones switch

Explanation:

hope this help's

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(2×100)+(9×10)+(6×1)+(3×1/10)+(8×1/100)= <br> In standard form
Sidana [21]
We know that standard form = the number that is not expanded.
This equation is expanded so standard form =

2*100=200
9*10=90
6*1=6
3*1/10= 0.3
8*1/100=0.08

200+90=290
290+6=296
296+0.3=296.3
296.3+0.08=296.38

Your answer is 296.38
Hope this helped!
5 0
3 years ago
Simiplify.<br> U^2 - 4 / U^2 - 2u
Papessa [141]
Remember
difference of 2 perfect squares
a^2-b^2=(a-b)(a+b)

and the undistributiv propeoryt
ab-ac=a(b-c)

so we have
\frac{u^{2}-4}{u^{2}-2u}
undistribute and factor
\frac{(u-2)(u+2)}{u(u-2)}
cancel out the one ( (u-2)/(u-2)=1)

\frac{(u+2)}{u} is result

3 0
3 years ago
Read 2 more answers
WILL MARK BRAINLIEST NEED IT ASAP
myrzilka [38]

Answer:

8

Step-by-step explanation:

6 x 9=54. 54+10=64. the square root of 64 = 8

8 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
Patriciodeposits $500 in the same account that pays a 1.5% simple interest he does not withdraw any money from the account any m
olga nikolaevna [1]

Answer: $537.5

Step-by-step explanation:

$500 at 1.5% for 5 years

p x  r x t = i

500 x 0.015 x 5= 37.5

37.5+500=537.5

7 0
3 years ago
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