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3 years ago

New math question! (screenshot attached)

Mathematics

2 answers:

Virty [35]3 years ago

4 0

Answer:

the answer would be D like the other person said.

Lyrx [107]3 years ago

3 0

Answer:

Step-by-step explanation:

hi again

because the cornets show that -4 and -2 are also by -4 and -6 so that would =x+1 and x-5

hope this helps have a good day

love, The chris mac

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Step-by-step explanation:

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3 years ago

An item is regularly priced at $75 It is now priced at a discount of 60% off the regular price. What is the price now?

mel-nik [20]

Answer:

Price is $ 30

Step-by-step explanation:

Regular price = $75

Discount = 60%

First we find the 60% of 75

( 75 × 60 ) ÷ 100 = 45

Now this answer with regular price

75 - 45 = 30

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2 years ago

Read 2 more answers

Find the solution to the differential equation dB/dt+4B=20 with B(1)=30

natita [175]

Answer:

The solution of the differential equation is $B=5+25e^{-4t+4}$

Step-by-step explanation:

The differential equation $\frac{dB}{dt}+4B=20$ is a first order separable ordinary differential equation (ODE). We know this because a separable first-order ODE has the form:

$y'(t)=g(t)\cdot h(y)$

where g(t) and h(y) are given functions.

We can rewrite our differential equation in the form of a first-order separable ODE in this way:

$\frac{dB}{dt}+4B=20\\\frac{dB}{dt}=20-4B\\\frac{dB}{dt}=4(5-B)\\\frac{1}{5-B}\frac{dB}{dt}=4$

Integrating both sides

$\frac{1}{5-B}\frac{dB}{dt}=4\\\frac{1}{5-B}\cdot dB=4\cdot dt\\\\\int\limits {\frac{1}{5-B}} \, dB=\int\limits {4} \, dt$

The integral of left-side is:

$\int\limits {\frac{1}{5-B}} \, dB\\\mathrm{Apply\:u-substitution:}\:u=5-B\\\int\limits {\frac{1}{5-B}} \, dB=\int\limits {\frac{1}{u}} \, dB\\\mathrm{du=-dB}\\-\int\limits {\frac{1}{u}} \,du\\\mathrm{Use\:the\:common\:integral}:\quad \int \frac{1}{u}du=\ln \left(\left|u\right|\right)\\-\int\limits {\frac{1}{u}} \,du =-\ln \left|u\right|\\\mathrm{Substitute\:back}\:u=5-B\\-\ln \left|5-B\right|\\\mathrm{Add\:a\:constant\:to\:the\:solution}\\-\ln \left|5-B\right|+C$