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3 years ago

Please help thank you so much

Mathematics

2 answers:

Alisiya [41]3 years ago

6 0

Answer:

Hello! Your answer would be, A)

Step-by-step explanation:

Hope I helped! Brainiest plz!♥ Have a nice afternoon. Hope you make a 100%! -Amelia

poizon [28]3 years ago

3 0

Answer:

it would be 2 or 1200$

Step-by-step explanation:

15% OF 1400$ is 210$.

1400 - 210 = 1190 almost 1200 so most likely 1200.

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Please I really need help! I will mark you as brainliest!

iogann1982 [59]

Answer:

-10 / -5 / -1 / 0 / 1 / 5 / 10

Step-by-step explanation:

I hope this helps!

5 0

3 years ago

Three times what equals 61.50

sergey [27]

Simply divide 61.50 by 3. You should be able to reach the answer that way

3 0

3 years ago

Read 2 more answers

Write the equation in slope-intercept form:<br> 3y = 2x +15

Sergeu [11.5K]

Answer:

y = 2/3x + 5

Step-by-step explanation:

Hope this helps! :)

3 0

3 years ago

The amount of a sample remaining after t days is given by the equation P(1) - Al2where A is the initial amount of

Bezzdna [24]

Answer:

The answer is actually 74 days. I know how to algebraically solve this, however I am unsure how to transpose my work onto the computer.

Step-by-step explanation:

I can also tell you that edgenuity will say 74 is correct.

7 0

3 years ago

Solve the initial value problems:<br> 1/θ(dy/dθ) = ysinθ/(y^2 + 1); subject to y(pi) = 1

ladessa [460]

Answer:

$-\theta cos\thsta+sin\theta = \frac{y^{2} }{2} + ln y + \pi - \frac{1}{2}$

Step-by-step explanation:

Given the initial value problem $\frac{1}{\theta}(\frac{dy}{d\theta} ) =\frac{ ysin\theta}{y^{2}+1 } \\$ subject to y(π) = 1. To solve this we will use the variable separable method.

Step 1: Separate the variables;

$\frac{1}{\theta}(\frac{dy}{d\theta} ) =\frac{ ysin\theta}{y^{2}+1 } \\\frac{1}{\theta}(\frac{dy}{sin\theta d\theta} ) =\frac{ y}{y^{2}+1 } \\\frac{1}{\theta}(\frac{1}{sin\theta d\theta} ) = \frac{ y}{dy(y^{2}+1 )} \\\\\theta sin\theta d\theta = \frac{ (y^{2}+1)dy}{y} \\integrating\ both \ sides\\\int\limits \theta sin\theta d\theta =\int\limits \frac{ (y^{2}+1)dy}{y} \\-\theta cos\theta - \int\limits (-cos\theta)d\theta = \int\limits ydy + \int\limits \frac{dy}{y}$

$-\theta cos\thsta+sin\theta = \frac{y^{2} }{2} + ln y +C\\Given \ the\ condition\ y(\pi ) = 1\\-\pi cos\pi +sin\pi = \frac{1^{2} }{2} + ln 1 +C\\\\\pi + 0 = \frac{1}{2}+ C \\C = \pi - \frac{1}{2}$

The solution to the initial value problem will be;

$-\theta cos\thsta+sin\theta = \frac{y^{2} }{2} + ln y + \pi - \frac{1}{2}$

5 0

3 years ago