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3 years ago

11

a system of two linear equations is graphed on a coordinate plane. if the system of equations has infinitely many solutions, whi

ch statement must be true

1 answer:

Tamiku [17]3 years ago

5 0

Answer:

Their slopes and y-intercepts are the same

Step-by-step explanation:

For example:

y=3x+5

y=3x+5

3x+5=3x+5

3x=3x

x=x

Thus, there's infinitely many solutions

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Answer:

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Step-by-step explanation:

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3 years ago

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Find the mean, median, and mode of the data set. Round to the nearest tenth. test scores on a math exam:

irinina [24]

Answer:

Mean: $\frac{89+93+76+89+68+80+89+83+88+87+63+86+73+74+67+93+68+95+66+99+78+100 }{22}$

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Step-by-step explanation:

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2 years ago

Select the correct answer.<br> Write (26 + 161) - (26 - 71 as a complex number in standard form.

slavikrds [6]

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Step-by-step explanation:

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3 years ago

HAVE A NICE DAY <br> WILL GIVE BRAINLST :)

nikitadnepr [17]

Answer:

A. 103 degrees

B. 77 degrees

C. 60 Degrees

Step-by-step explanation:

A. Triangles' interior angles add up to 180, so if you already have 2 angles you just need to put them together, and subtract them from 180.

47+30=77

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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