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OLEGan [10]
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
Mathematics
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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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}

Median: 63, 66, 67, 68, 73, 74, 76, 78, 80, 83, 86, 87, 88, 89, 89, 89, 93, 93, 95, 99, 100

\frac{86+83}{2}= 169/2

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Step-by-step explanation:

8 0
2 years ago
Select the correct answer.<br> Write (26 + 161) - (26 - 71 as a complex number in standard form.
slavikrds [6]

Answer:

23i

Step-by-step explanation:

EASy

4 0
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

180-77=103

B. Lines also are 180, so 103-180=77

C. Interior angles of a triangle add up to 180, so 180/3 is how you get your answer.

3 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
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