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2 years ago

Please help!!! geometry help needed! the other answer choices are 2x+2y = 166 and 2x + y = 173

Mathematics

1 answer:

Nadya [2.5K]2 years ago

8 0

Answer:

2x + y = 173

Step-by-step explanation:

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Explain how the words term and coefficient are used in talking about the algebraic expression.

cupoosta [38]

<h2>Answer:</h2>

The algebraic expressions are composed of several elements: Terms, Coefficients, Signs, Variables and Exponents. In this case we will focus on the first two.

The terms are composed of sign, coefficient, variable and exponent. The terms are separated from each other by the plus sign (+) or the minus sign (-).

For example, in the following expression:

2x-3x

We have two terms separated by the negative sign (-):

2x is the first term

-3x is the second term

Now, the coefficients are the numbers that multiply the variable. Note that if this coefficient is one (1) it is omitted .

For example in the following expression, the variable is x:

2x-3x+x

2 is the coefficient of x in the first term

-3 is the coefficient of x in the second term

1 is the coefficient of x in the third term, but is omitted

In order to understand this in a better way, see the figure attached, where the algebraic expression is composed by one term.

6 0

3 years ago

Find the solution of the given initial value problem in explicit form. y′=(1−5x)y2, y(0)=−12 Enclose numerators and denominators

Nata [24]

Answer:

The solution is $y=-\frac{12}{12x-30x^2+1}$ .

Step-by-step explanation:

A first order differential equation $y'=f(x,y)$ is called a separable equation if the function $f(x,y)$ can be factored into the product of two functions of $x$ and $y$ :

$f(x,y)=p(x)h(y)$

where $p(x)$ and $h(y)$ are continuous functions.

We have the following differential equation

$y'=(1-5x)y^2, \quad y(0)=-12$

In the given case $p(x)=1-5x$ and $h(y)=y^2$ .

We divide the equation by $h(y)$ and move $dx$ to the right side:

$\frac{1}{y^2}dy\:=(1-5x)dx$

Next, integrate both sides:

$\int \frac{1}{y^2}dy\:=\int(1-5x)dx\\\\-\frac{1}{y}=x-\frac{5x^2}{2}+C$

Now, we solve for $y$

$-\frac{1}{y}=x-\frac{5x^2}{2}+C\\-\frac{1}{y}\cdot \:2y=x\cdot \:2y-\frac{5x^2}{2}\cdot \:2y+C\cdot \:2y\\-2=2yx-5yx^2+2Cy\\y\left(2x-5x^2+2C\right)=-2\\\\y=-\frac{2}{2x-5x^2+2C}$