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3 years ago

Solve the system of equations below algebraically. 2x + 3y = 6 - 5x+2y=4

Mathematics

1 answer:

trapecia [35]3 years ago

4 0

Answer: The solution is (0,2)

Step-by-step explanation:

2x +3y = 6 We will solve the equations using the elimination method

-5x +2y= 4 but first we need to eliminate one of the variables and we will multiply the first equation by 5 and multiply the second equation by 2 to eliminate the x variable.

5(2x + 3y) =6(5) new equation : 10x +15y = 30

2(-5x + 2y) = 4(2) new equation ; -10x +4y =8

Now we have to new equations so we will eliminate the x term by adding

10x +15y = 30

-10x +4y =8

19y = 38 divide both sides by 19.

y= 2 Now using the solution for y plot it into one of the new equations and solve for x

10x + 15(2) = 30

10x + 30 = 30

-30 -30

10x = 0

x= 0

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Use the fraction strips to compare 2/4 and 5/8. use the drop-down menus to explain your comparison.

tia_tia [17]

Answer:

$\dfrac{4}{5}$

Step-by-step explanation:

We are given 2 fractional numbers:

$1.\ \dfrac{2}4\\2.\ \dfrac{5}8$

We have to use fraction strips to compare to the fractional numbers.

Let we are Comparing $\frac{2}{4}$ with the length of $x$ number of $\frac{1}4$ sections.

i.e.

$\dfrac{2}{4} = x \times \dfrac{1}{4}\\\Rightarrow x = \dfrac{2 \times 4}{4}\\\Rightarrow x = 2$

Let we are Comparing $\frac{5}{8}$ with the length of $y$ number of $\frac{1}8$ sections.

i.e.

$\dfrac{5}{8} = y \times \dfrac{1}{8}\\\Rightarrow y = \dfrac{5 \times 8}{8}\\\Rightarrow y = 5$

Now, let us have a look at 3rd part of question:

The sections of 2/4 is _____ the length of 5/8. Therefore, 2/4 < 5/8

Let the answer be $z$ .

So, the equation becomes:

$\dfrac{2}{4} = z \times \dfrac{5}{8}\\\Rightarrow z = \dfrac{2 \times 8}{4 \times 5}\\\Rightarrow z = \dfrac{2 \times 2}{5}\\\Rightarrow z = \dfrac{4}{5}$

So, the answers are:

$\dfrac{4}{5}$

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3 years ago

A line tangent to the curve f(x)=1/(2^2x) at the point (a, f(a)) has a slope of -1. What is the x-intercept of this tangent?

kirza4 [7]

Answer:

x-intercept = 0.956

Step-by-step explanation:

You have the function f(x) given by:

$f(x)=\frac{1}{2^{2x}}$ (1)

Furthermore you have that at the point (a,f(a)) the tangent line to that point has a slope of -1.

You first derivative the function f(x):

$\frac{df}{dx}=\frac{d}{dx}[\frac{1}{2^{2x}}]$ (2)

To solve this derivative you use the following derivative formula:

$\frac{d}{dx}b^u=b^ulnb\frac{du}{dx}$

For the derivative in (2) you have that b=2 and u=2x. You use the last expression in (2) and you obtain:

$\frac{d}{dx}[2^{-2x}]=2^{-2x}(ln2)(-2)$

You equal the last result to the value of the slope of the tangent line, because the derivative of a function is also its slope.

$-2(ln2)2^{-2x}=-1$

Next, from the last equation you can calculate the value of "a", by doing x=a. Furhtermore, by applying properties of logarithms you obtain:

$-2(ln2)2^{-2a}=-1 \\\\2^{2a}=2(ln2)=1.386\\\\log_22^{2a}=log_2(1.386)\\\\2a=\frac{log(1.386)}{log(2)}\\\\a=0.235$

With this value you calculate f(a):

$f(a)=\frac{1}{2^{2(0.235)}}=0.721$

Next, you use the general equation of line:

$y-y_o=m(x-x_o)$

for xo = a = 0.235 and yo = f(a) = 0.721:

$y-0.721=(-1)(x-0.235)\\\\y=-x+0.956$

The last is the equation of the tangent line at the point (a,f(a)).

Finally, to find the x-intercept you equal the function y to zero and calculate x:

$0=-x+0.956\\\\x=0.956$

hence, the x-intercept of the tangent line is 0.956

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Step 1: Substitute the value of x into the equation
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Step 2: Solve the equation
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Determine the singular points of the given differential equation. Classify each singular point as regular or irregular. (Enter y

Zina [86]

Answer:

Step-by-step explanation:

The given differential equation is:

$x^3y'' + 2x^2y' + 4y$

the main task here is to determine the singular points of the given differential equation and Classify each singular point as regular or irregular.

So, for a regular singular point ; $x=x_o$ is located at the first power in the denominator of P(x) likewise at the Q(x) in the second power of the denominator. If that is not the case, then it is termed as an irregular singular point.

Let first convert it to standard form by dividing through with x³

$y'' + \dfrac{2x^2y'}{x^3} + \dfrac{4y}{x^3} =0$

$y'' + \dfrac{2y'}{x} + \dfrac{4y}{x^3} =0$

The standard form of the differential equation is :

$\dfrac{d^2y}{dy} + P(x) \dfrac{dy}{dx}+Q(x)y =0$

Thus;

$P(x) = \dfrac{2}{x}$

$Q(x) = \dfrac{4}{x^3}$

The zeros of $x,x^3$ is 0

Therefore , the singular points of above given differential equation is 0

Classify each singular point as regular or irregular.

Let p(x) = xP(x) and q(x) = x²Q(x)

p(x) = xP(x)

p(x) = $x*\dfrac{2}{x}$

p(x) = 2

q(x) = x²Q(x)

q(x) = $x^2 * \dfrac{4}{x^3}$

q(x) = $\dfrac{4}{x}$

The function (f) is analytic if at a given point a it is represented by power series in x-a either with a positive or infinite radius of convergence.

Thus ; from above; we can say that q(x) is not analytic at x = 0

$Q(x) = \dfrac{4}{x^3}$ do not satisfy the condition,at most to the second power in the denominator of Q(x).

Thus, the point x =0 is an irregular singular point

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