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2 years ago

2 ACTIVITY: Estimating a Percents

Mathematics

2 answers:

Kobotan [32]2 years ago

8 0

Answer: 25%

Step-by-step explanation:

We know that 20 divided by 5 is 4, so 5 is one-fourth of 20 or 25%.

Natalija [7]2 years ago

3 0

The answer should be 25%

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Find an equation of the line that satisfies the given conditions.

8090 [49]

Answer:

$x=8$

Step-by-step explanation:

Given:

The equation of the given line is $y=9$

A point on the unknown line is (8, 5).

The unknown line is perpendicular to the line $y=9$ .

A line of the form $y=a$ where 'a' is a constant is a line parallel to the x axis. The line has a constant value for 'y' irrespective of the value of 'x'. The slope of such lines are equal to 0. So, slope of the known line is 0.

Now, when two lines are perpendicular, the product of their slopes is -1.

Let the slope of the unknown line be 'm', then:

$m\times 0 =-1\\m=-\frac{-1}{0}=$ undefined.

Therefore, the slope of the unknown line is undefined. A line parallel to y axis has undefined slope. So, the equation of the unknown line is of the form:

$x=b$ where, 'b' is a constant and is equal to the abscissa (x value) of the line.

As per question, (8, 5) lies on this line. Therefore, the value of 'x' is 8.

So, the equation of the unknown line is $x=8$

8 0

4 years ago

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Dennis went cross-country skiing for 6 hours

Ivanshal [37]

Dennis fell off a cliff and underwent surgery for his broken penis. i can’t answer this

5 0

3 years ago

Determine the solution, if it exists, for each system of linear equations. Verify your solution on the coordinate plane.

melisa1 [442]

Answer:

(x, y) = (2, 1)

Step-by-step explanation:

Adding the two equations gives ...

2y = 2

y = 1 . . . . . divide by 2

Subtracting the second equation from the first gives ...

0 = 6x -12

0 = x -2 . . . . divide by 6

2 = x

The solution is (x, y) = (2, 1).

The attached graph verifies this solution.

4 0

2 years ago

Find a particular solution to the differential equation y′′+2y′+y=2t2−t+3e−2t. y′′+2y′+y=2t2−t+3e−2t.

Jlenok [28]

$y''+2y'+y=2t^2-t+3e^{-2t}$

The characteristic equation is

$r^2+2r+1=(r+1)^2=0\implies r=-1$

so the characteristic solution is

$y_c=C_1e^{-t}+C_2te^{-t}$

For the particular solution, we can try looking for a solution of the form

$y_p=at^2+bt+c+de^{-2t}$
$\implies{y_p}'=2at+b-2de^{-2t}$
$\implies{y_p}''=2a+4de^{-2t}$

Substituting into the ODE, we have

$(2a+4de^{-2t})+2(2at+b-2de^{-2t})+(at^2+bt+c+de^{-2t})=2t^2-t+3e^{-2t}$
$at^2+(4a+b)t+(2a+2b+c)+de^{-2t}=2t^2-t+3e^{-2t}$
$\implies\begin{cases}a=2\\4a+b=-1\\2a+2b+c=0\\d=3\end{cases}\implies a=2,b=-9,c=14,d=3$

So the general solution to the ODE is

$y=y_c+y_p$
$y=C_1e^{-t}+C_2te^{-t}+2t^2-9t+14+3e^{-2t}$

4 0

3 years ago

Help?? I don't get it!

Gala2k [10]

The answer should be y=-4x^2

4 0

3 years ago

Read 2 more answers