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3 years ago

Ray had 60 football cards last year.This year he has 200. Find the percent of change

Mathematics

1 answer:

bulgar [2K]3 years ago

7 0

Answer:70%

Step-by-step explanation: no. of cards last year=60

no. of cards presently=200

increase in no. of cards=200-60

=140

%change=140/200 x 100

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Brainlyst for the best answer thank you!

densk [106]

Answer: B

We all know that area= LxW
The width that is being shown is 8ft. The length is 20ft. Therefore, you multiply 8 x 20 which gives you 160ft squared. Hope this this helps :)

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3 years ago

If W(-10, 4), X(-3, -1), and Y(-5, 11) classify ΔWXY by its sides. Show all work to justify your answer.

solniwko [45]

Given:

The vertices of ΔWXY are W(-10, 4), X(-3, -1), and Y(-5, 11).

To find:

Which type of triangle is ΔWXY by its sides.

Solution:

Distance formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using distance formula, we get

$WX=\sqrt{(-3-(-10))^2+(-1-4)^2}$

$WX=\sqrt{(-3+10)^2+(-5)^2}$

$WX=\sqrt{(7)^2+(-5)^2}$

$WX=\sqrt49+25}$

$WX=\sqrt{74}$

Similarly,

$XY=\sqrt{\left(-5-\left(-3\right)\right)^2+\left(11-\left(-1\right)\right)^2}=2\sqrt{37}$

$WY=\sqrt{\left(-5-\left(-10\right)\right)^2+\left(11-4\right)^2}=\sqrt{74}$

Now,

$WX=WY$

So, triangle is an isosceles triangles.

and,

$WX^2+WY^2=(\sqrt{74})^2+(\sqrt{74})^2$

$WX^2+WY^2=74+74$

$WX^2+WY^2=148$

$WX^2+WY^2=(2\sqrt{37})^2$

$WX^2+WY^2=WY^2$

So, triangle is right angled triangle.

Therefore, the ΔWXY is an isosceles right angle triangle.

3 0

2 years ago

Click an item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the corre

guajiro [1.7K]

Answer:

Step-by-step explanation:

Pythagorean formula

a²+b²=c²

in other words, sidea²+sideb²=hypotenuse

8²+15²=c²

64+225=c²

289=c²

√289=√c²

17=c

6 0

3 years ago

Find the volume with the explanation if possible <3

stellarik [79]

Answer: 905 cu. in.

Step-by-step explanation:

volume = (4/3) * π * r³

= (4/3) * π * 216

7 0

3 years ago

Consider the differential equation y'' − y' − 20y = 0. Verify that the functions e−4x and e5x form a fundamental set of solution

KIM [24]

Answer:

Therefore $e^{-4x}$ and $e^{5x}$ are fundamental solution of the given differential equation.

Therefore $e^{-4x}$ and $e^{5x}$ are linearly independent, since $W(e^{-4x},e^{5x})=9e^x\neq 0$

The general solution of the differential equation is

$y=c_1e^{-4x}+c_2e^{5x}$

Step-by-step explanation:

Given differential equation is

y''-y'-20y =0

Here P(x)= -1, Q(x)= -20 and R(x)=0

Let trial solution be $y=e^{mx}$

Then, $y'=me^{mx}$ and $y''=m^2e^{mx}$

$\therefore m^2e^{mx}-m e^{mx}-20e^{mx}=0$

$\Rightarrow m^2-m-20=0$

$\Rightarrow m^2-5m+4m-20=0$

$\Rightarrow m(m-5)+4(m-5)=0$

$\Rightarrow (m-5)(m+4)=0$

$\Rightarrow m=-4,5$

Therefore the complementary function is = $c_1e^{-4x}+c_2e^{5x}$

Therefore $e^{-4x}$ and $e^{5x}$ are fundamental solution of the given differential equation.

If $y_1$ and $y_2$ are the fundamental solution of differential equation, then

$W(y_1,y_2)=\left|\begin{array}{cc}y_1&y_2\\y'_1&y'_2\end{array}\right|\neq 0$

Then $y_1$ and $y_2$ are linearly independent.

$W(e^{-4x},e^{5x})=\left|\begin{array}{cc}e^{-4x}&e^{5x}\\-4e^{-4x}&5e^{5x}\end{array}\right|$

$=e^{-4x}.5e^{5x}-e^{5x}.(-4e^{-4x})$

$=5e^x+4e^x$

$=9e^x\neq 0$

Therefore $e^{-4x}$ and $e^{5x}$ are linearly independent, since $W(e^{-4x},e^{5x})=9e^x\neq 0$

Let the the particular solution of the differential equation is

$y_p=v_1e^{-4x}+v_2e^{5x}$

$\therefore v_1=\int \frac{-y_2R(x)}{W(y_1,y_2)} dx$

and

$\therefore v_2=\int \frac{y_1R(x)}{W(y_1,y_2)} dx$

Here $y_1= e^{-4x}$ , $y_2=e^{5x}$ , $W(e^{-4x},e^{5x})=9e^x$ ,and $R(x)=0$

$\therefore v_1=\int \frac{-e^{5x}.0}{9e^x}dx$

and

$\therefore v_2=\int \frac{e^{5x}.0}{9e^x}dx$

The the P.I = 0

The general solution of the differential equation is

$y=c_1e^{-4x}+c_2e^{5x}$

7 0

3 years ago