All categories

2 years ago

Which represents a side length of a square that has an area of 450 square inches?

Mathematics

1 answer:

amm18122 years ago

5 0

The formula of an area of a square:

$A=a^2$

a - a side

We have $A=450\ in^2$

substitute:

$a^2=450\to a=\sqrt{450}\\\\a=\sqrt{225\cdot2}\\\\a=\sqrt{225}\cdot\sqrt2\\\\\boxed{a=15\sqrt2\ in}$

Used:

$\sqrt{a\cdot b}=\sqrt{a}\cdot\sqrt{b}$

You might be interested in

-12 + 18. = 70 complete the square

FromTheMoon [43]

-12 + 18 = 70 - 64 or(-12+18)+(6*12)-2

8 0

2 years ago

Fully explain the solution set to the equation below. 1x = -1

Lady bird [3.3K]

Answer:

The answer for x should equal: -1.

Step-by-step explanation:

To find x, we would divide, our final value, by 1 to get x. In other words, you would divide: -1 ÷ 1 to get -1 as our unknown number.

6 0

2 years ago

How many numbers greater than zero and less than 50 equal the product of two consecutive positive integers?

spin [16.1K]

Consecutive numbers would be like 2 and 3, or 7 and 8.

All we need to do is keep multiplying pairs of consecutive numbers until we get above 50.

1 × 2 = 2 (that's one.)
2 × 3 = 6 (two)
3 × 4 = 12 (three)
4 × 5 = 20 (four)
5 × 6 = 30 (five)
6 × 7 = 42 (six...)
7 × 8 = 56 > 50
We have a total of 6 numbers that equal the product of 2 consecutive intergers

7 0

2 years ago

3y''-6y'+6y=e*x sexcx

Simora [160]

From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

$3r^2-6r+6=0\iff r^2-2r+2=0$

which has roots at $r=1\pm i$ . This admits the two fundamental solutions

$y_1=e^x\cos x$
$y_2=e^x\sin x$

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

$y_p=u_1y_1+u_2y_2$

where

$u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$

and $W(y_1,y_2)$ is the Wronskian of the fundamental solutions. We have

$W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}$

and so

$u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx$
$u_1=\dfrac13\ln|\cos x|$

$u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx$
$u_2=\dfrac13x$

Therefore the particular solution is

$y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

so that the general solution to the ODE is

$y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

7 0

2 years ago

Help how can I Write the given standard equation of the quadratic in intercept form

olya-2409 [2.1K]

Answer:

f(x) = (x - 10) (x + 10)

Step-by-step explanation:

The equation looks like the difference of squares so 5 and 20 are out

-10 x -10 = 100

-10 x 10 = - 100

so (x - 10) (x + 10)

7 0

2 years ago