You are 6 feet tall and cast a 10 foot shadow. At the same time of day, your friend

Asdhfj,..kjdsaDzfxhj,jhsaZ

RSB [31]

Answer:

ku hgkjfghdehfv bfd

Step-by-step explanation:

7 0

3 years ago

(10) Both red line and blue line trains left the station at 5:00 pm. If both

scoundrel [369]

Answer:

5:40

Step-by-step explanation:

This is a problem involving the least common difference.

If you know that the red and blue trains left at the same time at 5, you know that another red train will leave at 5:08. Another blue train at 5:10.

The way to solve this will be to write out the factors of 8 and 10 and find the smallest number that they overlap.

Red:

8, 16, 24, 32, 40, 48, 56, 64, 72, 80

Blue:

10, 20, 30, 40

You see that after 40 mnutes, they are both leaving the station again. After 40 minutes, at 5:40, they are both leaving.

6 0

3 years ago

Which expression is equivalent to w4−25w2+144

eduard

Sorry I’m just commenting to do it

8 0

3 years ago

A rancher wishes to build a fence to enclose a 2250 square yard rectangular field. Along one side the fence is to be made of hea

Bess [88]

Answer:

The least cost of fencing for the rancher is $1200

Step-by-step explanation:

Let x be the width and y the length of the rectangular field.

Let C the total cost of the rectangular field.

The side made of heavy duty material of length of x costs 16 dollars a yard. The three sides not made of heavy duty material cost $4 per yard, their side lengths are x, y, y. Thus

$C=4x+4y+4y+16x\\C=20x+8y$

We know that the total area of rectangular field should be 2250 square yards,

$x\cdot y=2250$

We can say that $y=\frac{2250}{x}$

Substituting into the total cost of the rectangular field, we get

$C=20x+8(\frac{2250}{x})\\\\C=20x+\frac{18000}{x}$

We have to figure out where the function is increasing and decreasing. Differentiating,

$\frac{d}{dx}C=\frac{d}{dx}\left(20x+\frac{18000}{x}\right)\\\\C'=20-\frac{18000}{x^2}$

Next, we find the critical points of the derivative

$20-\frac{18000}{x^2}=0\\\\20x^2-\frac{18000}{x^2}x^2=0\cdot \:x^2\\\\20x^2-18000=0\\\\20x^2-18000+18000=0+18000\\\\20x^2=18000\\\\\frac{20x^2}{20}=\frac{18000}{20}\\\\x^2=900\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{900},\:x=-\sqrt{900}\\\\x=30,\:x=-30$

Because the length is always positive the only point we take is $x=30$ . We thus test the intervals $(0, 30)$ and $(30, \infty)$

$C'(20)=20-\frac{18000}{20^2} = -25 < 0\\\\C'(40)= 20-\frac{18000}{20^2} = 8.75 >0$

we see that total cost function is decreasing on $(0, 30)$ and increasing on $(30, \infty)$ . Therefore, the minimum is attained at $x=30$ , so the minimal cost is

$C(30)=20(30)+\frac{18000}{30}\\C(30)=1200$

The least cost of fencing for the rancher is $1200

Here’s the diagram:

3 0

3 years ago

What values of c and d make the equation true?

saul85 [17]

Answer:

Third option.

Step-by-step explanation:

You need to cube both sides of the equation. Remember the Power of a power property:

$(a^m)^n=a^{mn$

$\sqrt[3]{162x^cy^5}=3x^2y(\sqrt[3]{6y^d})\\\\(\sqrt[3]{162x^cy^5})^3=(3x^2y(\sqrt[3]{6y^d}))^3\\\\162x^cy^5=27x^6y^36y^d$

According to the Product of powers property:

$(a^m)(a^n)=a^{(m+n)$

Then. simplifying you get:

$162x^cy^5=162x^6y^{(3+d)}$

Now you need to compare the exponents. You can observe that the exponent of "x" on the right side is 6, then the exponent of "x" on the left side must be 6. Therefore:

$c=6$

You can notice that the exponent of "y" on the left side is 5, then the exponent of "x" on the left side must be 5 too. Therefore "d" is:

$3+d=5\\d=5-3\\d=2$

4 0

3 years ago