What is the x-intercept of this function?

Pls find the surface area for me. I need help ASAP!

melomori [17]

Answer:

About 90.84 in^2

Step-by-step explanation:

The formula for the surface area of a pyramid is

$lw+l\sqrt{\frac{w}{2}^2+h^2}+w\sqrt{\frac{l}{2}^2+h^2}$ , which for this pyramid is equal to about 90.84 in^2. Hope this helps!

3 0

3 years ago

Which number line shows the solution to fx-5 > 3?

sp2606 [1]

Answer:

8

Step-by-step explanation:

send the -5 to where the +3 is then you can now add the value together.

7 0

3 years ago

What is the common ratio of 40,12,3.6

slava [35]

Answer:

0.3

Step-by-step explanation:

Divide the second term (12) with the first term (40), and you'll get 0.3

This is correct because you get the terms in the correct order when multiplying:
(First term: 40. Common ratio: 0.3)
40 x 0.3 = 12
12 x 0.3 = 3.6

40, 12, 3.6

3 0

2 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

Calcula y comprueba las ecuaciones: porfavor alguienn la nesesito pliss a) 2X = 6 b) 10 + Z = 20 c) P + 9 = 11 d) 3X + 8 = + 29

Alexeev081 [22]

Answer:

a) x = 3

b) z = 10

c) p = 2

d) x = 7

e) u = 1

Step-by-step explanation:

a) 2x = 6

Despejamos x dividiendo por 2 a amabos lados de la eacuacion.

(2/2)x = 6/2

x = 3

Si remplazamos x en la ecuación original:

2(3)=6

6 = 6

Queda demostrado.

b) 10 + z = 20

Despejamos z restando 10 en amabos lados de la eacuacion.

10-10+z = 20-10

z = 10

Si remplazamos z en la ecuación original:

10 + 10=20

20 = 20

Queda demostrado.

c) p + 9 = 11

Despejamos p restando 9 en amabos lados de la eacuacion.

p + 9 - 9 = 11-9

p = 2

Si remplazamos p en la ecuación original:

2 + 9 = 11

11 = 11

Queda demostrado.

d) 3x + 8 = 29

Despejamos x restando 8 en amabos lados de la eacuacion y luego divideindo por 3 en ambos lados de la ecuación.

3x+8-8 = 29-8

3x = 21

(3/3)x = 21/3

x = 7

Si remplazamos x en la ecuación original:

3(7) + 8 = 29

21 + 8 = 29

29 = 29

Queda demostrado

e) 2u + 8 = 10

Despejamos u restando 8 en amabos lados de la eacuacion y luego divideindo por 2 en ambos lados de la ecuación.

2u+8-8 = 10-8

2x = 2

(2/2)x = 2/2

x = 1

Si remplazamos x en la ecuación original:

2(1) + 8 = 10

2 + 8 = 10

10 = 10

Queda demostrado

Espero te haya sido de ayuda!

5 0

2 years ago