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3 years ago

Anyone??? Need help!!!

Mathematics

2 answers:

DochEvi [55]3 years ago

8 0

Your answer for this problem =45

hope this helps:)

yaroslaw [1]3 years ago

4 0

Answer:

x=45

Step-by-step explanation:

We know that this is a right isosceles triangle because it has a right angle (square) and a triangle has to add up to 180 degrees so 90+x+x=180

90+2x=180

2x=90

x=45

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Let f(x)=6(2)x−1+4

atroni [7]

G(x)=6(2)x+7+4 i think that is it

5 0

3 years ago

Indentify which values are solutions of 3x - 5 ≤ -2

azamat

Answer:

x ≤ 1

Step-by-step explanation:

First, add 5 to both sides (we want to isolate x).

3x ≤ 3

Now, we divide by 3.

x ≤ 1

That's the answer! All values equal to or less than 1 are solutions of 3x - 5 ≤ -2.

Hope this helps

8 0

3 years ago

If it costs $0.15 per square inch for the cardboard label, then what is the cost for plan B?

IceJOKER [234]

Answer:

a box of plan B cost roughly $50.00

it is pretty expensive but you don't need an I.D.

Step-by-step explanation:

6 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

Last one guys I promise lol. btw helpp

Marianna [84]

63 dollars 3 hours

21 dollars 1 hour

8 0

3 years ago