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2 years ago

Which is the net for the figure shown?

Mathematics

1 answer:

Nataliya [291]2 years ago

3 0

Answer:

The third one, with the sideways "z" shape.

Step-by-step explanation:

When folded, this net shape will become the new figure.

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Helppppppppppppppp lol

Sladkaya [172]

The given expression is:
(x/2) + 5

Let's look at each terms separately:
x/2 means that you you're getting half the value of a number
+5 means that you are adding 5
Combining both, you can conclude that you are adding 5 to half a certain given number.

Comparing this with the choices, the correct choice would be:
C. five more than half a number.

5 0

3 years ago

Read 2 more answers

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:

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3 years ago

Heather's school is switching from block scheduling to a more traditional schedule. She is trying to determine how the parents a

k0ka [10]

Had to look for the options for this question and here is my answer.
Based on the given scenario above regarding Heather school's scheduling, I can say that the question that would be the most appropriate for her to ask is "How do you feel about your child having to take summer school in order to graduate based on this traditional schedule?" Hope this helps.

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3 years ago

Which of the following correctly describes the interval shown?

adoni [48]

B is the correct one

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3 years ago

I need help with both of em

bazaltina [42]

The answer to the first one is C) 16. The second one is A) 3,072

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3 years ago

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