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3 years ago

How do i make a proof for this?

Mathematics

1 answer:

Lyrx [107]3 years ago

6 0

You could do ~CKW=DLK and the reason is because the are both 90 degrees.
Also don't worry about this on the regents I took it and was asked one of these.

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I need help with this

ozzi

Answer:

Minimum

Step-by-step explanation:

The equation is a quadratic function meaning the the shape is a parabola. The sign of x^2 is + so the graph open upward. Thus the vertex is the minimum point on the graph.

5 0

2 years ago

Solve for the variable: 2x - 28 = 16

svlad2 [7]

2x = 44
44/2 = x
X= 22

4 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:

3 0

3 years ago

What is the area of a parallelogram that has a base of 12 3/4 in. and a height of 2 1/2 in.?

EleoNora [17]

The area would be 24 inches

4 0

3 years ago

I really need help with this. can someone please help me

Andreyy89

Michael owns a small paper company called Michael Scott Paper Co. He has 5 clients and needs to sell them each a specific amount of paper. They have each ordered 3 3/10 worth of boxes. He has to find out how many paper boxes he has to sell and how much extra paper (in tenths) he has total.

IF you want to turn it into a question, add:

How many paper boxes is he selling?

5 0

3 years ago