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topjm [15]
3 years ago
6

Plot a rectangle with vertices (–1, –4), (–1, 6), (3, 6), and (3, –4).

Mathematics
2 answers:
11111nata11111 [884]3 years ago
5 0

Answer: 4 units.

Step-by-step explanation:

Once each point is plotted, you get the rectangle attached.

You can observe in the figure that the base of the rectangle is its longer side.

 Since the length of the base of the rectangle goes from x=-1 to x=3, you can say that the lenght of the base is the difference between 3 and -1.

So, you need to subtract this two coordinates, getting that the lenght of the base of the rectangle attached is:

lenght_{(base)}=3-(-1)\\\\lenght_{(base)}=3+1\\\\lenght_{(base)}=4\ units

aliina [53]3 years ago
4 0

Answer: 4 units

Step-by-step explanation:

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Working alone, Henry takes 3 hours less than Mary to clean the carpets in the entire office. Working together, they can clean th
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It would take Henry 3 hours to clean the office carpets if Mary were not there to help

Step-by-step explanation:

Assume that it takes Henry t hours to clean the carpets

∵ Henry takes 3 hours less than Mary to clean the carpets

∵ Henry takes t hours

- That means add 3 to t to find Mary time

∴ Mary takes t + 3 hours

∵ The rate of Henry is \frac{1}{t}

∵ The rate of Mary is \frac{1}{t+3}

∵ Working together, they can clean the carpets in 2 hours

- Add the two rates and equate the answer by \frac{1}{2} (their rate together)

∴ \frac{1}{t}+\frac{1}{t+3}=\frac{1}{2}

Multiply the two denominators and multiply each numerator by the other denominator to add the two fractions of the left hand side

∵ \frac{(1)(t+3)}{t(t+3)}+\frac{(1)t}{t(t+3)}=\frac{1}{2}

∴  \frac{(t+3)}{t(t+3)}+\frac{t}{t(t+3)}=\frac{1}{2}

- Add the two fractions in the left hand side

∴ \frac{t+3+t}{t(t+3)}=\frac{1}{2}

∴  \frac{2t+3}{t^{2}+3t}=\frac{1}{2}

Use the cross multiplication

∵ (t² + 3t)(1) = 2(2t + 3)

∴ t² + 3t = 4t + 6

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∴ t² - t = 6

- Subtract 6 from both sides

∴ t² - t - 6 = 0

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∴ (t - 3)(t + 2) = 0

Equate each factor by 0 to find t

∵ t - 3 = 0

- Add 3 to both sides

∴ t = 3

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∵ t + 2 = 0

- Subtract 2 from both sides

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∴ t = 3

∴ Henry would take 3 hours to clean the carpets alone

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how do you rewrite : y=x^2-5x+9 in standard form using complying the square? I keep getting to : x^2-5x-25/4 but I’m not allowed
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I presume with this you meant to say you want this equation in <u>general form</u>, also known as vertex form, which is y = a(x - h)² + k, instead of <u>standard form</u>, which is y = ax² + bx + c, which is what the original equation is already in.

So with completing the square, you first have to isolate the x-terms. To do this, subtract 9 on both sides of the equation:

y-9=x^2-5x

Next, we want to make the right side of the equation a perfect square. To find the constant of this soon-to-be perfect square, divide the x-coefficient by 2 and square that quotient. In this case:

-5\div 2 = -\frac{5}{2}\\\\(-\frac{5}{2})^2=-\frac{5}{2}\times-\frac{5}{2}=\frac{25}{4}

Now add 25/4 on both sides of the equation:

y-9+\frac{25}{4}=x^2-5x+\frac{25}{4}

Now to combine -9/1 and 25/4, they must have the same denominator, and to find it you must find the LCD, or lowest common denominator, of 1 and 4. To find it, list the multiples of both and the lowest one they share is their LCD. In this case, the LCD is 4. Multiply both sides of -9/1 by 4/4 and then add the numerators up:

-\frac{9}{1}\times \frac{4}{4}=-\frac{36}{4}\\\\-\frac{36}{4}+\frac{25}{4}=-\frac{11}{4}\\\\y-\frac{11}{4}=x^2-5x+\frac{25}{4}

Next, we need to factor the right side of the equation. Using this format of a^2-2ab+b^2=(a-b)^2 , we can factor it as:

y-\frac{11}{4}=(x-\frac{5}{2})^2

Now lastly, add both sides by 11/4, and <u>your final answer will be y=(x-\frac{5}{2})^2+\frac{11}{4}</u>

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