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Jet001 [13]
3 years ago
11

John has 12 pounds of dog food and is going to separate it into 3/4 pound portions.how many portions of dog food will he have

Mathematics
2 answers:
lisov135 [29]3 years ago
8 0
That would be 12 / 3/4   = 12 * 4/3  = 4*4 = 16 portions
Margarita [4]3 years ago
6 0

Answer:

16 portions.

Step-by-step explanation:

John has amount of dog food = 12 pounds

He wants to separate this food for his dogs weighing each portion = \frac{3}{4} pounds

Let x be the number of portions of dog food.

Then amount of dog food = \frac{3x}{4} pounds

Now we plug in the values in the equation

12 = \frac{3x}{4}

x = \frac{4\times 12}{3}

  = \frac{48}{3}

  = 16 portions

There will be 16 portions of the dog food weighing \frac{3}{4} pounds each.

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horsena [70]

Answer:

B. Perimeter of a square and

C. Side length of a square​

Step-by-step explanation:

if n= side length of square then

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  • diagonal length of a square is \sqrt{2} × n

Thus,

Perimeter of square can be expressed as

2\sqrt{2}×diagonal length of a square

Side length of a square can be expressed as

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but Area of square is

\frac{1}{\sqrt{2} }×n×diagonal length of a square

As a Result,  Area of square is <em>also dependent of the value n</em>, wheras in other cases it is <em>a proportion of diagonal length of a square</em>

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3 years ago
True or false.
valkas [14]

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Step-by-step explanation:no constant rate of change in y and x values

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3 years ago
Use the formula h(t)=−16t2+v0t+h0 , where v0 is the initial velocity and h0 is the initial height. How long will it take for the
Rama09 [41]

<em><u>Question:</u></em>

Britney throws an object straight up into the air with an initial velocity of 27 ft/s from a platform that is 10 ft above the ground. Use the formula h(t)=−16t2+v0t+h0 , where v0 is the initial velocity and h0 is the initial height. How long will it take for the object to hit the ground?

1s   2s   3s   4s

<em><u>Answer:</u></em>

It takes 2 seconds for object to hit the ground

<em><u>Solution:</u></em>

<em><u>The given equation is:</u></em>

h(t) = -16t^2 + v_0t+h_0

Initial velocity = 27 feet/sec

h_0 = 10\ feet

Therefore,

h(t) = -16t^2 +27t+10

At the point the object hits the ground, h(t) = 0

-16t^2 +27t+10 = 0\\\\16t^2-27t-10=0

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\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\mathrm{For\:}\quad a=16,\:b=-27,\:c=-10:\quad \\\\t=\frac{-\left(-27\right)\pm \sqrt{\left(-27\right)^2-4\cdot \:16\left(-10\right)}}{2\cdot \:16}\\\\t = \frac{27 \pm \sqrt{1369}}{32}\\\\t = \frac{27 \pm 37}{32}\\\\We\ have\ two\ solutions\\\\t = \frac{27+37}{32}\\\\t = \frac{64}{32}\\\\t = 2\\\\And\\\\t = \frac{27-37}{32}\\\\t = -0.3125

Ignore, negative value

Thus, it takes 2 seconds for object to hit the ground

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AURORKA [14]
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Explanation:

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Therefore, the other side is 3cm
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