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Ann [662]
3 years ago
9

Draw a square with a perimeter of 16 feet

Mathematics
1 answer:
Colt1911 [192]3 years ago
4 0
You need a ruler and a pencil.  Oh and paper.  Make sure that each side is equal to 4 inches.  Then make a scale of 1:12, meaning 1 inch=1 foot.
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What is the answer of a - 5 + -12
Pachacha [2.7K]
-5+-12 is -17
For lack of a better metaphor, it's like digging a pit deeper than it already is.
8 0
2 years ago
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HELP PLEASE HELP!!!!!!!
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If we are to write this equation in slope-intercept form, it will be in y = mx + b, where m is the slope of the line and b is the y intercept.  We need then to find the slope of the line using 2 points on the line and filling in the slope formula to find the slope.  One of the points we can use is (0, 3) which is also the y intercept.  The y-intercept is found where x = 0.  Where x = 0, y = 3.  So b = 3.  Now for the slope we will use (0,3) and (4,4):  m= \frac{4-3}{4-0} = \frac{1}{4}.  Using that m value and that b value we have the equation y= \frac{1}{4}x+3.  There you go!
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3 years ago
Could you explain what the answer is and why? I am not sure of how to do this
lidiya [134]
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8 0
3 years ago
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Write an equation for the n th term of the arthemeric sequence 6,3,0,-3
s344n2d4d5 [400]

Answer:

x nth term =6–3(n-1)

Step-by-step explanation:

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3 0
3 years ago
You are looking at the New York ball drop on New Year’s Eve at a distance of 100 m away from the base of the structure. If the b
Fofino [41]

The question is an illustration of related rates.

The rate of change between you and the ball is 0.01 rad per second

I added an attachment to illustrate the given parameters.

The representations on the attachment are:

\mathbf{x = 100\ m}

\mathbf{\frac{dy}{dt} = 2\ ms^{-1}} ---- the rate

\mathbf{\theta = \frac{\pi}{4}}

First, we calculate the vertical distance (y) using tangent ratio

\mathbf{\tan(\theta) = \frac{y}{x}}

Substitute 100 for x

\mathbf{y = 100\tan(\theta) }

\mathbf{\tan(\theta) = \frac{y}{100}}

Differentiate both sides with respect to time (t)

\mathbf{ \sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot \frac{dy}{dt}}

Substitute values for the rates and \mathbf{\theta }

\mathbf{ \sec^2(\pi/4) \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 2}

This gives

\mathbf{ (\sqrt 2)^2 \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 2}

\mathbf{ 2 \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 2}

Divide both sides by 2

\mathbf{ \frac{d\theta}{dt} = \frac{1}{100} }

\mathbf{ \frac{d\theta}{dt} = 0.01 }

Hence, the rate of change between you and the ball is 0.01 rad per second

Read more about related rates at:

brainly.com/question/16981791

8 0
2 years ago
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