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algol [13]
2 years ago
9

Guided Practice

Mathematics
1 answer:
tatiyna2 years ago
3 0

Answer:

A.

1.005

Step-by-step explanation:

Answer:

b=1.005

Step-by-step explanation:

b=(1+0.5%)

=1+0.005

=1.005

...............................................................................................................................................

Answer:

b=1.005

Step-by-step explanation:

b=(1+0.5%)

=1+0.005

=1.005

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Write an equation for a circle with a diameter that has endpoints at (–4, –7) and (–2, –5). Round to the nearest tenth if necess
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since we know the endpoints of the circle, we know then that distance from one to another is really the diameter, and half of that is its radius.

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\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[-2-(-4)]^2+[-5-(-7)]^2}\implies d=\sqrt{(-2+4)^2+(-5+7)^2} \\\\\\ d=\sqrt{2^2+2^2}\implies d=\sqrt{2\cdot 2^2}\implies d=2\sqrt{2}~\hfill \stackrel{~\hfill radius}{\cfrac{2\sqrt{2}}{2}\implies\boxed{ \sqrt{2}}} \\\\[-0.35em] ~\dotfill

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{-2-4}{2}~~,~~\cfrac{-5-7}{2} \right)\implies \left( \cfrac{-6}{2}~,~\cfrac{-12}{2} \right)\implies \stackrel{center}{\boxed{(-3,-6)}} \\\\[-0.35em] ~\dotfill

\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-3}{ h},\stackrel{-6}{ k})\qquad \qquad radius=\stackrel{\sqrt{2}}{ r} \\[2em] [x-(-3)]^2+[y-(-6)]^2=(\sqrt{2})^2\implies (x+3)^2+(y+6)^2=2

4 0
3 years ago
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Write a function rule for the area of a triangle whose base is 4 ft more than the height. What is the area of the triangle when
solong [7]

Answer:

A) Area as function of H ,

     F'(a) = 2 +H

B) Area of triangle with height 6ft = 30 ft²      

Step-by-step explanation:

Given as for a triangle ,

Let the height of triangle = H ft

And base = 4 ft + H ft

Now area of triangle = \frac{1}{2} × height  × base

                          F (a) =  \frac{1}{2} × H × ( 4ft + H ft)

Or,                      F (a) =  \frac{1}{2} × ( 4H + H² )

Now,   here Area is function of height ,

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Hence Area as function of H ,

                          F'(a) = 2 +H

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Area of triangle = \frac{1}{2} × height × base

                          = \frac{1}{2} × H × ( 4ft + H ft)

                          = \frac{1}{2} × ( 4H + H² )

                          =  \frac{1}{2} × ( 24 + 36 )

                          = \frac{1}{2} × ( 60 )

                          = 30 ft²

Hence Area of triangle with height 6ft = 30 ft²      Answer

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