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3 years ago

HELP ME ASAP!!! I never understood the concept of finding the missing length/side... Can anyone help?

Mathematics

2 answers:

Levart [38]3 years ago

8 0

Hello, I am most likely wrong but, I thought I would try to help!
since the bigger one is 15 and the smaller one is 5 so its 2 times bigger? I think 3cm or 2cm.

sattari [20]3 years ago

7 0

18? Sorry but I may get it wrong.

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Math question please show work due today last day to submit Assignments or I fail please help

irina [24]

Answer:

$\$128.43$

Step-by-step explanation:

Since the cost for border paper is given in terms of feet and the dimensions of the gym are given in meters, we want to convert them to the same unit. Let's choose to convert feet to meters since that would be only one conversion:

$15\:\mathrm{ft}=4.572\:\mathrm{m}$

Therefore, the cost of border paper is $6.99 pet 15 feet or $6.99 pet 4.572 meters.

Since the gym is rectangular, the gym's perimeter is equal to:

$p=15+27+15+27,\\p=2(15)+2(27)=84\:\mathrm{m}$

Therefore, the cost of the total border paper for the gym is equal to:

$\text{cost}=\frac{84}{4.572}\cdot 6.99\approx \boxed{\$128.43}$

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2 years ago

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Hi. I need help. This is for homework I would appreciate it if you helped.

Triss [41]

Answer:

Ik this isn't an answer but wanted to say good luck im battling stuff too :(

Step-by-step explanation:

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2 years ago

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Can someone please help me with this one? Oh and in the boxes I need to put a number NOT a check mark.

kari74 [83]

Grade 7 : Yes = 2 , No = 3
Grade 8 : Yes = 1 , No = 4

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3 years ago

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

Zinaida [17]

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.

we can also find the midpoint of those two endpoints and we'll be landing right on the center of the circle.

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

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

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

4 0

3 years ago

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Pls help me asap :(

Andreas93 [3]

x^2-12x+(6)^2 is correct answer

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3 years ago