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Strike441 [17]
3 years ago
6

Need some help. Its hard.

Mathematics
1 answer:
MArishka [77]3 years ago
7 0
Obvious. All you have to do is divide the equilified factor from the inblified decimal, and incorperate the imberal number from the outside jenquizical factor.  so, all you are left with is 2.05 and 1.72. then you have to divide the terms of them, and you are done.  Your answer is c
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When -2(x − 4) + 1 = 7 is solved, the result is:
Sergeu [11.5K]

Answer:

1

Step-by-step explanation:

-2(x-4)+1=7

First Distribute the -2.

-2x+8+1=7

Subtract the 8 and 1 from the whole equation.

-2x=-2

Divide both sides of the equation by -2.

x=1

I hope this helps!

6 0
2 years ago
Read 2 more answers
Find the perimeter of the polygon with vertices A(-6,-4), B(-3,6), C(4,0), and D(2,-1)?
EleoNora [17]
Check the picture below.

so the perimeter of the polygon is the sum of all its sides, namely, AB + BC + CD + DA.

now, let's check how long each side is,

\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\ \quad \\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&A&(~{{ -6}} &,&{{ -4}}~) 
%  (c,d)
&B&(~{{ -3}} &,&{{ 6}}~)
\end{array}
\\\\\\
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}\\\\
-------------------------------\\\\
AB=\sqrt{[-3-(-6)]^2+[6-(-4)]^2}
\\\\\\
AB=\sqrt{(-3+6)^2+(6+4)^2}
\\\\\\
AB=\sqrt{3^2+10^2}\implies \boxed{AB=\sqrt{109}}\\\\
-------------------------------

\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\ \quad \\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&B&(~{{ -3}} &,&{{6}}~) 
%  (c,d)
&C&(~{{ 4}} &,&{{ 0}}~)
\end{array}
\\\\
-------------------------------\\\\
BC=\sqrt{[4-(-3)]^2+[0-6]^2}\implies BC=\sqrt{(4+3)^2+(0-6)^2}
\\\\\\
BC=\sqrt{7^2+(-6)^2}\implies \boxed{BC=\sqrt{85}}\\\\
-------------------------------

\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\ \quad \\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&C&(~{{ 4}} &,&{{0}}~) 
%  (c,d)
&D&(~{{ 2}} &,&{{ -1}}~)
\end{array}
\\\\
-------------------------------\\\\
CD=\sqrt{(2-4)^2+(-1-0)^2}\implies CD=\sqrt{(-2)^2+(-1)^2}
\\\\\\
\boxed{CD=\sqrt{5}}\\\\
-------------------------------

\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\ \quad \\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&D(~{{ 2}} &,&{{-1}}~) 
%  (c,d)
&A&(~{{ -6}} &,&{{ -4}}~)
\end{array}\\\\
-------------------------------\\\\
DA=\sqrt{[-6-2]^2+[-4-(-1)]^2}\\\\\\ DA=\sqrt{(-6-2)^2+(-4+1)^2}
\\\\\\
DA=\sqrt{(-8)^2+(-3)^2}\implies \boxed{DA=\sqrt{73}}

sum those sides up, and that's the perimeter of the polygon.

6 0
2 years ago
What is the price called at which the quantity demanded is equal to the quantity supplied?
sladkih [1.3K]
Equilibrium Price 
Hope This Helps 
8 0
3 years ago
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M∠COX = m∠AOX + m∠<br><br> Answers:<br> COD<br> COE<br> COX<br> COA
postnew [5]

Answer:

ok

Step-by-step explanation:

8 0
2 years ago
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What is the scale factor of triangle ABC to triangle DEF. <br> A: 2<br> B: 1/2<br> C: 1/6<br> D: 6
Fudgin [204]
Little tip ** if u r going from a larger shape to a smaller shape, the scale factor is going to be a proper fraction....a number less then 1 but greater then 0

BC = 66......EF = 11.......66 times what is 11
66x = 11
x = 11/66 reduces to 1/6
so 66 * 1/6 = 11

now compare the other sides...
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so ur multiplying every number in the big triangle by 1/6 , and getting the measures of the little triangle....so ur scale factor is 1/6 <==

8 0
2 years ago
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