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Artyom0805 [142]
3 years ago
5

Solve for y in terms of x

Mathematics
1 answer:
AnnZ [28]3 years ago
7 0
-8x+ 4y= -16
4y= 8x - 16
divide it by 4
y= 2x -4
if x is 0 what is y?
y = 2 (0)-4
y = 4
just put an x value in the equation u will get your y value

1 (0,4)
2 (1,-2)
3 (2,0)
4 (3,2)
5 (4,4)
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Como se daca el area de un triangulo​
Ivahew [28]
Area= base x altura dividido entre 2
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3 years ago
Let U = {1, 2, 3, 4, 5, 6, 7}, A= {1, 3, 4, 6}, and B= {3, 5, 6}. Find the set A’ U B’
Art [367]

Answer:

Step-by-step explanation:

A'={2,5,7}

B'={1,2,4,7}

A'UB'={1,2,4,5,7}

8 0
3 years ago
Pls can some one help if you like
Snezhnost [94]
Slope is rise/run
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rise=4m
run=3sec
slope=4/3

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4 0
3 years ago
A circle passes through points A(7,4), B(10,6), C(12,3). Show that AC must be the diameter of the circle.
Artist 52 [7]

so we have three points, A, B and C, if indeed AC is the diameter of the circle, then half the distance of AC is its radius, and the midpoint of AC is the center of the circle, morever, since B is also on the circle, the distance from B to the center must be the same radius distance.

in short, half the distance of AC must be equals to the distance of B to the midpoint of AC, if indeed AC is the diameter.

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{7}~,~\stackrel{y_1}{4})\qquad C(\stackrel{x_2}{12}~,~\stackrel{y_2}{3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{12+7}{2}~~,~~\cfrac{3+4}{2} \right)\implies \left( \cfrac{19}{2}~~,~~\cfrac{7}{2} \right)=M\impliedby \textit{center of the circle}

now, let's check the distance from say A to the center, and check the distance of B to the center, if it's indeed the center, they'll be the same and thus AC its diameter.

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{7}~,~\stackrel{y_1}{4})\qquad M(\stackrel{x_2}{\frac{19}{2}}~,~\stackrel{y_2}{\frac{7}{2}})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ AM=\sqrt{\left( \frac{19}{2}-7 \right)^2+\left( \frac{7}{2}-4 \right)^2} \\\\\\ AM=\sqrt{\left( \frac{5}{2}\right)^2+\left( -\frac{1}{2} \right)^2}\implies \boxed{AM\approx 2.549509756796392} \\\\[-0.35em] ~\dotfill

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ B(\stackrel{x_1}{10}~,~\stackrel{y_1}{6})\qquad M(\stackrel{x_2}{\frac{19}{2}}~,~\stackrel{y_2}{\frac{7}{2}}) \\\\\\ BM=\sqrt{\left( \frac{19}{2}-10 \right)^2+\left( \frac{7}{2}-6 \right)^2} \\\\\\ BM=\sqrt{\left( -\frac{1}{2}\right)^2+\left( -\frac{5}{2} \right)^2}\implies \boxed{BM\approx 2.549509756796392}

6 0
3 years ago
One kind of hard candy sells for $.89 per kilogram another sells for $1.10 per kilogram how many kilograms of each kind a need t
Furkat [3]

Answer:

20kg of $0.89 candy

10kg of $1.10 candy

Step-by-step explanation:

Candy 1 = 0.89 per kg

Candy 2 = 1.10 per kg

Total kilogram, kg = 30

Let candy 1 = x ; candy 2 = (30 - x) ;

0.89x + 1.10(30 - x) = 0.96(30)

0.89x + 33 - 1.10x = 28.8

0.89x - 1.10x = 28.8 - 33

-0.21x = - 4.2

x = 4.2 / 0.21

x = 20

20kg of $0.89 candy

(30 - x) = (30 - 20) = 10kg

10kg of $1.10 candy

4 0
3 years ago
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