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kenny6666 [7]
4 years ago
15

Y=-5x+3 in standard form

Mathematics
1 answer:
vazorg [7]4 years ago
6 0
Standard form =
ax + by = c

y = -5x + 3
Add 5x to both sides.

5x + y = 3
And this is your answer.
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The Jayden family eats at a restaurant that is having a 15% discount promotion. Their meal costs $78.25 before the discount, and
soldi70 [24.7K]

Answer:

$82.16

Step-by-step explanation:

78.25 * .20 = 15.65 as a tip,

78.25 *.15 = 11.74 as the discount, subtract from initial value:

78.25 - 11.74 = 66.51

then add the tip: 15.65 + 66.51 = $82.16

8 0
3 years ago
Suppose that angle ABC and angle DEF are both supplementary to angle XYZ, and angle XYZ is a right angle. Name all of the remain
yarga [219]
ABC n DEF COULD BE ANY ONE OF THE OPTIONS.<span>

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6 0
3 years ago
Read 2 more answers
PLEASE HELP ASAP MATH
Scilla [17]

T h e   a n s w e r  i s :

A.) 3x-y

7 0
3 years ago
What is the rate of change between the interval x= pi and x= 3pi/2
Romashka [77]

Answer:

\frac{6}{\pi } or 1.9099

Step-by-step explanation:

Look for y values at each of those given values of "x" and apply the slope formula.  When x = pi. y = -1 so the coordinate is (\pi,-1).  When x = 3pi/2, y = 2 so the coordinate is (\frac{3\pi }{2},2)

Plug those values into the slope formula:

\frac{2-(-1)}{\frac{3\pi }{2}-\pi}

You need a common denominator of pi:

\frac{3}{\frac{3\pi-2\pi}{2} }=\frac{3}{\frac{\pi }{2} }

Do the math on that to get a slope of \frac{6}{\pi } =1.9099

8 0
3 years ago
Use cross products to find the area of the triangle in the xy-plane defined by (1, 2), (3, 4), and (−7, 7).
Free_Kalibri [48]

I love these. It's often called the Shoelace Formula. It actually works for the area of any 2D polygon.


We can derive it by first imagining our triangle in the first quadrant, one vertex at the origin, one at (a,b), one at (c,d), with (0,0),(a,b),(c,d) in counterclockwise order.


Our triangle is inscribed in the a \times d rectangle. There are three right triangles in that rectangle that aren't part of our triangle. When we subtract the area of the right triangles from the area of the rectangle we're left with the area S of our triangle.


S = ad - \frac 1 2 ab -  \frac 1 2 cd - \frac 1 2 (a-c)(d-b) = \frac 1 2(2 ad - ab -cd - ad +ab +cd -bc) = \frac 1 2(ad -bc)


That's the cross product in the purest form. When we're away from the origin, a arbitrary triangle with vertices A(x_1, y_1), B(x_2, y_2), C(x_3, y_3) will have the same area as one whose vertex C is translated to the origin.


We set a=x_1 - x_3, b= y_1  - y_3, c=x_2 - x_3, d=y_2- y_3


S= ad-bc=(x_1 - x_3)(y_2 - y_3) -(x_2-x_3)(y_1 - y_3)


That's a perfectly useful formula right there. But it's usually multiplied out:


S= x_1y_2 - x_1 y_3  - x_3y_2 + x_3 y_3 - x_2 y_1 + x_2y_3 + x_3 y_1 - x_3 y_3


S= x_1 y_2 - x_2 y_1  + x_2y_3 - x_3y_2   + x_3 y_1 - x_1 y_3


That's the usual form, the sum of cross products. Let's line up our numbers to make it easier.


(1, 2), (3, 4), (−7, 7)

(−7, 7),(1, 2), (3, 4),


[tex]A = \frac 1 2 ( 1(7)-2(-7) + 3(2)-4(1) + -7(4) - (7)(3)

8 0
4 years ago
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