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Oduvanchick [21]
3 years ago
13

Problem

Mathematics
2 answers:
aniked [119]3 years ago
7 0

Answer:

13 units

Step-by-step explanation:

Fiesta28 [93]3 years ago
3 0

Answer:

do you have a photo of the figure?

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Can someone help me with this one question???
Hitman42 [59]
Z and 113° are supplementary, they add to make up 180. So z + 113 = 180, z = 67°. x and 113° are equivelant, so 8x + 41 = 113, x = 9.
3 0
3 years ago
Snappys, Inc. is a company that produces digital cameras. They are planning to release a new line of camera.
Vaselesa [24]

Answer:

1334

Step-by-step explanation:

$250-$100 = $150

$200,000/ $150

1333.333

Round up

1334

5 0
3 years ago
Differentiate tan inverse of x÷a<br><br>​
ArbitrLikvidat [17]

Answer:

Expression Derivatives

y = tan-1(x / a) dy/dx = a / (a2 + x2)

y = cot-1(x / a) dy/dx = - a / (a2 + x2)

y = sec-1(x / a) dy/dx = a / (x (x2 - a2)1/2)

y = cosec-1(x / a) dy/dx = - a / (x (x2 - a2)1/2)

8 0
3 years ago
Find the missing angle on the triangle pls need help
Maslowich
Answer:
Angle B is 44 degrees.
This is a right triangle.

Steps:
The sum of interior angles in a triangle equals 180 degrees. The given triangle is a right triangle.

So A + B + C = 180
We are given A and C.
46 + B + 90 = 180
B + 136 = 180
180 - 136 = 44
4 0
1 year ago
Read 2 more answers
write the equation of the perpendicular bisector that goes through the line segment with end points of A -1,-2 and B -2,-8
Dennis_Churaev [7]

Answer:

  2x +12y = -63 . . . in standard form

  y = -1/6x -21/4 . . . in slope-intercept form

Step-by-step explanation:

It is useful to find the midpoint of the segment. That is the average of the end points:

  M = ((-1, -2) +(-2, -8))/2 = ((-1-2)/2, (-2-8)/2) = (-3/2, -5)

It is also useful to find the changes in coordinates from B to A:

  Δ = A-B = (-1-(-2), -2-(-8)) = (1, 6)

From here, there are a couple of ways you can write the equation of the perpendicular line.

__

One way is to use the Δ values to compute the slope of the segment. The perpendicular line will have a slope that is the negative reciprocal of that.

  Δy/Δx = 6/1 = 6

  m = -1/6 . . . . . slope of the perpendicular line

Now we have a point and a slope for the desired line, so we can use a point-slope form of the equation for a line:

  y = m(x -h) +k

  y = (-1/6)(x -(-3/2)) +(-5)

  y = (-1/6)x -21/4 . . . . . . . . eliminate parentheses; point-slope form

__

Another way to write the perpendicular line is to use the Δ values directly as coefficients in the standard form equation:

  Δx(x -h) +Δy(y -k) = 0

  1(x -(-3/2)) + 6(y -(-5)) = 0 . . . substitute values

  x +6y +31.5 = 0 . . . . . . . . . . .collect terms

  2x +12y = -63 . . . . . . . . . . . . multiply by 2, put in standard form

7 0
3 years ago
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