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DiKsa [7]
2 years ago
7

What type of triangle must ANB be?

Mathematics
1 answer:
allochka39001 [22]2 years ago
4 0
The answer is isosceles triangle
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An electric company charges an initial monthly fee for service and a fixed rate per kilowatt-hours (kWh) used. This table shows
Vlada [557]

<em>Here we are required to determine the initial monthly fee charged by the electric company.</em>

The initial fee charged by the electric company is; C = $10

To solve this, we need to evaluate the slope and intercepts of the equation of the straight line graph of the relation.

y = mx + c.

  • where m = slope of the relation.

  • and c = <em>intercept = the initial fee charged by the electric company</em>.

  • y = <em>Monthly charge at each time</em>.

  • x = Usage

To find the slope;

  • m = (y2 - y1)/(X2 - x1)

  • m = (28 -22)/(150 - 100)

  • m = 6/50

  • m = 0.12.

By substituting m into the equation y = mx + c, alongside a pair of values of usage and monthly charge, we can obtain the intercept, c (i.e the initial fee charged).

Therefore, m = 0.12 , y = 82 and X = 600;

we then have;

  • 82 = 0.12(600) + c.

  • C = 82 -72

  • C = $10.

Therefore, the initial fee charged by the electric company is; C = $10.

Read more:

brainly.com/question/22163485

8 0
2 years ago
Help help its pre pre agelbra
grandymaker [24]
I believe Line segment
3 0
2 years ago
Directions: Graph the equation. Label the points where the line crosses the axes.
kolezko [41]

Answer:

Both the equations landed on point ( 15.167 , 8.267 )

Step-by-step explanation:

7 0
2 years ago
Which is not an equation of the line that passes through the points (1, 1) and (5, 5)?
nydimaria [60]

Answer:

Any equation of the line that is different from y=x is the solution to the problem.

Step-by-step explanation:

step 1

Find the slope of the line that passes through the points (1, 1) and (5, 5)

m=(5-1)/(5-1)=1

step 2

Find the equation of the line into slope point form

y-y1=m(x-x1)

we have m=1

(x1,y1)=(1,1)

substitute

y-1=(1)(x-1)

y=x-1+1

y=x

therefore

Any equation of the line that is different from y=x is the solution to the problem.

5 0
3 years ago
Please help me asap<br>​
Lera25 [3.4K]

Answer:

Step-by-step explanation:

From the picture attached,

∠4 = 45°, ∠5 = 135° and ∠10 = ∠11

Part A

∠1 = ∠4 = 45°  [Vertically opposite angles]

∠1 + ∠3 = 180° [Linear pair of angles]

∠3 = 180° - ∠1

     = 180° - 45°

     = 135°

∠2 = ∠3 = 135°  [Vertically opposite angles]

∠8 = ∠5 = 135° [Vertically opposite angles]

∠5 + ∠6 = 180° [Linear pair of angles]

∠6 = 180° - 135°

∠6 = 45°

∠7 = ∠6 = 45° [Vertically opposite angles]

By triangle sum theorem,

m∠4 + m∠7 + m∠10 = 180°

45° + 45° + m∠10 = 180°

m∠10 = 180° - 90°

m∠10 = 90°

m∠10 = m∠12 = 90°  [Vertically opposite angles]

m∠10 = m∠11 = 90° [Given]

Part B

1). ∠1 ≅ ∠4  [Vertically opposite angles]

2). ∠7 + ∠5 = 180° [Linear pair]

3). ∠9 + ∠10 = 180° [Linear pair]

7 0
3 years ago
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