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Serhud [2]
3 years ago
5

Define slope intersect

Mathematics
2 answers:
Delicious77 [7]3 years ago
5 0
A line that goes through the x axis or y axis
Vanyuwa [196]3 years ago
4 0

I think you meant intercept and not intersect but the definition is the equation of a straight line in the form y = mx + b where m is the slope of the line and b is its y-intercept



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Please only one question I need help
postnew [5]
Answer:

H= g + 15

15. 30
30. 45
35 50
45 60

Hope this helps :)
6 0
3 years ago
Someone helppp pls due in 5 min <br> I will thank
patriot [66]
Angel 3 is adjacent to Angel 4 therefore they are equal.
5 0
3 years ago
8. Which of the following correctly multiplies f(x) x g(x) if f(x) = 3x-1 and g(x) = 2x + 1?
raketka [301]

Answer:

C) f(x) x g(x) = 6x2 + x-1

Step-by-step explanation:

Multiplication of functions:

Two multiply two polynomial functions, we apply the distributive property(multiply each two terms and add).

In this question, we have that:

f(x) = 3x - 1, g(x) = 2x + 1

The multiplication is:

f(x) \times g(x) = (3x - 1)(2x+1) = 3x*2x + 3x*1 -1*2x -1*1 = 6x^2 + 3x - 2x - 1 = 6x^2 + x - 1

The correct answer is given by option C.

4 0
3 years ago
Answer for a lot of points!
earnstyle [38]

Given :

  • ZC = 90°

  • CD is the altitude to AB.

  • \angleA = 65°.

To find :

  • the angles in △CBD and △CAD if m∠A = 65°

Solution :

In Right angle △ABC,

we have,

=> ACB = 90°

=> \angleCAB = 65°.

So,

=> \angleACB + \angleCAB+\angleZCBA = 180° (By angle sum Property.)

=> 90° + 65° + \angleCBA = 180°

=> 155° +\angleCBA = 180°

=> \angleCBA = 180° - 155°

=> \angleCBA = 25°.

In △CDB,

=> CD is the altitude to AB.

So,

=> \angle CDB = 90°

=> \angleCBD = \angleCBA = 25°.

So,

=> \angleCBD + \angleDCB = 180° (Angle sum Property.)

=> 90° +25° + \angleDCB = 180°

=> 115° + \angleDCB = 180°

=> \angleDCB = 180° - 115°

=> \angleDCB = 65°.

Now, in △ADC,

=> CD is the altitude to AB.

So,

=> \angleADC = 90°

=>\angle CAD =\angle CAB = 65°.

So,

=> \angleADC + \angleCAD +\angleDCA = 180° (Angle sum Property.)

=> 90° + 65° + \angleDCA = 180°

=> 155° +\angleDCA = 180°

=> \angleDCA = 180° - 155°

=> \angleDCA = 25°

Hence, we get,

  • \angleDCA = 25°
  • \angleDCB = 65°
  • \angleCDB = 90°
  • \angleACD = 25°
  • \angleADC = 90°.
7 0
3 years ago
19
allsm [11]

Answer:

Use the information given to write an equation in standard form.  

Slope is -3, and (0, -4) is on the line.  

Step-by-step explanation:

8 0
3 years ago
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