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Korolek [52]
4 years ago
7

Please help!

Mathematics
2 answers:
Alekssandra [29.7K]4 years ago
6 0
1. B
2. C
3. A
4. B
5. C
Nikolay [14]4 years ago
3 0
1.B 
2.C (5*20)+3
3.A 22/7= 3 r.1
4.B
5.C (3*8)+3
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The total cost for bowling includes the fee for shoe rental plus fee per game. The cost of each game increases the price by $4.
satela [25.4K]

Answer:


Step-by-step explanation:

A: 3*4+2

B: The shoe rental is 2 dollar. Since the cost per game is 4 dollar you will need to multiply by the 3 game you pay then subtract 14-12 and get 2 which is the rental fee

3 0
3 years ago
F(x) = x² + 10 <br>What is the average rate of change of f over the interval [ -2,-1?​
KIM [24]

Answer:

-3

Step-by-step explanation:

f(-2) = (-2)^2 + 10 = 14 -> (-2, 14)

f(-1) = (-1)^2 + 10 = 11 -> (-1, 11)

(11-14)/(-1-(-2)) = -3/1 = -3

8 0
3 years ago
A student turned in an 18-page paper which is
arsen [322]

Answer:

i believe its around like a 4.1% or 4.2%

HOPE IT HELPS!!

4 0
3 years ago
What is the value of x in the equation 2(x – 3) + 9 = 3(x + 1) + x?<br><br> x =
V125BC [204]

Answer:

The answer is 0.

Step-by-step explanation:

when solving  you'll have two equations that would not make sense but if you put the x as 0 then both equations are 3.

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
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