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Rudik [331]
3 years ago
5

Marc has 3 fewer than 4 times the amount of ringtones than Melanie has. Write

Mathematics
1 answer:
Marrrta [24]3 years ago
6 0

Answer:

4r-3

Step-by-step explanation:

3 fewer means subtracting 3 and 4 represents the r

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1 + 5 • 2 + ( 1 + 3)2
Sergeeva-Olga [200]

Answer:

1+5(2)+2(1+3)

1+10+2+6

=19

5 0
3 years ago
Read 2 more answers
A glass is
fenix001 [56]

Step-by-step explanation:

x=volume of the glass

(1/3)x+63=(5/8)x

63=(5/8)x-(1/3)x

63=(15/24)x-(8/24)x

63=(7/24)x

multiply both sides by 24/7

(24/7)(63)=(24/7)(7/24)x

216=x

216 cm3 is the volume (answer)

1/3 of 216 is 72

72+63=135

135/216=0.625=5/8

4 0
3 years ago
Two stabilizing wires extend from the top of a pole to the ground, forming right triangles on
FrozenT [24]

Answer: the height of the pole is 16.1 m.

Step-by-step explanation:

The scenario is illustrated in the attached photo.

x represents the height of the pole.

y represents the distance from the foot of one stabilizing wire to the foot of the pole.

30 - y represents the distance from the foot of the other stabilizing wire to the foot of the pole.

In solving the triangles, we would apply the tangent trigonometric ratio which is expressed as

Tan θ, = opposite side/adjacent side.

Considering triangle ACD,

Tan 60 = x/y

x = ytan60 = y × 1.732

x = 1.732y- - - - - - - - -1

Considering triangle BCD,

Tan 38 = x/(30 - y)

x = (30 - y)tan38 = 0.781(30 - y)

x = 23.43 - 0.781y- - - - - - - - -2

Substituting equation 1 into equation 2, it becomes

1.732y = 23.43 - 0.781y

1.732y + 0.781y = 23.43

2.513y = 23.43

y = 23.43/2.513

y = 9.3

x = 1.732y = 1.732 × 9.3

x = 16.1 m

6 0
3 years ago
How many groups of 2/3 are in 3/5
amid [387]

To find the answer, we will need to divide 3/5 by 2/3.

3/5 / 2/3

3/5 x 3/2

9 / 10

There are 9/10 groups of 2/3 in 3/5.

Hope this helps!! :)

5 0
2 years ago
What is the least common multiple of 8 and 6?
Degger [83]

Answer:

24

Step-by-step explanation:

8 16 24 32 40...

6 12 18 24 30 36....

24 and 24.

4 0
3 years ago
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