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Artemon [7]
2 years ago
8

Can someone help me with this 6th grade math problem? 1 1/30 + 1/4 = ?

Mathematics
2 answers:
Nonamiya [84]2 years ago
8 0

Answer:

1 17/60 or 1.34 ( not sure if it was 11/30 or one and 1/30 )

Step-by-step explanation:

1)

1 \frac{1}{30} =

+

\frac{1}{4} =

2)

\frac{31}{30} = 124/120

+

\frac{1}{4} = 30/120

----------

154/120

1 34/120

3) 1 17/60

Nuetrik [128]2 years ago
7 0

Answer:

1.2833333333....

Step-by-step explanation:

Just add it

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Taya2010 [7]

Answer:

B.

Step-by-step explanation:

You must first subtract 12 from 27, which marks out two answers. A, and D.

Then, if NAthan traveled 3 times as far s philip, then you would multiply by three. leaving you with the only option being,

B.

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Solve the following system of equations below x-2y=3 2x-3y=9
Artyom0805 [142]

x - 2y = 3 \\ 2x - 3y = 9 \\  \\  - 2x + 4y =  - 12 \\  \: 2x - 3y = 9 \\  \\ y =  - 3 \\ 2x + 9 = 9 \\ 2x = 0 \\ x = 0 \\ x = 0 \:  \: y =  - 3
7 0
3 years ago
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C is the midpoint of line segment AB Find the value of x, if line segment AB = 5x and line segment AC = 20.
Softa [21]
D.8

A and C are out because they would get you ether lower or equal to 20, which is equal to AC and your trying to find X in AB which is so posed to be the what equals AC+BC.
AB=AC+BC
5x=20+BC
5(8)=20+BC
40=20+BC
40=20+20

AC equals 20. AB will have to equal AC+ CB. If you put in A.4 in AB 5X you will get twenty and that not what you want. You want to get a number that will get you the an answer greater to AC. So 5(8)=40. 
5 0
3 years ago
Carla uses a compass and straightedge to construct perpendicular lines. Daryl uses technology to construct perpendicular lines.
devlian [24]
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3 years ago
In the circle below, suppose m VUX = 152° and mZUVW = 77º. Find the following.
bezimeni [28]

Given:

In the circle, m(\overarc{VUX})=152^\circ and m(\angle MUV)=77^\circ.

To find:

The following measures:

(a) m\angle VUX

(b) m\angle UXW

Solution:

According to the central angle theorem, the central angle is always twice of the subtended angle intercepted on the same same arc.

m(VUX)=2\times m\angle VWX

152^\circ=2\times m\angle VWX

\dfrac{152^\circ}{2}=m\angle VWX

76^\circ=m\angle VWX

In a cyclic quadrilateral, the opposite angles are supplementary angles.

UVWX is a cyclic quadrilateral. So,

m\angle VUX+m\angle VWX=180^\circ          [Opposite angles of a cyclic quadrilateral]

m\angle VUX+76^\circ=180^\circ

m\angle VUX=180^\circ-76^\circ

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m\angle UXW+m\angle UVW=180^\circ          [Opposite angles of a cyclic quadrilateral]

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m\angle UXW=103^\circ

Therefore, m\angle VUX=104^\circ  and m\angle UXW=103^\circ .

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