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Gala2k [10]
3 years ago
8

I need help pls I’m on a test

Mathematics
2 answers:
emmasim [6.3K]3 years ago
6 0

Answer:

D. No solutions

Step-by-step explanation:

I just took a test and this was the answer :)

Alecsey [184]3 years ago
3 0

Answer:

No solution (i'm not 100% sure this is correct)

Step-by-step explanation:

3(0.5x-4)=3/2x-1.2

1.5x-12=3/2x-1.2

     +12         +12

1.5x=3/2x-13.2

-3/2x  -3/2x

0x=-13.2

x=0

no solution

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Pls, help! (NO LINKS!)<br> Thank you!
german

Answer:

18÷25 =0.72 convert proper fraction to decimal ,divide the numerator by the denominator.

3-0.72=3.72. add this decimal number to the whole number .final answer is 3.72

8 0
3 years ago
The base of an isosceles triangle is 24 cm and its area is 192 cm^2 . Find its perimeter.
Dmitriy789 [7]

Step-by-step explanation:

Area of Traingle=1/2×b×h

192cm²=1/2×24×h

192cm²=12×h

h=192/12

h=16cm

Hypotenuse ²=Base²+Altitude ²

Hypotenuse ²=12²+16²

Hypotenuse ²=144+256

Hypotenuse ²=400

Hypotenuse =20cm

Perimeter=20+20+24

=64cm

5 0
3 years ago
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6 0
3 years ago
Signs for science project displays are cut from pieces of poster board that measure 1 yard on each side. Each sign is LaTeX: \fr
iren [92.7K]

Answer:

27

Step-by-step explanation:

Given that :

Dimension of poster board = 1 yd by 1 yd

Dimension of each poster sign = 1/3 yd by 1/9 yd

Number of poster signs that can be cut :

Area of poster sign = 1/3 * 1/9 = 1/27 yard²

Area of poster board = 1 yard²

Number of poster signs that can be cut :

Area of poster board / area of poster sign

1 yard² ÷ (1/27) yard²

1 ÷ 1/27

1 * 27/1

= 27 poster signs

4 0
3 years ago
In parallelogram ABCD , diagonals AC¯¯¯¯¯ and BD¯¯¯¯¯ intersect at point E, AE = 3x - 5, and CE = x + 11.
german

Answer:

38

Step-by-step explanation:

For the case of a parallelogram, when the diagonals ( in our case AC & BD) are intersected, they are basically bisected ( divided into two equal halves ). Therefore when the diagonals intersect at point E, we can say that the diagonal AC is divided in two equal halves which in our case are AE and CE. since AE and CE are equal , we can say that,

AE= CE \\or\\3x-5 = x+11 ---- (1)\\now  solving (1)  for x, \\3x-x=11+5\\2x=16\\x=8-----> (2)\\\\as      \\ AC = AE+CE\\AC = 3x-5+x+11\\AC = 4x+6\\\\\\from (2)\\\\\\AC= 4(8)+6\\AC = 32+6\\AC = 38\\\\since AC = BD ( Diagonals     are    of same Length )\\therefore \\BD =38

5 0
4 years ago
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