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eduard
3 years ago
13

Pls send me step by step pic​

Mathematics
1 answer:
Thepotemich [5.8K]3 years ago
7 0

To solve, simply do this:

(\frac{1}{2}^3)-(\frac{3}{4} )^2\\\\\frac{1}{2} *\frac{1}{2} *\frac{1}{2} *\frac{1}{2} \\\\\frac{1}{8} -(\frac{3}{4})^2\\\\\frac{1}{8}-\frac{9}{16}\\\\=\frac{-7}{16}

Then you'll get the answer, -7/16

You might be interested in
What two ratios are equivalent to 48:156
anastassius [24]
If there are no answer choices, I'll give you a few to work with:
16:52 (divide each number by 3 to get this)
8:26 (divide the above by 2, both sides)
4:13 (divide the above by 2, both sides)
96:312 (multiply the original by 2, both sides)
I think you get the gist.
~Hope this helps!
3 0
3 years ago
a cone has a radius of 6cm and height 9 calculate a)the volume and curved surface area b) total surface area
Orlov [11]

Answer:

V = 339.3 cm³; L = 203.9 cm²; A = 317.0 cm²  

Step-by-step explanation:

a) Volume

The formula for the volume (V) of a cone is

V = ⅓πr²h

V = ⅓π × 6² × 9

V = ⅓π × 36 × 9

V = 108π cm³

V ≈ 339.3 cm³

=====

Curved surface area  

The formula for the lateral surface area (L) of a cone is

L = πr√(r² + h²)

L = π×6√(36 + 81)

L = 6π√[9(4 + 9)]

L = 6π√(9 × 13)

L = 18π√13 cm²

L ≈ 203.9 cm²

===============

b) Base surface area

The base is a circle, so the formula for base surface area (B) is

B = πr²

B = π×6²

B = 36π cm²

B ≈ 113.1 cm²

=====

Total surface area

A = L + B

A = 18π√(13 +36π)

A = 18π(2 + √13) cm²

A ≈ 203.9 + 113.1

A ≈ 317.0 cm²

3 0
3 years ago
A chocolate chip cookie recipe asks for three and one half times as much flour as chocolate chips. If one and three quarters cup
Ganezh [65]
3.5 times as much flour as chocolate chips....
f = 3.5c
f = 1 3/4 (or 1.75)

1.75 = 3.5c
1.75 / 3.5 = c
0.5 = c......chocolate chips needed would be 1/2 a cup


6 0
3 years ago
Subtract (x^2+3x-4) -(4x-2x^2-3)
Feliz [49]

Answer:

3 x^2 - x + -1

Step-by-step explanation:

Simplify the following:

-(4 x - 2 x^2 - 3) + x^2 + 3 x - 4

Factor -1 out of -2 x^2 + 4 x - 3:

--(2 x^2 - 4 x + 3) + x^2 + 3 x - 4

(-1)^2 = 1:

2 x^2 - 4 x + 3 + x^2 + 3 x - 4

Grouping like terms, 2 x^2 + x^2 + 3 x - 4 x - 4 + 3 = (x^2 + 2 x^2) + (3 x - 4 x) + (-4 + 3):

(x^2 + 2 x^2) + (3 x - 4 x) + (-4 + 3)

x^2 + 2 x^2 = 3 x^2:

3 x^2 + (3 x - 4 x) + (-4 + 3)

3 x - 4 x = -x:

3 x^2 + -x + (-4 + 3)

3 - 4 = -1:

Answer: 3 x^2 - x + -1

4 0
4 years ago
Which series of transformations will not map figure H onto itself
7nadin3 [17]

Answer:

D

Step-by-step explanation:

Given a square with vertices at points (2,1), (1,2), (2,3) and (3,2).

Consider option A.

1st transformation (x+0,y-2) will map vertices of the square into points

  • (2,1)\rightarrow (2,-1);
  • (1,2)\rightarrow (1,0);
  • (2,3)\rightarrow (2,1);
  • (3,2)\rightarrow (3,0).

2nd transformation = reflection over y = 1 has the rule (x,2-y). So,

  • (2,-1)\rightarrow (2,3);
  • (1,0)\rightarrow (1,2);
  • (2,1)\rightarrow (2,1);
  • (3,0)\rightarrow (3,2)

These points are exactly the vertices of the initial square.

Consider option B.

1st transformation (x+2,y-0) will map vertices of the square into points

  • (2,1)\rightarrow (4,1);
  • (1,2)\rightarrow (3,2);
  • (2,3)\rightarrow (4,3);
  • (3,2)\rightarrow (5,2).

2nd transformation = reflection over x = 3 has the rule (6-x,y). So,

  • (4,1)\rightarrow (2,1);
  • (3,2)\rightarrow (3,2);
  • (4,3)\rightarrow (2,3);
  • (5,2)\rightarrow (1,2)

These points are exactly the vertices of the initial square.

Consider option C.

1st transformation (x+3,y+3) will map vertices of the square into points

  • (2,1)\rightarrow (5,4);
  • (1,2)\rightarrow (4,5);
  • (2,3)\rightarrow (5,6);
  • (3,2)\rightarrow (6,5).

2nd transformation = reflection over y = -x + 7 will map vertices into points

  • (5,4)\rightarrow (3,2);
  • (4,5)\rightarrow (2,3);
  • (5,6)\rightarrow (1,2);
  • (6,5)\rightarrow (2,1)

These points are exactly the vertices of the initial square.

Consider option D.

1st transformation (x-3,y-3) will map vertices of the square into points

  • (2,1)\rightarrow (-1,-2);
  • (1,2)\rightarrow (-2,-1);
  • (2,3)\rightarrow (-1,0);
  • (3,2)\rightarrow (0,-1).

2nd transformation = reflection over y = -x + 2 will map vertices into points

  • (-1,-2)\rightarrow (4,3);
  • (-2,-1)\rightarrow (3,4);
  • (-1,0)\rightarrow (2,3);
  • (0,-1)\rightarrow (3,2)

These points are not the vertices of the initial square.

5 0
3 years ago
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