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3 years ago

What is (2 × 2) to the 2nd power, minus 4?

2 answers:

Ivan3 years ago

7 0

Well first, you would do 2 x 2 which equals 4. Then you make 4 to the 2nd power, which is 16. You subtract 4 from 16, and it's 12.

Umnica [9.8K]3 years ago

6 0

$((2)(2))^2 - 4 \\ \\ 4^2 - 4 \\ \\ 16 - 4 \\ \\ 12$

So, 12 is the answer.

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frutty [35]

The answer is: 98

Cause is you add everything you will get that

3 0

2 years ago

Read 2 more answers

Using the diagram, calculate the values of the unknown variables: a, b, and c.

Lady_Fox [76]

Answer:

a = 82

b = 130

c = 118

Step-by-step explanation:

A quadrilateral is inscribed in a circle. So, it is a cyclic quadrilateral.

Opposite angles of a cyclic quadrilateral are supplementary.

Therefore,

a° + 98° = 180°

a° = 180° - 98°

a° = 82°

a = 82

By inscribed angle theorem:

a° = 1/2(b° + 34°)

82° * 2 = b° + 34°

164° - 34° = b°

b° = 130°

b = 130

Again by inscribed angle theorem:

76° = 1/2(c° + 34°)

76° *2 = c° + 34°

152° - 34° = c°

c° = 118°

c = 118

3 0

3 years ago

A certain bridge arch is in the shape of half an ellipse 106 feet wide and 33.9 feet high. At what horizontal distance from the

nata0808 [166]

Answer:

The horizontal distance from the center is 49.3883 feet

Step-by-step explanation:

The equation of an ellipse is equal to:

$\frac{x^2}{a^{2} } +\frac{y^2}{b^{2} } =1$

Where a is the half of the wide, b is the high of the ellipse, x is the horizontal distance from the center and y is the height of the ellipse at that distance.

Then, replacing a by 106/2 and b by 33.9, we get:

$\frac{x^2}{53^{2} } +\frac{y^2}{33.9^{2} } =1\\\frac{x^2}{2809} +\frac{y^2}{1149.21} =1$

Therefore, the horizontal distances from the center of the arch where the height is equal to 12.3 feet is calculated replacing y by 12.3 and solving for x as:

$\frac{x^2}{2809} +\frac{y^2}{1149.21} =1\\\frac{x^2}{2809} +\frac{12.3^2}{1149.21} =1\\\\\frac{x^2}{2809}=1-\frac{12.3^2}{1149.21}\\\\x^{2} =2809(0.8684)\\x=\sqrt{2809(0.8684)}\\x=49.3883$