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Rzqust [24]
3 years ago
6

What is the ratio for 4, 8, 12, 14 (r)

Mathematics
1 answer:
olasank [31]3 years ago
7 0

answer for this question is 2 is to 4 is to6 is to 7

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Does it lie inside, outside, or on the circle?​
FrozenT [24]

Answer:

<u>Given circle:</u>

  • (x - 4)² + y² = 25

The center is at (4, 0) and the radius is 5 units

<u>Lets find the distance from (7, 2) to the center of the circle:</u>

  • <u />d = \sqrt{(7 - 4)^2 + (2 - 0)^2}  = \sqrt{9+4} = \sqrt{13} < 5

Since d < 5, the point (7, 2) lies <u>inside</u> the circle

8 0
3 years ago
Multiply the polynomial (12x^4 - 4x^3 + 5) by (-4x^2) *
murzikaleks [220]

Answer:

<em>- 48 {x}^{6}  + 16 {x}^{5}  - 20 {x}^{2}</em>

<em>Option</em><em> </em><em>D</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>correct</em><em> </em><em>option</em><em>.</em>

<em>Sol</em><em>ution</em><em>,</em>

<em>- 4 {x}^{2} (12 {x}^{4}  - 4 {x}^{3}  + 5) \\  =  - 4 {x}^{2}  \times 12 {x}^{4}  + 4 {x}^{2}  \times 4 {x}^{3}  - 4 {x}^{2}  \times 5 \\  =  - 48 {x}^{6}  + 16 {x}^{5}  - 20 {x}^{2}</em>

<em>hope</em><em> </em><em>this </em><em>helps</em><em>.</em><em>.</em><em>.</em>

<em>Good</em><em> </em><em>luck</em><em> on</em><em> your</em><em> assignment</em><em>.</em><em>.</em><em>.</em>

7 0
3 years ago
How to simplify (81^-0.25)^3
IRISSAK [1]
(81^-0.25)^3 = ( 1 / (81^0.25) )^3 

<span>81^.025 is the 4th root of 81 which is 3 </span>
<span>Therefor </span>
<span>( 1 / (81^0.25) )^3 = (1/3)^3 </span>

<span>(1/3)^3 = 1/27 <-----

Hope I Helped You!!! :-)

Have A Good Day!!!</span>
4 0
3 years ago
Solve the equation 3sin2 theta = sin e, for values of e between Oº and 360°
Reil [10]

Answer:

[B] 0, 19.5, 160.5, 180, 360

Step-by-step explanation:

3 sin²θ = sin θ

3 sin²θ − sin θ = 0

sin θ (3 sin θ − 1) = 0

sin θ = 0 or sin θ = ⅓

If sin θ = 0, θ = 0°, 180°, 360°.

If sin θ = ⅓, θ = 19.5°, 160.5°.

5 0
3 years ago
Can anyone pls help me with this I'll mark as brainlist for correct answer please do explain in detail each answers Ty! (I'm pre
densk [106]

<u>Question 8</u>

a^2 + 7a + 12

= (a+3)(a+4)

When factorising a quadratic, the product of the two factors should equal the constant term (12), and the sum of the two factors should equal the linear term (7). To find the two factors, list out the factors of 12 (1x12, 2x6, 3x4) and identify the pair that adds up to 7 (3+4).

An alternative method if you get stuck during your exam would be to solve it algebraically using the quadratic formula and then write it in the factorised form.

a = (-7 +or- sqrt(7^2 - 4(1)(12)) / 2(1)

= (-7 +or- sqrt(1))/2

= -3 or -4

These factors are the negative of the values that would go in the brackets when written in factorised form, as when a = -3 the factor (a+3) would equal 0. (If it were positive 3 instead, then in the factorised form it would be a-3).

<u>Question 10</u>

-3(x - y)/9 + (4x - 7y)/2 - (x + y)/18

Rewrite each fraction with a common denominator so you can combine the fractions into one.

= -6(x - y)/18 + 9(4x - 7y)/18 - (x + y)/18

= (-6(x - y) + 9(4x - 7y) - (x + y)) /18

Expand the brackets and collect like terms.

= (-6x + 6y + 36x - 63y - x - y)/18

= (29x - 58y)/18

= 29/18 x - 29/9 y

7 0
2 years ago
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