Mathematics, 19.10.2021 14:50 larissaday33

I'ma keep posting this question

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Answer from: khadythiam6957

3/5

Step-by-step explanation:

An easy way of dividing fractions is keep change flip

Meaning, keep the first fraction, change divide to multiply, and then flip the second fraction.

2/5 ÷ 2/3 = 2/5 × 3/2 = 3/5

Answer from: MyaMya12

3/5

Step-by-step explanation:

So you are gonna need the reciprocal of 2/3, thats 3/2 next you have to multiply 2/5 and 3/2 and you will get 6/10. now you have to simplify. so you divide by 2 and get 3/5

Answer from: Quest

Intersecting secant-tangent theorem states that if a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.so to break it down: 1. pq acts as the tangent and ps as the secant with an intersection at p.sr = 21rp = 3x + 3pq = 4x + 4following to equation: tangent squared = the two segments multiplied by each otherso (4x + 4) ^ 2 = (3x + 3) • (21) simplifies to 16x^2 - 31x - 47 = 0 or (16x - 47)(x + 1) x = -1 or x = 47/162. pq acts as the tangent and ps as the secant with an intersection at p (again! )sr = 16rp = xpq = 15(15) ^ 2 = 16 • x225 = 16xx = 225/163. the interesting chords are proportional to each other so a ratio is possible to set up: ap dp 10 3 + x = = —— = pc pb 8 xcross multiple to get 10x = (3 + x)(8)x = 12

Answer from: Quest

x= -3 and $x=\frac{-3+/-3i\sqrt{3} }{2}$

step-by-step explanation:

to find the roots of a cubic, use division or factoring to find the factors then set equal to 0. as this is a difference of cubes, it factors into

$(x^3-a^3)=$ $(x-a)(x^2+ax+a^2)$

where a is the cube root of $a^3$ .

here this is 27 and the cube root of 27 is 3. so we write using the form:

$(x-3)(x^2+(3)x+3^2)\\(x-3)(x^2+3x+9)$ .

set each factor to 0 and solve for x.

x-3=0

x=3

the quadratic expression does not factor and must be solved using the quadratic formula.

$x=\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

where a = 1, b=3, and c=9

$x=\frac{-3+/-\sqrt{3^2-4(1)(9)} }{2(1)} \\x=\frac{-3+/-\sqrt{9-36)} }{2} \\x=\frac{-3+/-\sqrt{-27)} }{2}\\x=\frac{-3+/-3i\sqrt{3} }{2}$

this cubic has two complex roots.

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