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2 years ago

I need help with this math problem!

Mathematics

1 answer:

kherson [118]2 years ago

4 0

Answer: (50, 70)

The answer is (50, 70).

Step-by-step explanation:

Hope this helps =)

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Please help meeeeeeeee

Daniel [21]

Answer: 36

Step-by-step explanation:

30*6/5=36

you can check this by seeing that 5/6 of 36 is indeed 30.

5 0

3 years ago

Read 2 more answers

The 3rd term of the geometrical sequence is larger than the 2nd

notsponge [240]

Answer:the first term is 81

Step-by-step explanation:

The formula for determining the nth term of a geometric sequence is expressed as

Tn = ar^(n - 1)

Where

Tn represents the nth term

n represents the number of terms

r represents the common ratio

The 3rd term of the geometrical sequence is larger than the 2nd

term by 36. This means that

T3 - T2 = 36 - - - - - - - - - - 1

The product of these two terms is -243. This means that

T2 × T3 = - 243

T2 = - 243/T3

Substituting T2 = - 243/T3 into equation 1, it becomes

T3 - (- 243/T3) = 36

T3 + 243/T3 = 36

T3^2 + 243 = 36T3

T3^2 - 36T3 + 243 = 0

T3^2 - 27T - 9T3 + 243 = 0

T3(T3 - 27) - 9(T3 - 27) = 0

T3 - 9 = 0 or T3 - 27 = 0

T3 = 9 or T3 = 27

Substituting T3 = 9 or T3 = 27 into

T2 = - 243/T3 = 27, it becomes

T2 = - 243/9 or - 243/27

T2 = - 27 or T2 = - 9

Therefore,

The 3rd term is 9

The 2nd term is - 27

The common ratio, r would be

T3/T2 = 9/-27 = - 1/3

The expression for the third term would be

9 = a × - 1/3^(3 - 1)

9 = a × (- 1/3)^2

9 = a × 1/9

a = 9 × 9 = 81

3 0

3 years ago

Quadrilateral ABCD is inscribed in circle 0. Chords BA and CD are extended to intersect at point E. A tangent at B intersects li

irina [24]

Check the picture below.

let's notice the "white" ∡1 is an inscribed angle with an intercepted arc of (x-32), and the "green" ∡5 is also an inscribed angle with an intercepted arc of (2x).

the ∡6 and ∡2 are both external angles, however they intercepted two arcs, a "far arc" and a "near arc", thus we'll use the far arc - near arc formula, as you see in the picture, and we'll use the inscribed angle theorem for the other two.

$\bf \measuredangle 1=\cfrac{x-32}{2}\implies \measuredangle 1 =\cfrac{32}{2}\implies \measuredangle 1 = 16 \\\\[-0.35em] ~\dotfill\\\\ \measuredangle 5 =\cfrac{2x}{2}\implies \measuredangle 5 = x\implies \measuredangle 5 = 64 \\\\[-0.35em] ~\dotfill$

$\bf \measuredangle 2 = \cfrac{(2x+8)~~-~~(x-32)}{2}\implies \measuredangle 2=\cfrac{2x+8-x+32}{2} \\\\\\ \measuredangle 2=\cfrac{x+40}{2}\implies \measuredangle 2=\cfrac{104}{2}\implies \measuredangle 2=52 \\\\[-0.35em] ~\dotfill\\\\ \measuredangle 6=\cfrac{[(2x+8)+(x)]~~-~~(2x)}{2}\implies \measuredangle 6=\cfrac{3x+8-2x}{2}\implies \measuredangle 6=\cfrac{x+8}{2} \\\\\\ \measuredangle 6=\cfrac{72}{2}\implies \measuredangle 6=36$