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Tanzania [10]
2 years ago
13

PLEASE HELP

Mathematics
1 answer:
stira [4]2 years ago
5 0

Answer:

The values of x and y are x = 3 and y = 11

Step-by-step explanation:

The isosceles triangle has two equal sides in length, and its base angles are equal in measures

In ΔJKL

∵ ΔJKL is an isosceles triangle with vertex K

→ That means the equal sides are JK and KL

∴ JK = KL

∵ JK = 8x and KL = 13x - 15

→ Equate them

∴ 13x - 15 = 8x

→ Add 15 to both sides

∵ 13x - 15 + 15 = 8x + 15

∴ 13x = 8x + 15

→ Subtract 8x from both sides

∵ 13x - 8x = 8x - 8x + 15

∴ 5x = 15

→ Divide both sides by 5 to find x

∴ x = 3

∵ K is the vertex of the ΔJKL

∴ ∠KJL and ∠KLJ are the base angles

∵ The base angles are equal in measures

∴ m∠KJL = m∠KLJ

∵ m∠KJL = 5y - 3

∴ m∠KLJ = 5y - 3

∵ The sum of the measures of the angles of a Δ is 180°

∴ m∠KJL + m∠KLJ + m∠JKL = 180°

∵ m∠JKL = 76°

→ Substitute the measures of the 3 angles in the equation

∴ 5y - 3 + 5y - 3 + 76 = 180

→ Add the like terms on the left side

∵ (5y + 5y) + (76 - 3 - 3) = 180

∴10y + 70 =180

→ Subtract 70 from both sides

∵ 10y + 70 - 70 = 180 - 70

∴ 10y = 110

→ Divide both sides by 10 to find y

∴ y = 11

∴ The values of x and y are x = 3 and y = 11

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My name is Ann [436]
<span>From the message you sent me:

when you breathe normally, about 12 % of the air of your lungs is replaced with each breath. how much of the original 500 ml remains after 50 breaths

If you think of number of breaths that you take as a time measurement, you can model the amount of air from the first breath you take left in your lungs with the recursive function

b_n=0.12\times b_{n-1}

Why does this work? Initially, you start with 500 mL of air that you breathe in, so b_1=500\text{ mL}. After the second breath, you have 12% of the original air left in your lungs, or b_2=0.12\timesb_1=0.12\times500=60\text{ mL}. After the third breath, you have b_3=0.12\timesb_2=0.12\times60=7.2\text{ mL}, and so on.

You can find the amount of original air left in your lungs after n breaths by solving for b_n explicitly. This isn't too hard:

b_n=0.12b_{n-1}=0.12(0.12b_{n-2})=0.12^2b_{n-2}=0.12(0.12b_{n-3})=0.12^3b_{n-3}=\cdots

and so on. The pattern is such that you arrive at

b_n=0.12^{n-1}b_1

and so the amount of air remaining after 50 breaths is

b_{50}=0.12^{50-1}b_1=0.12^{49}\times500\approx3.7918\times10^{-43}

which is a very small number close to zero.</span>
5 0
3 years ago
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Leno4ka [110]

Answer:

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\frac{10}{22}

3 0
2 years ago
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Guided Practice
elena-s [515]

Answer:

see below

Step-by-step explanation:

The experimental probability of heads is

P( heads) = 28/50 = 14/25

The theoretical probability is

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Answer:

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