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Amiraneli [1.4K]
3 years ago
6

The base of a ladder is 10 feet from the side of a building. The top of the ladder reaches a window that is 24 feet above the gr

ound, What is the length of the ladder? A. 14 ft B. 22 ft C. 26 ft D. 34 ft​
Mathematics
2 answers:
zubka84 [21]3 years ago
8 0

\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}

Actually Welcome to the concept of Pythagorus theorem,

here, the perpendicular height = 24 ft and the base = 10ft ,

hence then length of the ladder is the Hypotenuse.

so we get as,

l \:  =  \sqrt{ {24}^{2}  +  {10}^{2} }

l \:  =  \sqrt{576 + 100}

l \:  =  \sqrt{676}

hence, l = 26 ft

so the length of the ladder is 26 feets.

Amanda [17]3 years ago
7 0

Answer: 26

Step-by-step explanation:

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marta [7]
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lets start of with a regular cos(x) graph.  This starts on its maximum instead of minimum so we have to multiply it by -1 to get -cos(x) which does start on its minimum.

-cos(x) has an amplitude of 1 instead of 10, to fix that we multiply it by 10 to get -10cos(x) which has an amplitude of 10.
 
-10cos(x) has a period of 2π instead of 10, to fix this we multiply the x by 2π/10 to get -10cos((π/5)x) which now has a period of 10.

-10cos((π/5)x) has a minimum of -10 and maximum of 10 instead of a minimum of -2 and maximum of 18, to fix this we add 8 to -10cos((π/5)x) to get -10cos((π/5)x)+8 which does have a minimum of -2 and maximum of 18.

Therefore the answer is y=-10cos((π/5)x)+8. x being time in minutes and and y being the height in feet.  

I hope this helps.  Let me know if anything is unclear.
8 0
3 years ago
Evaluate the limit assuming that limx→−5f(x)=17limx→−5f(x)=17 and limx→−5g(x)=22limx→−5g(x)=22. (use symbolic notation and fract
Gemiola [76]

In this question it is given that

\lim_{x->-5}f(x)=17, \lim_{x->-5}g(x)=22

And we have to find the value of the given limit

\lim_{x->-5}(23f(x)+3g(x))

Using properties of limit, first we separate the two functions, that is

23\lim_{x->-5}f(x)+3\lim_{x->-5}g(x)

Substituting the values of the given limit,

23(17)+3(22)=457

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For a circle with a radius of 6 meters, what is the measurement of a central angle (in degrees) subtended by an arc with a lengt
lesantik [10]
I hope this helps you




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8 0
3 years ago
Read 2 more answers
A certain company sends 40% of its overnight mail parcels by means of express mail service A1. Of these parcels, 4% arrive after
Harrizon [31]

Answer:

(a) The probability that a randomly selected parcel arrived late is 0.026.

(b) The probability that a parcel was late was being shipped through the overnight mail service A₁ is 0.615.

(c) The probability that a parcel was late was being shipped through the overnight mail service A₂ is 0.192.

(d) The probability that a parcel was late was being shipped through the overnight mail service A₃ is 0.192.

Step-by-step explanation:

Consider the tree diagram below.

(a)

The law of total probability sates that: P(A)=P(A|B)P(B)+P(A|B')P(B')

Use the law of total probability to determine the probability of a parcel being late.

P(L)=P(L|A_{1})P(A_{1})+P(L|A_{2})P(A_{2})+P(L|A_{3})P(A_{3})\\=(0.04\times0.40)+(0.01\times0.50)+(0.05\times0.10)\\=0.026

Thus, the probability that a randomly selected parcel arrived late is 0.026.

(b)

The conditional probability of an event A provided that another event B has already occurred is:

P(A|B)=\frac{P(B|A)P(A)}{P(B)}

Compute the probability that a parcel was late was being shipped through the overnight mail service A₁ as follows:

P(A_{1}|L)=\frac{P(L|A_{1})P(A_{1})}{P(L)} \\=\frac{0.04\times 0.40}{0.026} \\=0.615

Thus, the probability that a parcel was late was being shipped through the overnight mail service A₁ is 0.615.

(c)

Compute the probability that a parcel was late was being shipped through the overnight mail service A₂ as follows:

P(A_{2}|L)=\frac{P(L|A_{2})P(A_{2})}{P(L)} \\=\frac{0.01\times 0.50}{0.026} \\=0.192

Thus, the probability that a parcel was late was being shipped through the overnight mail service A₂ is 0.192.

(d)

Compute the probability that a parcel was late was being shipped through the overnight mail service A₂ as follows:

P(A_{3}|L)=\frac{P(L|A_{3})P(A_{3})}{P(L)} \\=\frac{0.05\times 0.10}{0.026} \\=0.192

Thus, the probability that a parcel was late was being shipped through the overnight mail service A₃ is 0.192.

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Eduardwww [97]
 Your answer is 19.53$ if you do add.
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