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LiRa [457]
3 years ago
14

The sum of Pete's and Sam's ages is 30. Five years ago, Pete was 3 times as old as Sam. How old is Sam?

Mathematics
2 answers:
abruzzese [7]3 years ago
5 0

Answer : The equation complete the system is, P-5=3S-15 and the Sam is 10 years old.

Step-by-step explanation :

As we are given that sum of Pete's and Sam's ages is 30. Thus, the equation will be:

P+S=30     ..............(1)

P=30-S        ..............(A)

And, five years ago, Pete was 3 times as old as Sam. Thus, the equation will be:

(P-5)=3\times (S-5)     ..............(2)

P-5=3S-15

Now substituting equation A in equation 2, we get:

(P-5)=3\times (S-5)

(30-S-5)=3\times (S-5)

30-S-5=3S-15

30+15-5=3S+S

40=4S

S=\frac{40}{4}

S=10

Now put the value of 'S' in equation A, we get the value of 'P'.

P=30-S

P=30-10

P=20

The age of Pete is, 20 and the age of Sam is, 10

Thus, the equation complete the system is, P-5=3S-15 and the Sam is 10 years old.

RUDIKE [14]3 years ago
4 0
P-5=3s-15 because Pete is 20 and Sam is 10. 20-5= 15 and 3(10)-15 also equals 15. The answer is C.
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A tower stand on a horizontal plane. P is the top and Q is the bottom of the tower. A and B are two points on this plane such th
pochemuha

Answer:

32√5

Step-by-step explanation:

We have the right triangles PQA and PQB as well as the given right triangle QAB.

cot(PAQ) = 2/5 = QA/PQ

cot(PBQ) = 3/5 = QB/PQ

cot(PAQ) / cot(PBQ) = (2/5) / (3/5) = 2/3

cot(PAQ) / cot(PBQ) = (QA/PQ) / (QB/PQ) = QA / QB

QA / QB = 2/3

QA = (2/3) QB

QB = (3/2) QA

By the Pythagorean Theorem we have:

(QA)² + 32² = (QB)²

(QA)² + 32² = (3/2 QA)²

(QA)² + 1024 = (9/4) (QA)²

(5/4) (QA)² = 1024

(QA)² = (4/5)1024 = 4096/5

QA = 64/√5

Solve for PQ.

cot(PAQ) = QA/PQ

PQ = QA / cot(PAQ)

PQ = (64/√5) / (2/5) = 32√5

The height of the tower is 32√5.

7 0
3 years ago
Which pair of funtions is not a pair of inverse functions? please help!!
antiseptic1488 [7]

Answer:

f(x)=\frac{x}{x+20} , g(x)=\frac{20x}{x-1}

Step-by-step explanation:

we know that

To find the inverse of a function, exchange variables x for y and y for x. Then clear the y-variable to get the inverse function.

we will proceed to verify each case to determine the solution of the problem

<u>case A)</u> f(x)=\frac{x+1}{6} , g(x)=6x-1

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

x=\frac{y+1}{6}

Isolate the variable y

6x=y+1

y=6x-1

Let

f^{-1}(x)=y

f^{-1}(x)=6x-1

therefore

f(x) and g(x) are inverse functions

<u>case B)</u> f(x)=\frac{x-4}{19} , g(x)=19x+4

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

x=\frac{y-4}{19}

Isolate the variable y

19x=y-4

y=19x+4

Let

f^{-1}(x)=y

f^{-1}(x)=19x+4

therefore

f(x) and g(x) are inverse functions

<u>case C)</u> f(x)=x^{5}, g(x)=\sqrt[5]{x}

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

x=y^{5}

Isolate the variable y

fifth root both members

y=\sqrt[5]{x}

Let

f^{-1}(x)=y

f^{-1}(x)=\sqrt[5]{x}

therefore

f(x) and g(x) are inverse functions

<u>case D)</u> f(x)=\frac{x}{x+20} , g(x)=\frac{20x}{x-1}

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

x=\frac{y}{y+20}

Isolate the variable y

x(y+20)=y

xy+20x=y

y-xy=20x

y(1-x)=20x

y=20x/(1-x)

Let

f^{-1}(x)=y

f^{-1}(x)=20x/(1-x)

\frac{20x}{1-x}\neq \frac{20x}{x-1}

therefore

f(x) and g(x) is not a pair of inverse functions

7 0
2 years ago
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5 0
3 years ago
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