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stiks02 [169]
3 years ago
7

y varies jointly as x and z, and y = 32 m, x = 6 m, and z = 8 m. What is the value of y when x = 5 m and z = 12 m

Mathematics
2 answers:
elixir [45]3 years ago
5 0

Answer:

y = 40

Step-by-step explanation:

If y varies jointly as x and z, the equation is

y = k* xz

We know y  x  and z  so we can substitute them in

32 = k * 6*8

32 = k*48

Divide by 48

32/48 = k*48/48

32/48 =k

2/3 =


So the equation is y =2/3  x*z

Since we know x=5 and z =12, we can find y

y = 2/3 * 5 * 12

y = 40

Fantom [35]3 years ago
3 0
We are given y = kxz
We also know 32 = k×6×8, so k = 32/48 = 2/3.
We are then asked to solve:

y =  \frac{2}{3}  \times 5 \times 12 = 40
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in justin's school,0.825 of students participate in a sport. if there are one thousand students in justin's school, how many pat
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Write three consecutive multiples of 13 in a general form.
S_A_V [24]

Answer:

answer given in step by step...

Step-by-step explanation:

13x+13(x+1)+13(x+2)=312

13x+13x+13+13x+26=312

39x+39=312

39x=312-39

39x=273

x=7

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5 0
3 years ago
00:36:46
miskamm [114]

Question:

In some code, letters a, b, c, d and e represent numbers 2, 4, 5, 6 and 10. We just do not know which letter represents which number. Consider the following relationships:

a + c = e,

b – d = d and

e + a = b

Which of the following statements is true?

(A) b = 4, d = 2

(B) a = 4, e = 6

(C) b = 6, e = 2

(D) a = 4, c = 6

Answer:

(B) a = 4, e = 6 is the correct answer.

Explanation:

(A) b = 4, d = 2

a + c = e => a + c = e

b – d = d => 4 - 2 = 2

e + a = b => e + a = 4

We are not able to find out what will be at place of a , c , e by which all other relationships are not getting satisfied. And if we apply numbers by hit and trial they are not getting satisfied.

(B) a = 4, e = 6

a + c = e => 4 + c = 6 => 4 + 2 = 6

b – d = d => b - d = d => 10 - 5 = 5

e + a = b => 6 + 4 = 10

We were able to find all a , b , c , d , e and it satisfies all relationships given.

(C) b = 6, e = 2

a + c = e => a + c = 2

b – d = d => 6 - d = d

e + a = b => e + a = 6

We are not able to find out what will be at place of a , c , d by which all other relationships are not getting satisfied. And if we apply numbers by hit and trial they are not getting satisfied.

(D) a = 4, c = 6

a + c = e => 4 + 6 = 10

b – d = d => b - d = d

e + a = b => e + 4 = b => 10 + 4 = 14(this is not from the numbers as given in question so b can't be equal to 14)

We got stuck at the value of b which was 14 and not given in question so, we can leave this option in between as it is not satisfying the relationships.

Therefore, (B) a = 4, e = 6 is the correct answer.

4 0
3 years ago
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