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Stolb23 [73]
3 years ago
11

y is directly proportional to x. When y = 30, x = 6 a) Work out an equation connecting y and x. b) Work out the value of y when

x = 12.​
Mathematics
1 answer:
erica [24]3 years ago
5 0
Answer:
y=5x (your equation)

y=60 (your answer)

Step by step explanation:
You are given two numbers, 6 and 30. Since 6 goes into 30, 5 times, then y(which is equal to 30) is equal to x(which is equal to 6) multiplied by 5, so you get your equation,

Y=5X

then you have to solve for y, so just substitute

Y=5(12)

And simplify

Y=60

And there’s your answer
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Determine whether or not 630 is a triangular number. start with the formula t n = n ( n + 1 ) 2 and then use the quadratic formu
ziro4ka [17]
Triangular sequence = n(n + 1)/2

If 630 is a triangular number, then:

n(n + 1)/2  = 630

Then n should be a positive whole number if 630 is a triangular number.

n(n + 1)/2  = 630

n(n + 1)  = 2*630

n(n + 1)  = 1260

n² + n = 1260

n² + n - 1260 = 0

By trial an error note that 1260 = 35 * 36

n² + n - 1260 = 0

Replace n with 36n - 35n

n² + 36n - 35n - 1260 = 0

n(n + 36) - 35(n + 36) = 0

(n + 36)(n - 35) = 0

n + 36 = 0   or   n - 35 = 0

n = 0 - 36,   or  n = 0 + 35

n = -36, or 35

n can not be negative. 

n = 35 is valid.

Since n is a positive whole number, that means 630 is a triangular number.

So the answer is True.
7 0
3 years ago
Which property is shown by -3 x (6+5) = -3 x 6+ -3 x 5
aniked [119]

Answer:

The distributive property of multiplication

Step-by-step explanation:

we know that

The <u>distributive property of multiplication</u>  states that the product of a number by a sum, is equal to multiply each addend by the number (This is called distributing the number) and then, you can add the products

so

a*(b+c)=a*b+a*c

in this problem we have

-3*(6+5)=-3*6+-3*5 ----> the number -3 is distributed

therefore

We have the distributive property of multiplication

4 0
4 years ago
Use f (x)=7 and g (x)=x+5 to evaluate (f+g)(3)=
dangina [55]
(f+g)(3)=f(3)+g(3)

f(x)=7 so f(3)=7
g(x)=x+5 so g(3)=3+5=8

(f+g)(3)=f(3)+g(3)=7+8=15
4 0
3 years ago
Mrs. Welch and her 7 children are shopping at a local grocery store. Each of the children will be allowed to select one box of c
Sholpan [36]

Answer: I am pretty sure the answer is 40,320

Step-by-step explanation:

If you do 8x7x6x5x4x3x2 it equals 40,320

7 0
3 years ago
A particle moves according to a law of motion s = f(t), t ? 0, where t is measured in seconds and s in feet.
Usimov [2.4K]

Answer:

a) \frac{ds}{dt}= v(t) = 3t^2 -18t +15

b) v(t=3) = 3(3)^2 -18(3) +15=-12

c) t =1s, t=5s

d)  [0,1) \cup (5,\infty)

e) D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46

And we take the absolute value on the middle integral because the distance can't be negative.

f) a(t) = \frac{dv}{dt}= 6t -18

g) The particle is speeding up (1,3) \cup (5,\infty)

And would be slowing down from [0,1) \cup (3,5)

Step-by-step explanation:

For this case we have the following function given:

f(t) = s = t^3 -9t^2 +15 t

Part a: Find the velocity at time t.

For this case we just need to take the derivate of the position function respect to t like this:

\frac{ds}{dt}= v(t) = 3t^2 -18t +15

Part b: What is the velocity after 3 s?

For this case we just need to replace t=3 s into the velocity equation and we got:

v(t=3) = 3(3)^2 -18(3) +15=-12

Part c: When is the particle at rest?

The particle would be at rest when the velocity would be 0 so we need to solve the following equation:

3t^2 -18 t +15 =0

We can divide both sides of the equation by 3 and we got:

t^2 -6t +5=0

And if we factorize we need to find two numbers that added gives -6 and multiplied 5, so we got:

(t-5)*(t-1) =0

And for this case we got t =1s, t=5s

Part d: When is the particle moving in the positive direction? (Enter your answer in interval notation.)

For this case the particle is moving in the positive direction when the velocity is higher than 0:

t^2 -6t +5 >0

(t-5) *(t-1)>0

So then the intervals positive are [0,1) \cup (5,\infty)

Part e: Find the total distance traveled during the first 6 s.

We can calculate the total distance with the following integral:

D= \int_{0}^1 3t^2 -18t +15 dt + |\int_{1}^5 3t^2 -18t +15 dt| +\int_{5}^6 3t^2 -18t +15 dt= t^3 -9t^2 +15 t \Big|_0^1 + t^3 -9t^2 +15 t \Big|_1^5 + t^3 -9t^2 +15 t \Big|_5^6

And if we replace we got:

D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46

And we take the absolute value on the middle integral because the distance can't be negative.

Part f: Find the acceleration at time t.

For this case we ust need to take the derivate of the velocity respect to the time like this:

a(t) = \frac{dv}{dt}= 6t -18

Part g and h

The particle is speeding up (1,3) \cup (5,\infty)

And would be slowing down from [0,1) \cup (3,5)

5 0
3 years ago
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