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sergeinik [125]
4 years ago
6

Which expression is equivalent to −1/6y+1/3 ?

Mathematics
1 answer:
Reil [10]4 years ago
3 0
Hello,

Which expression is equivalent to −1/6y+1/3 ?

−1/6(−y+1/3) = 1/6 y - 1/18 no

1/6(−y+2) = -1/6 y + 1/3 yes

−1/6(y+2) = -1/6 y - 1/3 no

1/6(−y+1/3) = -1/6 y + 1/18 no
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5 raised by 2- 5 (68-60)= <br> (with explanation plz)
Arlecino [84]

Step-by-step explanation:

2-5(68-60)=2-5×8=2-40=38

6 0
3 years ago
A radio is being discounted 40%. If the sale price of the radio is $31.02, what was the original price of the radio?
kobusy [5.1K]

Answer:

$51.70 is the original price

Step-by-step explanation:

Step 1:

Turn 40% into a decimal which makes it 0.40

ex. 50% would be 0.50, 5% would be 0.05 etc.

Step 2:

Put the numbers into this formula: Original Price (x) - Percent Off times Original Price = Sales Price

x - 0.40x = 31.02  this is basically 1x - 0.40x = 31.02 so,

Solving the left side of our equation gives us 1x - 0.40x = 0.60x

so now you have: 0.60x = 31.02

Step 3:

Divide both sides by 0.60 which gets x by itself and gives us the original price which is $51.70

8 0
3 years ago
If x and y are odd numbers such that 10&lt;x&lt;y14, what is value of x+y?
wel
I will suppose that your queation is 10<x<y<14
there are only two odd between 10 and 14
which are 11 and 13
since x is smaller than y
x=11 and y=13
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5 0
3 years ago
How many sq. ft. are in an acre?
Ainat [17]
43560 sq. ft
Answer till 20 characters 
8 0
3 years ago
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The position of an object along a vertical line is given by s(t) = −t3 + 3t2 + 7t + 4, where s is measured in feet and t is meas
saw5 [17]

Answer:

The maximum velocity of the object in the time interval [0, 4] is 10 ft/s.

Step-by-step explanation:

Given : The position of an object along a vertical line is given by s(t) = -t^3+3t^2+7t +4, where s is measured in feet and t is measured in seconds.

To find : What is the maximum velocity of the object in the time interval [0, 4]?

Solution :

The velocity is rate of change of distance w.r.t time.

Distance in terms of t is given by,

s(t) = -t^3+3t^2+7t +4

Derivate w.r.t. time,

v(t)=s'(t) = -3t^2+6t+7

It is a quadratic function so its maximum is at vertex of the function.

The x point of the function is given by,

x=-\frac{b}{2a}

Where, a=-3, b=6 and c=7

t=-\frac{6}{2(-3)}

t=-\frac{6}{-6}

t=1

As 1 lie between interval [0,4]

Substitute t=1 in the function,

v(t)= -3(1)^2+6(1)+7

v(t)= -3+6+7

v(1)=10

Th maximum velocity is 10 ft/s.

Therefore, the maximum velocity of the object in the time interval [0, 4] is 10 ft/s.

8 0
3 years ago
Read 2 more answers
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