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jenyasd209 [6]
3 years ago
7

Jay evaluated the expression 3x(3+12 dived by 3)-4. For his first step, he added 3+12 to get 15. What was Jay's error? Find the

correct answer.
Mathematics
2 answers:
sp2606 [1]3 years ago
6 0
Bidmas, division before addition3x(3+12/3)-4 = 3x(3+4)-4 = 21-4 = 17
Naya [18.7K]3 years ago
6 0
He would get 11x because 3t12equels 15 devided by 3 eqals 5 x 3x eqales 15x minus 4 eqauls 11x
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What is the equation of a line that passes through (-2, 4) and has a slope of 3?
devlian [24]

Equation,

y - y1 = m(x - x1)

Plugging in our values,

x1 =  - 2, \: y1 = 4, \: m = 3

y - 4 = 3(x - ( - 2))

y - 4 = 3(x + 2)

So the equation is,

y - 4 = 3(x + 2)

6 0
3 years ago
What is the area of a circle with a diameter of 16?
tigry1 [53]

Answer:

<h2>64π</h2>

Step-by-step explanation:

A = π r2

diameter : 2 = r

16 : 2 = 8

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3 0
3 years ago
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Aubree practices the piano 1197 minutes in 3 weeks. At what rate did she practice, in minutes per day?
Norma-Jean [14]

Answer:

57

Step-by-step explanation:

We need to create a ratio. so Aubree practices the piano for 1197 min: in 3 weeks. So for 1 week, we have to divide 1197 by 3, so 1197/=399. And so we know that 1 week is 7 days, just divide 399 by 7! 399/7=57. The rate that she practiced per day is 57 minutes. So one last final step just to check if this answer is correct is to divide 57 by 60, and that is 0.95. 0.95 out of 1.00 minutes per day is 95/100. Cross multiply with x/60, and you will get your answer, which is 57.

4 0
2 years ago
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Find the area of the triangle that divides the parallelogram in half.​
nignag [31]

equation: 1/2bh

workings...

13x9= 117

117/2=58.5

answer: 58.5cm^2

4 0
4 years ago
The class midpoint of a class is
stiv31 [10]

Answer:

The center of that class, which is the sum of its largest and smallest values divided by 2 ⇒ E

Step-by-step explanation:

* Lets explain what is the class mid-point

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- The class midpoint is the lower class limit plus the upper class limit

 divided by 2

- The easiest way to find the class mid-point is to add the upper

  and lower boundary and divide your answer by two

- The lower limit for every class is the smallest value in that class.

- The upper limit for every class is the greatest value in that class

* <u><em>Lets solve the problem</em></u>

- It is not the largest value of that class minus the class width

- Its not the difference between the largest and smallest values of

 that class

- It is not the difference between the largest and smallest values of

that class divided by 2

- It is the center of that class, which is the sum of its largest and

 smallest values divided by 2

3 0
3 years ago
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