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Lorico [155]
4 years ago
15

Use Order of Operations to simplify. 42 + 5[61 – (5x6)]

Mathematics
1 answer:
NNADVOKAT [17]4 years ago
6 0

Answer:

BEDMAS/PEMDAS - [61 - (5x6)] x 42 + 5

Step-by-step explanation:

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How do you add integers with the same sign? <br><br><br><br>PLEASE I NEED IT NOW AND THANK YOU
dedylja [7]
When you add integers, you keep the sign on the sum.

For example:
-9 + -9 = -18 (kept the sign)
9 + 9 = 18 (kept the "sign")
3 0
3 years ago
The area of a triangle is 80. The length is 2 more than the width. Find the dimensions
Jobisdone [24]

Answer:

The width is 19 units. The length is 21 units.

Step-by-step explanation:

W=width: L=length=W+2; P=perimeter=80

P=2(L+W)

80=2(W+2+W)

40=2W+2

38=2W

19=W answer 1: The width is 19 units.

L=W+2=21 ANSWER 2: The length is 21 units.

CHECK:

P=2(L+W)

80=2(21+19)

80=2(40)

80=80

4 0
3 years ago
What's the addition property of 23 + 4 + 109 = 4 + 23 + 109
Grace [21]

Answer:

I am confused on this question too

Step-by-step explanation:

8 0
3 years ago
Give the equation of the line through the point (4, 5) with a slope of - 9/2
kodGreya [7K]

Answer:

Step-by-step explanation:

y - y1 = m(x - x1)

y - 5 = -9/2*(x - 4)

y -5  = -9/2x + 9*4/2

y - 5 = -9x/2 +9*2

y-5=-9x/2 + 18

y= -9x/2 + 18 +5

y= -9x/2 + 23    {y =mx+c}

4 0
3 years ago
Read 2 more answers
Help me pls need help​
frez [133]

Problem 1

w = width

w+36 = length, because it's 36 feet longer compared to the width

=======================================================

Problem 2

The area w(w+36) is less than 2040, and it's also larger than 0.

We can write that as 0 \le w(w+360) \le 2040 which is the same as 0 \le w^2+360w \le 2040

If you wanted to drop the first part, then you can say either w(w+360) \le 2040 or w^2+360w \le 2040

I used the formula area = length*width

=======================================================

Problem 3

There are many possibilities here. Let's say w = 10 feet. If so then the length would be w+36 = 10+36 = 46 feet. This 10 by 46 rectangle has an area of 10*46 = 460 square feet which is under the 2040 sq ft limit.

Another possibility is that w = 20 ft and w+36 = 56. This 20 by 56 rectangle has an area of 20*56 = 1120 sq ft.

It turns out you can pick any value of w between 1 and 30, assuming you only restrict yourself to integers. You can find the largest possible value of w by solving the equation w(w+36) = 2040. The positive solution to this equation is roughly w = 30.62

Side note: even though something like w = 1 is possible, it's not very realistic. A conference hall that's only 1 ft wide won't be able to fit a person comfortably unless they don't mind being sandwiched between two walls and have to walk sideways.

=======================================================

Problem 4

If you solved w(w+36) = 360 with a graphing calculator or the quadratic formula, then you would find the positive solution for w is roughly 8.15

If your teacher is considering positive real numbers, then this is realistic; however, if they are only considering positive integers, then something like w = 8.15 isn't possible.

5 0
3 years ago
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