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Kryger [21]
1 year ago
11

Evaluate the expression 2x-(4y-3)+5xz, when x=-3,y=2, and z=-1

Mathematics
1 answer:
AVprozaik [17]1 year ago
3 0

Answer: 4

Step-by-step explanation:

2x - (4y - 3) + 5xz

Let  x = -3, y = 2, and z = ‐1.

2(-3) -( 4(2) -3) +5(-3)(-1)

-6 - (8-3) +15

-6 - 5 +15

-11 +15

4

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How can you use number patterns to find the least common factor
Aleks04 [339]

Answer:

The correct answer is: 360.

Explanation:

First we can express 120 as follows:

2 * 2 * 2 * 3 * 5 = 120

You can get the above multiples as follows:

120/2 = 60

60/2 =30

30/2 = 15

15/3 = 5 (Since 15 cannot be divisible by 2, so we move to the next number)

5/5 = 1

Take all the terms in the denominator for 120, you would get: 2 * 2 * 2 * 3 * 5 --- (1)

Second we can express 360 as follows:

360/2 = 180

180/2 = 90

90/2 =45

45/3 = 15 (Since 45 cannot be divisible by 2, so we move to the next number)

15/3 = 5

5/5 = 1

Take all the terms in the denominator for 360, you would get: 2 * 2 * 2 * 3 * 3 * 5 --- (2)

Now in (1) and (2) consider the common terms once and multiple that with the remaining:

2*2*2*3*5 = Common between the two

3 = Remaining

Hence (2*2*2*3*5) * (3) = 360 = LCM (answer)

5 0
3 years ago
What is x^4+25 factored
OLga [1]

(x^2+5)(x^2-5)

I hope this helps :)

3 0
3 years ago
Read 2 more answers
11. |6- 15| - |5(-4)|= ?<br> A.-29<br> B. -11<br> C. 1<br> D. 11<br> E. 29
Burka [1]

Answer:

B

Step-by-step explanation:

|6- 15| - |5(-4)| =

|-9| - |-20| =

(9)- (20) =

-11

4 0
2 years ago
2. Please answer the question on exponential growth or decay. Round to
Ivan

Exponential growth

Equation:

y=748(1+.03)^10


Solve…

Answer: 1005 minutes.

Like if it helped! :)
4 0
1 year ago
It is 70cm long and has a circumference of 40cm what is the radius
Aleksandr [31]
The legnth is irrelivant

we know that
Circumference=2 times pi time radius

c=40
40=2 times pi times radius
divide both sides by 2
20=pi times radius
divide both sides b pi
20/pi=radius

the radius is 20/pi cm or aproximately 6.366199048 cm
4 0
3 years ago
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