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shusha [124]
3 years ago
11

Find the difference. (-ab+3a-3) - (3ab+2)

Mathematics
2 answers:
lesantik [10]3 years ago
8 0

Answer:

-4ab+3a-5

Step-by-step explanation:

Hi there!

(-ab+3a-3) - (3ab+2)

Open up the parentheses

-ab+3a-3- 3ab-2

Combine like terms

-ab- 3ab+3a-3-2\\-4ab+3a-5

I hope this helps!

anyanavicka [17]3 years ago
6 0

Answer:

3a-4ab-5

Step-by-step explanation:

(-ab+3a-3) - (3ab+2)

Distribute the minus sign

(-ab+3a-3) - 3ab-2

Combine like terms

3a-4ab-5

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A lottery offers one $1000 prize, two $600 prizes, two $300 prizes, and five $200 prizes. One thousand tickets are sold at $7 ea
romanna [79]

Answer:

-$9.6

Step-by-step explanation:

X : ___ 1000 ____600 ____300 ___200

P(x): __ 1/1000 _ 2/1000 _ 2/1000 _ 5/1000

Expected winning per ticket :

Σ(X * p(X)) = [(1000 * 1/1000) + (600 * 2/1000) + (300 * 2/1000) + (200 * 5/1000) - price per ticket

= 1 + 1.2 + 0.6 + 1 - 7

= 3.8 - 7

= - $3.2

Expextwd winning if 3 tickets Is purchased :

3 * - 3.2= - $9.6

8 0
3 years ago
A. 140°<br><br> B. 300°<br><br> C. 220°<br><br> D. 180°<br><br> E. 320°
kvasek [131]

Answer:

140

Step-by-step explanation:

3 0
3 years ago
What is the value of x given the following image?
BlackZzzverrR [31]

Answer:

27

Step-by-step explanation:

CDE is a right angle so the sum of CDF and FDE is equal to 90

2x + x + 9 = 90 add like terms

3x + 9 = 90 subtract 9 from both sides

3x = 81 divide both sides by 3

x = 27

8 0
3 years ago
Explain why the x-coordinates of the points where the graphs of the equations y = 4^−x and y = 2^x + 3 intersect are the solutio
lawyer [7]
The point where 2 lines intersect contain (x,y) coordinates that are equal for both equations.
By setting the 2 equations equal to each other, it is same as equating the 2 y-values, which we know they are equal at point of intersection.
4 0
3 years ago
Draw a diagram in which segment AB intersects segment CD equals segment CB
Contact [7]
The <u>correct diagram</u> is attached.

Explanation:

Using technology (such as Geogebra), first construct a line segment.  Name the endpoints C and D.

Construct the perpendicular bisector of this segment.  Label the intersection point with CD as B, and create another point A above it.

Measure the distance from C to B and from B to D.  They will be the same.

Measure the distance from A to B.  If it is not the same as that from C to B, slide A along line AB until the distance is the same.

Using a compass and straightedge:

First construct segment CD, being sure to label the endpoints.

Set your compass a little more than halfway from C to D.  With your compass set on C, draw an arc above segment CD.

With your compass set on D (the same distance as before) draw an arc above segment CD to intersect your first arc.  Mark this intersection point as E.

Connect E to CD using a straightedge; mark the intersection point as B.

Set your compass the distance from C to B.  With your compass on B, mark an arc on EB.  Mark this intersection point as A.

AB will be the same distance as CB and BD.

5 0
3 years ago
Read 2 more answers
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