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Law Incorporation [45]
3 years ago
10

HELP ME PLEASEEE! Ty:)

Mathematics
1 answer:
Jlenok [28]3 years ago
5 0
I can help you if you mark me as braì LG
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Expression A Expression B Expression C (−5)(−0.4)(1.8) (-45)(-19)(-112) −1(5+17) Expression D Expression E Expression F (−5)(−0.
kramer

Answer:

Expressions A, E, and F

Step-by-step explanation:

In general, when negative integers are multiplied; if the negative integers are even, then the product would be positive. But if the negative integers are odd, then the product would be negative.

Thus,

A. (−5)(−0.4)(1.8) = (2.0)(1.8)

                           = 3.6

B. (-45)(-19)(-112) = (855)(-112)

                           = -95760

C. −1(5+17) = -1(22)

                 = -22

D. (−5)(−0.4)(1.8)(−3.25) = (2.0)(-5.85)

                                      = -11.7

E. (-49)(-34)(-12)(-12) = (1666)(144)

                                = 239904

F. −1(3.5−7) = -1(-3.5)

                  = 3.5

Thus expressions A, E and F have positive products.

4 0
3 years ago
What is the equation for n in 5n+10=n+4
In-s [12.5K]

Answer:

-3/2 or 1.5

Step-by-step explanation:

5n + 10 = n + 4

5n - n = -10 + 4

4n = -6

n = -6/4

n = -3/2 (or -1.5)



3 0
4 years ago
Can anyone help with this?
son4ous [18]
Plug in x.

y= 0.1(3)^3

y=0.1(27)

y= 2.7

"C" is your answer.

I hope this helps!
~kaikers
6 0
4 years ago
Answer 4 and 5 plzzzzzzzz
Stolb23 [73]

Answer:

#4:so you find out the y and the x next to the 2 and if the answer is 10 you got it right  then you add it to this to the graph.

#5: i think i need help on this one

Step-by-step explanation:

hope this helps

5 0
3 years ago
Read 2 more answers
Find the equation of a circle with a center at (7,2) and a point on the circle at (2,5)?
monitta

Answer:

(x-7)^2+(y-2)^2=34

Step-by-step explanation:

We want to find the equation of a circle with a center at (7, 2) and a point on the circle at (2, 5).

First, recall that the equation of a circle is given by:

(x-h)^2+(y-k)^2=r^2

Where (<em>h, k</em>) is the center and <em>r</em> is the radius.

Since our center is at (7, 2), <em>h</em> = 7 and <em>k</em> = 2. Substitute:

(x-7)^2+(y-2)^2=r^2

Next, the since a point on the circle is (2, 5), <em>y</em> = 5 when <em>x</em> = 2. Substitute:

(2-7)^2+(5-2)^2=r^2

Solve for <em>r: </em>

<em />(-5)^2+(3)^2=r^2<em />

Simplify. Thus:

25+9=r^2

Finally, add:

r^2=34

We don't need to take the square root of both sides, as we will have the square it again anyways.

Therefore, our equation is:

(x-7)^2+(y-2)^2=34

4 0
3 years ago
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