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IceJOKER [234]
3 years ago
8

Which of the expressions below has a 4 in the ones place of the sum? Select all that apply.

Mathematics
2 answers:
lara [203]3 years ago
8 0

Answer:

its C or D

Step-by-step explanation:

i think it's both !

Molodets [167]3 years ago
6 0
Answer is D because 91.45 + 13.23= 104.68 4 is in the ones place
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HELP!!!!!!!!! ASP
Anarel [89]

Answer:

D

Step-by-step explanation:

What you need to do is find the slope of both lines. Perpendicular lines are the negative inverse of each other. The slope of the lie between points R and F is  (2-4)/(1+9)=-1/5. Now you need to find the line where the slop is 5. If you look at D, the slope of the line those points lie on is (25-15)/(4-2)=5 which makes D the answer.

3 0
3 years ago
Read 2 more answers
Which matrix is equal to
polet [3.4K]
The one that look exactly the same it the third one
3 0
3 years ago
Please help me!!!!!!!!!!!!!!!
ycow [4]

Answer:

  7)  (f+g)(x) = 4^x +5x -5

  8)  (f-g)(x) = 4^x +x +5

Step-by-step explanation:

7) add the two expressions.

  (f+g)(x) = f(x) +g(x) = (4^x +3x) +(2x -5)

  (f+g)(x) = 4^x +5x -5

__

8) subtract g(x) from f(x).

  (f-g)(x) = f(x) -g(x) = (4^x +3x) -(2x -5) = 4^x +3x -2x +5

  (f-g)(x) = 4^x +x +5

7 0
3 years ago
Simplify ..............
larisa86 [58]

Answer:

The third response

Step-by-step explanation:

4 0
3 years ago
Is the given point interior , exterior, or on the circle k (x+2)2 + (y-3)2 =18 P (8,4)
Mnenie [13.5K]
One way would be to find the distance from the point to the center of the circle and compare it to the radius

for
(x-h)^2+(y-k)^2=r^2
the center is (h,k) and the radius is r

and the distance formula is
distance between (x_1,y_1) and (x_2,y_2) is
D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}


r=radius
D=distance form (8,4) to center

if r>D, then (8,4) is inside the circle
if r=D, then (8,4) is on the circle
if r<D, then (8,4) is outside the circle


so
(x+2)^2+(y-3)^2=18
(x-(-2))^2+(y-3)^2=(\sqrt{18})^2
(x-(-2))^2+(y-3)^2=(3\sqrt{2})^2

the radius is 3\sqrt{2}
center is (-2,3)

find distance between (8,4) and (-2,3)

D=\sqrt{(8-(-2))^2+(4-3)^2}
D=\sqrt{(8+2)^2+(1)^2}
D=\sqrt{10^2+1}
D=\sqrt{100+1}
D=\sqrt{101}




r=3\sqrt{2}≈4.2
D=\sqrt{101}≈10.04

do r<D

(8,4) is outside the circle

6 0
3 years ago
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