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sammy [17]
4 years ago
13

There's the question :))​

Mathematics
1 answer:
Brut [27]4 years ago
7 0

Answer:

Should be A

Step-by-step explanation:

You going back 3 spaces but than having that three be divided by a positive so it's going to keep going because a negative will make a negative so that positive will be turned into a negative. -3 - 1.5 = -4.5

You might be interested in
Determine the first five multiples for the following number 22
iragen [17]

Lets find

\\ \sf\longmapsto M_n=22n

\\ \sf\longmapsto M_1=22(1)=22

\\ \sf\longmapsto M_2=22(2)=44

\\ \sf\longmapsto M_3=22(3)=66

\\ \sf\longmapsto M_4=22(4)

\\ \sf\longmapsto M_4=88

\\ \sf\longmapsto M_5=22(5)=110

3 0
3 years ago
What is the answer to the equation 6x-x/x+1
avanturin [10]

Answer:

5x/x+1

Step-by-step explanation:

Numerator: 6x minus an x equals 5x.

Denominator: Since x and 1 are not like terms, they can't be combined. You just leave them as they were.

Hope this helps!

6 0
4 years ago
2x + 7= -11 please help
Dvinal [7]

Answer:

x = -9

Step-by-step explanation:

You are solving for x. Note the equal sign, what you do to one side, you do to the other. Do the opposite of PEMDAS.

First, subtract 7 from both sides:

2x + 7 (-7) = -11 (-7)

2x = -11 - 7

2x = -18

Next, divide 2 from both sides:

(2x)/2 = (-18)/2

x = -18/2

x = -9

x = -9 is your answer.

~

6 0
3 years ago
Read 2 more answers
4.
marta [7]

Answer:

a) (x-2)(x-6) + (y-5)(y-11) =0\Rightarrow (x-4)^{2}+(y+3)^{2}=68\\C(4,-3) \:r=\sqrt{68} b) (a,b) (c, d) As long as (a,b) and (c,d) are the endpoints of of the diameter.

Step-by-step explanation:

a)

1) The reduced formula of the Circumference is given by:

(x-a)^{2}}+(y-b)^{2}=r^{2}

2) Let's expand the factored one into one closer to the pattern above:

x^{2}-8x+12+y^{2}+6y-55=0\Rightarrow x^{2}-8x+y^{2}+6y=43

3) Completing the square for both trinomials:

(x-4)^{2}+(y+3)^{2}=43+16+9\Rightarrow (x-4)^{2}+(y+3)^{2}=68

4) In the Reduced Formula, (x-a)^{2}}+(y-b)^{2}=r^{2},

C(a,b) \Rightarrow C(4,-3) \:the\:radius\:is\:r=\sqrt{68} \Rightarrow r=2\sqrt{17}

b) Using the previous example to show this:

When we  factor this way

(x-a)(x -c) + (y -b)(y-d) = 0

We are indeed, naming "a" and "b", the coordinates of (a, b) of the first endpoint and "b" and "d" the second endpoint as well.Id est, D (2, 5) and B (6,-11).

The radius, is \sqrt{68} \cong 8.25

So yes, the equation of the circle can be written as

(x-a)(x -c) + (y -b)(y-d) = 0

As long as (a,b) and (c,d) are the endpoints of of the diameter.

d_{AC}=d_{BC}=R

3 0
3 years ago
Help me with this, I don’t understand it. It’s angles that share a vertex.
Serga [27]
63 + 55 = 118
360- 118 = 242
6 0
3 years ago
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