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slega [8]
2 years ago
6

2×__=9+9 Complete the equation

Mathematics
2 answers:
Hitman42 [59]2 years ago
6 0
The answer you are looking for is 9

Elenna [48]2 years ago
5 0

Answer:

2×9=9+9

Step-by-step explanation:

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Which relation is a function?
Oksanka [162]
It’s B because the domain(x) doesn’t repeat like the other options.
5 0
2 years ago
I
iren2701 [21]

Answer:

The distance between points A and B is 13

Step-by-step explanation:

We need to find distance between points A and B

Looking at the graph,

Point A : (2,6) and Point B: (-3,-6)

We need to find the distance between these points.

The formula used is: Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\

We have Points A : (2,6) and  B: (-3,-6)

So, x_1=2, y_1=6, x_2=-3, y_2=-6

Putting values in formula and finding distance

Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\\\Distance=\sqrt{(-3-2)^2+(-6-6)^2} \\Distance=\sqrt{(-5)^2+(-12)^2} \\Distance=\sqrt{25+144} \\Distance =\sqrt{169}\\Distance = 13

So, the distance between points A and B is 13

6 0
3 years ago
Can somebody PLEASE HELP ME
svet-max [94.6K]

Answer:

150.796π square meters

maybe!

Step-by-step explanation:

4 0
2 years ago
The number y of hits a new website receives each month can be modeled by y = 4070ekt, where t represents the number of months th
makvit [3.9K]

Answer:

<u><em>k = 0.2645</em></u>

Step-by-step explanation:

Given model:

y = 4070 e^{kt}

y = no. of hits website received = 9000 (in 3rd month)

t= no. of months website has been operational = 3

put in the above equation:

9000 = 4070  e^{3k}

\frac{9000}{4070} = e^{3k}

\frac{900}{407}=e^{3k}

<u><em>Taking natural logarithm on both sides, we get:</em></u>

ln\frac{900}{407}=ln(e^{3k})

ln\frac{900}{407}= 3k ln<em>e</em>

ln<em>e</em>=1

ln \frac{900}{407}= 3k

or k = \frac{1}{3}ln\frac{900}{(407)}

k =\frac{1}{3}(0.7936)

<em>k = 0.2645</em>

<em />

7 0
2 years ago
Solve the equation -31=-6z-4z
jok3333 [9.3K]
<span>-31=-6z-4z
Subtract 4z from -6z
-31=-10z
Divide both sides by -10
Final Answer: 3.1 or 3 1/10</span>
8 0
2 years ago
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