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yan [13]
3 years ago
6

Help Plz I will mark Brainliest

Mathematics
2 answers:
olya-2409 [2.1K]3 years ago
8 0

Answer:

c

Step-by-step explanation:

olasank [31]3 years ago
5 0
It’s c ........……’

D C was was r
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Two-thirds Times a number plus 7 equals 7 minutes the number. Find the number
skelet666 [1.2K]
2/3x + 7 = x-7
         +7      +7
---------------------
2/3x + 14 = x
-2/3x        -2/3x
----------------------
14 = 1/3x
*3     *3
-------------
42 = x

Hope this helps! :)


3 0
3 years ago
What's the solution for y+1/2=3/4
creativ13 [48]
\frac{y+1}{2} = \frac{3}{4}

2(y+1) = 3        LCD is 4

2*y + 2*1 = 3

2y + 2 = 3

2y = 3 - 2

2y = 1

y = 1 / 2


6 0
3 years ago
Please help with #2 I’m struggling and can someone please explain and give answer??!!!!
MatroZZZ [7]

Check the picture below.

8 0
2 years ago
The slope of curve zz at point c is approximately
lana66690 [7]
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4 0
3 years ago
find the angle between the vectors. (first find the exact expression and then approximate to the nearest degree. ) a=[1,2,-2]. B
SashulF [63]

Answer:

\theta = cos^{-1} (\frac{10}{\sqrt{9} \sqrt{25}})=cos^{-1} (\frac{10}{15}) = cos^{-1} (\frac{2}{3}) = 48.190

Since the angle between the two vectors is not 180 or 0 degrees we can conclude that are not parallel

And the anfle is approximately \theta \approx 48

Step-by-step explanation:

For this case first we need to calculate the dot product of the vectors, and after this if the dot product is not equal to 0 we can calculate the angle between the two vectors in order to see if there are parallel or not.

a=[1,2,-2], b=[4,0,-3,]

The dot product on this case is:

a b= (1)*(4) + (2)*(0)+ (-2)*(-3)=10

Since the dot product is not equal to zero then the two vectors are not orthogonal.

Now we can calculate the magnitude of each vector like this:

|a|= \sqrt{(1)^2 +(2)^2 +(-2)^2}=\sqrt{9} =3

|b| =\sqrt{(4)^2 +(0)^2 +(-3)^2}=\sqrt{25}= 5

And finally we can calculate the angle between the vectors like this:

cos \theta = \frac{ab}{|a| |b|}

And the angle is given by:

\theta = cos^{-1} (\frac{ab}{|a| |b|})

If we replace we got:

\theta = cos^{-1} (\frac{10}{\sqrt{9} \sqrt{25}})=cos^{-1} (\frac{10}{15}) = cos^{-1} (\frac{2}{3}) = 48.190

Since the angle between the two vectors is not 180 or 0 degrees we can conclude that are not parallel

And the anfle is approximately \theta \approx 48

3 0
3 years ago
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