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Anvisha [2.4K]
3 years ago
14

Bdjdnndndndndndkkdkdkd

Mathematics
1 answer:
andrew-mc [135]3 years ago
8 0
I think the answer is C
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Which of the following is an equation for the plane through the point (1, 1, −1) and parallel
Bas_tet [7]

The given plane, x-y+z=3, has normal vector \vec n = \langle1,-1,1\rangle. Any plane parallel to this one has the same normal vector.

Let (x,y,z) be any point in the plane we want. The plane contains the point (1, 1, -1), so an arbitrary vector in this plane is

\langle x,y,z\rangle - \langle 1,1,-1\rangle = \langle x-1, y-1, z+1 \rangle

and this is perpendicular to \vec n.

So the equation of the plane is

\langle x-1, y-1, z+1 \rangle \cdot \vec n = 0 \implies (x-1) - (y-1) + (z+1) = 0 \implies \boxed{x - y + z = -1}

or equivalently,

\boxed{-x + y - z = 1}

6 0
2 years ago
1. In triangle ABC, if AB = 3cm, BC = 4cm and AC = 5cm Find
Veseljchak [2.6K]

Answer:

Step-by-step explanation:

A) The diagram of the triangle is shown in the attached photo

The lengths of the given triangle form a Pythagoras triple. This means that the square of the longest side equals the sum of the squares of the two shorter sides.

Area = 1/2 × base × height

Height = 3 cm

Base = 4 cm

Area = 1/2 × 3 × 4 = 6 cm²

B) The altitudes can be AB or BC. Therefore, the length of the shortest altitude is AB = 3 cm

The diagonal is the same as the hypotenuse which is also the longest side of the triangle. Thus,

Length of diagonal = 5 cm

8 0
3 years ago
HELP A.S.A.P!!!! The data to represent average test scores for a class of 16 students includes an outlier value of 78. If the ou
kolezko [41]

When the outlier is removed The mean would increase.

<u>Step-by-step explanation:</u>

Outliers are the value that lies outside the set of data. i.e. the data doesn't fit in the group.

Removing outliers can affect mean more than median. Because outlier removal sometimes affect the median but in a small percentage or sometimes it will remain the same.

Let us prove, consider 16 values one with 78 and other 15 with values nearly 85, 84 that will give the mean value as 84.

78, 84, 82, 86, 84, 85, 83, 86, 82, 85, 84, 83, 86, 85, 86, 85.

Mean =\frac{Sum of the values}{Number of values}.

Mean =\frac{78+84+82+86+84+85+83+86+82+85+84+83+86+85+86+85}{16}.

Mean =\frac{1344}{16}.

Mean = 84.

Remove 78 from the data set then the data set is 84, 82, 86, 84, 85, 83, 86, 82, 85, 84, 83, 86, 85, 86, 85 and number of data is reduced to 15.

Mean =\frac{84+82+86+84+85+83+86+82+85+84+83+86+85+86+85}{15}.

Mean =\frac{1266}{15}.

Mean = 84.4.

The mean is increased.

8 0
3 years ago
Solve for x. Would like if both could be answered if not just answer one.
ratelena [41]
328 is the answer sir
3 0
3 years ago
Read 2 more answers
Choose all of the statements that correctly compare the two functions shown.
Kisachek [45]
The 1st one is true
the 2nd one is false
the 3rd one is false
the 4th one is true
the 5th one is true
8 0
3 years ago
Read 2 more answers
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