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3 years ago

Can someone please answer. There is a picture. It is only one question. Thank you so much!

Mathematics

1 answer:

monitta3 years ago

5 0

The legs of a right triangle are the sides that are connected to the 90 degree (right) angle. The remaining side is called the hypotenuse.

Another way to look at it is the two legs of a right triangle are the two shorter sides. The longest side is the hypotenuse.

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How do you solve 4x^6/2x-6

Tanzania [10]

Divide 4 and 2 like you normally would then x^6 / x^-6 would be x^6-(-6)

=x^6+6 = x^12

so your answer should be
2x^12

6 0

3 years ago

The frist three terms of a geometric sequence are as follows 80,40,20

olga2289 [7]

80,40,20,10 its dividing by 2

7 0

3 years ago

What are the coordinates of the point on the directed line segment from (- 10, - 10) to (6, - 2) that partitions the segment int

Nina [5.8K]

Answer:

Denote:

A(-10, -10)

B(6, -2)

vector AB(6 - -10, -2 - -10) = (16, 8)

(3/5)vector AB = (16 x 3/5, 8 x 3/5) = (9.6, 4.8)

=> Point M that partitions the directed line segment AB into a ratio of 3 to 5:

M = A + (3/5)vector AB = (-10 + 9,6, -10 + 4.8) = (-0.4, -5.2)

Hope this helps!

6 0

3 years ago

Read 2 more answers

What is the maximum value of the picture below?

Cloud [144]

Answer:

y = 4

Step-by-step explanation:

The maximum value occurs at the vertex

vertex = (2, 4 )

when x = 2 the maximum value is y = 4

4 0

3 years ago

Find a nonzero vector orthogonal to the plane through the points: ????=(0,0,1), ????=(−2,3,4), ????=(−2,2,0).

ser-zykov [4K]

Answer:

The nonzero vector orthogonal to the plane is <-9,-8,2>.

Step-by-step explanation:

Consider the given points are P=(0,0,1), Q=(−2,3,4), R=(−2,2,0).

$\overrightarrow {PQ}==$

$\overrightarrow {PR}==$

The nonzero vector orthogonal to the plane through the points P,Q, and R is

$\overrightarrow n=\overrightarrow {PQ}\times \overrightarrow {PR}$

$\overrightarrow n=\det \begin{pmatrix}i&j&k\\ \:\:\:\:\:-2&3&3\\ \:\:\:\:\:-2&2&-1\end{pmatrix}$

Expand along row 1.

$\overrightarrow n=i\det \begin{pmatrix}3&3\\ 2&-1\end{pmatrix}-j\det \begin{pmatrix}-2&3\\ -2&-1\end{pmatrix}+k\det \begin{pmatrix}-2&3\\ -2&2\end{pmatrix}$

$\overrightarrow n=i(-9)-j(8)+k(2)$

$\overrightarrow n=-9i-8j+2k$

$\overrightarrow n=$

Therefore, the nonzero vector orthogonal to the plane is <-9,-8,2>.

8 0

3 years ago