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Ainat [17]
3 years ago
9

Help Me With These Please

Mathematics
1 answer:
anzhelika [568]3 years ago
6 0
10×sine of 60 degrees=8.66 which is the same as B. you can check by finding the square root of 3 and then multiplying that by 5.
You might be interested in
Just Number 15? Please
Damm [24]
15)
an exterior angle equals the sum of the remote interior angles
<span>so
m<XYT = m<X + m<Z
4x - 25 = x + 15 + 50 - 3x
4x - 25 = -2x + 65
6x = 90
  x = 15

answer
x = 15



</span>
8 0
3 years ago
Find the three angles of the triangle with the given vertices: P(1,1,1), ????(1,−3,2), and ????(−3,2,5).
velikii [3]

Answer:

The three angles of the triangle are 90, 35.67 and 54.33 degrees.

Step-by-step explanation:

One way to find the angles of the triangle with vertices (1, 1, 1), (1, -3, 2), (-3, 2, 5) is using the definition of the <em>dot product</em> of two vectors, defined as:

a . b = |a| |b| cosФ [1]

Where |a| and |b| are the norms of vectors <em>a</em> and b, and cosФ is the cosine of the angle between either vector <em>a</em> and <em>b</em>.

If a = [ \\ a_{1}, a_{2}, a_{3} ] and b = [\\ b_{1}, b_{2}, b_{3}], then the dot product is simply a number (not a vector) obtained from:

a . b = \\ a_{1}*b_{1} + a_{2}*b_{2} + a_{3}*b_{3}

The norm of a vector (its length) is, for instance, |a| = \\ \sqrt{{a_{1}^2 + {a_{2}}^2 + {a_{3}}^2}, for a vector in \\ R^{3}.

Having all that into account, we can determine the angles of the triangle for each vertex using equation [1] and solving it for Ф.

<h3>Angle of the triangle for vertex in (1, 1, 1)</h3>

The vectors which form an angle from this vertex are the result of subtracting the vertex (1, 1, 1) to any of the remaining points (1, -3, 2) and (-3, 2, 5):

v(1, 1, 1) - v(1, -3, 2) = \\ (1 - 1, 1 - (-3), 1 - 2) = (0, 1 + 3, -1) = (0, 4, -1)

v(1, 1, 1) - v(-3, 2, 5) = \\ (1 - (-3), 1 - 2, 1 - 5) = (1 + 3, -1, -4) = (4, -1, -4)

The <em>dot product</em> for these vectors is:

[0, 4, -1] . [4, -1, -4] = [0 * 4 + 4 * -1 + -1 * -4] = 0 - 4 + 4 = 0

The norm for each vector is:

|(0, 4, -1)| = \\ \sqrt{0^2 + 4^2 + -1^2} = \sqrt{0 + 16 + 1} = \sqrt{17}

|(4, -1, -4)| = \\ \sqrt{4^2 + -1^2 + -4^2} = \sqrt{16 + 1 + 16} = \sqrt{33}

So

a . b = |a| |b| cosФ

\\ 0 = \sqrt{17} * \sqrt{33} * cos{\theta}

\\ \frac{0}{\sqrt{17} * \sqrt{33}} = cos{\theta}

\\ cos^{-1}{0}} = cos^{-1}(cos{\theta})

\\ 90 = \theta

In vertex (1, 1, 1) the angle of the triangle is 90 degrees. We have here a right triangle.

We have to follow the same procedure for finding the vectors for angles in vertices (1, -3, 2) and (-3, 2, 5), or better, after finding one of the previous angles, we find the remaining angle subtracting the sum of two angles from 180 degrees to finally obtaining the three angles in question.

Therefore, the other angles are 35.67 degrees and 180 - (90 + 35.67) = 180 - 125.67 = 54.33 degrees.

5 0
3 years ago
A sandwich shop is ordering apples and grapes to make chicken salad. Apples cost $2.19 per pound and grapes cost $2.60 per pound
Aleksandr-060686 [28]

Answer:

Less than 13 pounds of grapes,

as 14 pounds of grape exceeds $35.80,

If you order as low as zero pounds of grapes, then the price of apple pounds would be $35.80/20 =$1.79 per pound

Step-by-step explanation:

Let

Quantity of Apple and grapes=20

Cost of apple=$2.19

Cost of grape=$2.60

Total cost of apple and grape=$35.80

a+g=20

a=20-g

PaQa+PgQg=35.80

2.19(20-g)+2.60g=35.80

43.8-2.19g+2.60g=35.80

0.41g=35.80-43.8

0.41g=-8

g=-8/0.41

g= -19.51

PaQa+PgQg=35.80

2.19a+2.60(-19.51)=35.80

2.19a-50.726=35.80

2.19a=35.80+50.726

2.19a=86.526

a=86.526/2.19

=39.51

PaA+PgG=35.80

Pa(20-G)+2.60G=35.80

35.80/2.60 = 13.77 pounds of grapes, but no money will be left to buy apples, unless Pa=0

G<13.77

4 0
3 years ago
PLEASE HELP ME!!!!! I NEED HELP!!!!!!
Shalnov [3]
The length of the sides is 6
4 0
4 years ago
Read 2 more answers
Gina wrote the following paragraph to prove that the segment joining the midpoints of two sides of a triangle is parallel to the
beks73 [17]
Hello,
Please, see the attached files.
Thanks.

5 0
3 years ago
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