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3 years ago

13

Raheem made a net of a triangular pyramid as shown.

1 answer:

Cerrena [4.2K]3 years ago

5 0

Answer:

the answer is b

Step-by-step explanation:

i took the test it is b. The sum of the areas of faces A, B, and C.

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Solve the system by elimination.(show your work)

PilotLPTM [1.2K]

Answer:

x = 1 , y = 1 , z = 0

Step-by-step explanation by elimination:

Solve the following system:

{-2 x + 2 y + 3 z = 0 | (equation 1)

-2 x - y + z = -3 | (equation 2)

2 x + 3 y + 3 z = 5 | (equation 3)

Subtract equation 1 from equation 2:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x - 3 y - 2 z = -3 | (equation 2)

2 x + 3 y + 3 z = 5 | (equation 3)

Multiply equation 2 by -1:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+3 y + 2 z = 3 | (equation 2)

2 x + 3 y + 3 z = 5 | (equation 3)

Add equation 1 to equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+3 y + 2 z = 3 | (equation 2)

0 x+5 y + 6 z = 5 | (equation 3)

Swap equation 2 with equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y + 6 z = 5 | (equation 2)

0 x+3 y + 2 z = 3 | (equation 3)

Subtract 3/5 × (equation 2) from equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y + 6 z = 5 | (equation 2)

0 x+0 y - (8 z)/5 = 0 | (equation 3)

Multiply equation 3 by 5/8:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y + 6 z = 5 | (equation 2)

0 x+0 y - z = 0 | (equation 3)

Multiply equation 3 by -1:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y + 6 z = 5 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Subtract 6 × (equation 3) from equation 2:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y+0 z = 5 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Divide equation 2 by 5:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Subtract 2 × (equation 2) from equation 1:

{-(2 x) + 0 y+3 z = -2 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Subtract 3 × (equation 3) from equation 1:

{-(2 x)+0 y+0 z = -2 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Divide equation 1 by -2:

{x+0 y+0 z = 1 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Collect results:

Answer: {x = 1 , y = 1 , z = 0

6 0

3 years ago

Read 2 more answers

Based on the information marked in the diagram, ∆XYZ and ∆RST must be congruent.

saveliy_v [14]

The answer is A true

7 0

3 years ago

Pls tell me how to do it.

kkurt [141]

Lets get started :)

The ratio of juice is as follows:
orange : mango : apple
3 : 5 : 4
↑
35 bottles

5x = 35

$\frac{5x}{5} = \frac{35}{5}$
5 and 5 cancels out
x = 7

3x + 5x + 4x = total
Subsitute the value you go for x, x = 7
3(7) + 5(7) + 4(7) = total
21 + 35 + 28 = 84

Total of 84 bottles of juices :)
Your final answer will be option D

6 0

3 years ago

Please help I will give Brainliest please!

WITCHER [35]

Part (a)

The domain is the set of allowed x inputs of a function.

The graph shows that x = 0 is not allowed because of the vertical asymptote located here. It seems like any other x value is fine though.

<h3>Domain: set of all real numbers but $x \ne 0$ </h3>

To write this in interval notation, we can say $(-\infty, 0) \cup (0, \infty)$ which is the result of poking a hole at 0 on the real number line.

--------------

The range deals with the y values. The graph makes it seem like it stretches on forever in both up and down directions. If this is the case, then the range is the set of all real numbers.

<h3>Range: Set of all real numbers</h3>

In interval notation, we would say $(-\infty, \infty)$ which is almost identical to the interval notation of the domain, except this time of course we aren't poking at hole at 0.

=======================================================

Part (b)

<h3>The x intercepts are x = -4 and x = 4</h3>

We can compact that to the notation $x = \pm 4$

These are the locations where the blue hyperbolic curve crosses the x axis.

=======================================================

Part (c)

<h3>Answer: There aren't any horizontal asymptotes in this graph.</h3>

Reason: The presence of an oblique asymptote cancels out any potential for a horizontal asymptote.

=======================================================

Part (d)

The vertical asymptote is located at x = 0, so the equation of the vertical asymptote is naturally x = 0. Every point on the vertical dashed line has an x coordinate of zero. The y coordinate can be anything you want.

<h3>Answer: x = 0 is the vertical asymptote</h3>

=======================================================

Part (e)

The oblique or slant asymptote is the diagonal dashed line.

It goes through (0,0) and (2,6)

The equation of the line through those points is y = 3x

If you were to zoom out on the graph (if possible), then you should notice the branches of the hyperbola stretch forever upward but they slowly should approach the "fencing" that is y = 3x. The same goes for the vertical asymptote as well of course.

<h3>Answer: Oblique asymptote is y = 3x</h3>

5 0

3 years ago

What number must you add to complete the square x^2+10x=8<br><br> A. 25<br> B. 5<br> C. 15<br> D. 10

Ilia_Sergeevich [38]

To complete the square you take the coefficient of the 'b' term (10), divide it by 2 (10/2), then square it (5^2 = 25). This is the number you add.
Letter C 25

8 0

3 years ago