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3 years ago

PLEASE HELP ASAP 25 PTS + BRAINLIEST TO RIGHT/BEST ANSWER

Mathematics

2 answers:

elixir [45]3 years ago

8 0

Answer:

It's (5,-2,0)

Step-by-step explanation:

1. 5 - -2+2(0)= 7 = 7 true

2. 2(5)+-2+0= 8 = 8 true

3. 5-0=5 = 5 true

avanturin [10]3 years ago

8 0

$\left\{\begin{array}{ccc}x-y+2z=7&(1)\\2x+y+z=8&(2)\\x-z=5&(3)\end{array}\right\\\\(3)\qquad x-z=5\qquad\text{add z to both sides}\\.\qquad x=5+z\\\\\text{Substitute it to (1) and (2):}\\\\\left\{\begin{array}{ccc}(5+z)-y+2z=7\\2(5+z)+y+z=8\end{array}\right\\\\\left\{\begin{array}{ccc}5+z-y+2z=7&\text{subtract 5 from both sides}\\10+2z+y+z=8&\text{subtract 10 from both sides}\end{array}\right$

$\underline{+\left\{\begin{array}{ccc}3z-y=2\\3z+y=-2\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad6z=0\qquad\text{divide both sides by 6}\\.\qquad\boxed{z=0}\\\\\text{Put the value of z to the second equation and to (3)}\\\\3(0)+y=-2\to\boxed{y=-2}\\\\x=5+0\to\boxed{x=5}\\\\Answer:\ \boxed{x=5,\ y=-2,\ z=0\to(5,\ -2,\ 0)}\to\boxed{C}$

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The average of five distinct scores has the same value as the median of the five scores. The sum of the five scores is 420. What

givi [52]

Answer:

The sum of five scores that is not the median is, 336

Step-by-step explanation:

Average defined as the the sum of the observation to the total number of the observation.

Given: The sum of the five scores is 420

Then, by definition of average;

Average of five distinct score = $\frac{420}{5} = 84$

It is also, given that the average of five distinct scores has the same value as the median of the five scores.

⇒Median of five scores = 84

Since, scores are distinct, therefore

Sum of all scores that is not the median = Sum of five scores - median = 420-84 = 336

therefore, the sum of five scores that is not the median is, 336

3 0

3 years ago

How to find the length of a triangle with only one side non right triangle?

castortr0y [4]

The trigonometry of non-right triangles

So far, we've only dealt with right triangles, but trigonometry can be easily applied to non-right triangles because any non-right triangle can be divided by an altitude* into two right triangles.

Roll over the triangle to see what that means →

Remember that an altitude is a line segment that has one endpoint at a vertex of a triangle intersects the opposite side at a right angle. See triangles.

Customary labeling of non-right triangles

This labeling scheme is comßmonly used for non-right triangles. Capital letters are anglesand the corresponding lower-case letters go with the side opposite the angle: side a (with length of a units) is across from angle A (with a measure of A degrees or radians), and so on.

Derivation of the law of sines

Consider the triangle below. if we find the sines of angle A and angle C using their corresponding right triangles, we notice that they both contain the altitude, x.

The sine equations are

We can rearrange those by solving each for x(multiply by c on both sides of the left equation, and by a on both sides of the right):

Now the transitive property says that if both c·sin(A) and a·sin(C) are equal to x, then they must be equal to each other:

We usually divide both sides by ac to get the easy-to-remember expression of the law of sines:

We could do the same derivation with the other two altitudes, drawn from angles A and C to come up with similar relations for the other angle pairs. We call these together the law of sines. It's in the green box below.

The law of sines can be used to find the measure of an angle or a side of a non-right triangle if we know:

two sides and an angle not between them ortwo angles and a side not between them.

Law of Sines

Examples: Law of sines

Use the law of sines to find the missing measurements of the triangles in these examples. In the first, two angles and a side are known. In the second two sides and an angle. Notice that we need to know at least one angle-opposite side pair for the Law of Sines to work.

Example 1

Find all of the missing measurements of this triangle:

The missing angle is easy, it's just

Now set up one of the law of sines proportions and solve for the missing piece, in this case the length of the lower side:

Then do the same for the other missing side. It's best to use the original known angle and side so that round-off errors or mistakes don't add up.

Example 2

Find all of the missing measurements of this triangle:

First, set up one law of sines proportion. This time we'll be solving for a missing angle, so we'll have to calculate an inverse sine:

Now it's easy to calculate the third angle:

Then apply the law of sines again for the missing side. We have two choices, we can solve

Either gives the same answer,

Derivation of the law of cosines

Consider another non-right triangle, labeled as shown with side lengths x and y. We can derive a useful law containing only the cosine function.

First use the Pythagorean theorem to derive two equations for each of the right triangles:

Notice that each contains and x2, so we can eliminate x2 between the two using the transitive property:

Then expand the binomial (b - y)2 to get the equation below, and note that the y2 cancel:

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Substituting c·cos(A) for y, we get

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The law of cosines can be used to find the measure of an angle or a side of a non-right triangle if we know:

two sides and the angle between them orthree sides and no angles.

We could again do the same derivation using the other two altitudes of our triangle, to yield three versions of the law of cosines for any triangle. They are listed in the box below.

Law of Cosines

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Examples: Law of cosines

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3 0

3 years ago

Please help me <br>ineed it now

yanalaym [24]

Answer:

look at explanation

Step-by-step explanation:

To solve all of these questions you just need to scale. While scaling you can divide or multiply.

1. $\frac{180 km}{4 hrs}$ (divide by 4 to get unit rate)

unit rate = $\frac{45km}{1hr}$

2. If it takes one hour to travel 40 miles, then we don't have to scale or anything. We know $\frac{40 miles}{1 hr}$ , the answer is...

"It'll only take one hour to travel 40 miles. This is correct because the unit rate is $\frac{40 miles}{1 hr}$ . "

3. Knowing that the constant speed of the plane is $\frac{800 km}{1 hour\\}$ , we can scale then add half of the unit rate to get our answer.

$\frac{800 km}{1 hour\\}$ x 3 = $\frac{2400 km}{3 hrs}$ 800 x 3 = 2400

then add half of unit rate which is $\frac{400 km}{30 mins}$ $\sqrt[2]{800}$ = 400

so, our answer is $\frac{2800 km}{3.5 hours}$

4. $\frac{3km}{30 mins}$ <--- ( divided by 2) $\frac{6 km}{1 hr}$ ( multiplied x 2.5) ---> $\frac{15 km}{2.5 hrs}$

To get an easier way to find the answer, if possible you can scale back farther than the 1 hour mark.

Elmer would take 2 and a half hours to ride 15 km.

5. The boys speed is $\frac{4 km}{1 hr}$ . All we need to do is divide by 2. $\sqrt[2]{8}$ = 4

3 0

3 years ago

21.) A rectangle has an area of 36 square units. As the length and width change, what do you know about their

natima [27]

Answer:

the length and the width are inversely proportional to each other ( $l=\frac{36}{w}$ or $w=\frac{36}{l}$ )

Step-by-step explanation:

Let

l units = length of the rectangle,

w units = width of the rectangle.

The area of the rectangle is

$\text{Area}=\text{Length}\cdot \text{Width}$

A rectangle has an area of 36 square units, so

$l\cdot w=36$

Complete the table with some values of l and w that fit the previous formula:

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As you can see, the length and the width are inversely proportional to each other ( $l=\frac{36}{w}$ or $w=\frac{36}{l}$ )

4 0

3 years ago

Write an equation of the line that passes through (2 -5) and is parallel to the line 2y=3x+10

andrey2020 [161]

Step-by-step explanation:

Given equation is

2y = 3x + 10

3x - 2y + 10 = 0 .....i)

Any line parallel to line I) is

3x - 2y + k = 0 ......ii)

As the line two passes through the point ( 2 , - 5 ) Now substituting the values

3 * 2 - 2 * ( - 5) + k = 0

6 + 10 + k = 0

16 + k = 0

k = - 16

Now putting the value of k in equation two

3x - 2y + 16 = 0 is the required equation.

Hope it will help :)❤

8 0

3 years ago