(6x + 7) + (13x + 21) = 180 x =

If a > b > c > d, then which is larger, a+c or b+d ? Can we tell from a > b > c > d which of a+d and b+c is la

victus00 [196]

Answer:

1. a+c is larger than b+d

2. No way to tell whether a+d or b+c is larger.

Step-by-step explanation:

1. Which is larger, a+c or b+d?

Let a, b, c, and d be any numbers such that $a > b > c > d$ .

Specifically, note that $a > b$ , and subtracting b from both sides of the inequality, observe that $a-b > 0$ .

Similarly, $c > d$ , and subtracting d from both sides of the inequality, observe that $c-d > 0$ .

From this, add "a-b" (a positive number, as proven above) to both sides of the inequality.

$(a-b)+(c-d) > (a-b)+0$

Addition by zero (the additive identity) doesn't change anything, so the right side remains "a-b"...

$(a-b)+(c-d) > a-b$

... and "a-b" is positive...

$(a-b)+(c-d) > a-b > 0$

... so, by the transitive property of inequality...

$(a-b)+(c-d) > 0$

Recall that subtraction is addition by a negative number...
$a+(-b)+c+(-d) > 0$

...and that addition is associative and commutative, so things can be added in any order, so the middle two terms on the left side can be rearranged...

$a+c+(-b)+(-d) > 0$

Adding b + d to both sides of the inequality

$(a+c+(-b)+(-d))+(b+d) > 0+(b+d)$

... and simplifying

$a+c > b+d$

So, a+c is larger than b+d.

2. Which is larger, a+d or b+c?

Consider the following two examples:

Example 1

Suppose a=10; b=3; c=2; d=1.

Note that $a > b > c > d$ ( $10 > 3 > 2 > 1$ ) and, also observe that $a+d=(10)+(1)=11$ , and $b+c=(3)+(2)=5$ , so a+d is larger than b+c.

Example 2

However, suppose a=10; b=9; c=8; d=1.

Note that $a > b > c > d$ ( $10 > 9 > 8 > 1$ ) but that $a+d=(10)+(1)=11$ , and $b+c=(9)+(8)=17$ , so a+d is smaller than b+c.

So, in one example, a+d is bigger, and in the other, a+d is smaller. Therefore, there is no way to tell which of a+d or b+c is larger from only the given information.

5 0

2 years ago

Jake is flying his kite. About how far above his hands,x,is the

andreev551 [17]

Answer: 91.65 ft

Step-by-step explanation:

(200*200) + (x) = 220*220

40000 + x = 48400

48400 - 40000 = x^2

8400 = x^2

91.65 = x

7 0

3 years ago

Help please will mark brainliest

makvit [3.9K]

I will help the answer is 55 b

8 0

3 years ago

If A = 2 −2 7 4 6 3 ; find matrix X such that X+A=O where O is a null matrix.

Juliette [100K]

Answer:

I think this is the correct answer for that question

4 0

3 years ago

Describe three ways to determine the measure of segment YZ.

mixas84 [53]

Using 3 different methods, we can find that YZ = 25m

The complete image of a triangle is attached with the answer below.

<h3>What is trigonometry?</h3>

The branch of mathematics sets up a relationship between the sides and the angles of the right-angle triangle is termed trigonometry.

For a given triangle rectangle with a known angle θ and a hypotenuse H, we can write the trigonometric relations

sin(θ) = (opposite cathetus)/H

cos(θ) =(adjacent cathetus)/H

tan(θ) =(opposite cathetus)/(adjacent cathetus)

Now, in the given image we can see that:

θ = 30°

H = 50m

XY = adjacent cathetus

YZ = opposite cathetus.

We want to find YZ, so with the known things, we can use the first relation:

Sin 30 = YZ / 50

YZ = 50 x Sin30 = 25 meters

Now let's try another way.

We know that the sum of the internal angles of a triangle is always equal to 180°

In a triangle rectangle, we always have an angle equal to 90°.

Then the missing angle for our triangle can be computed from:

Z + 90° + 30° = 180°

Z = 180° - 90° - 30° = 60°

From this angle, the side YZ is the adjacent cathetus, then we can use the second trigonometric relation:

Cos 60 = YZ / 50

YZ = 50 x Cos 60 = 25 meters

Third way.

Whit the same angle we could find the side XY, the opposite cathetus, with:

Sin 60 = XY / 50

XY = 50Sin 60 = 43.3 meters

Now that we know one of the catheti, we can use the last trigonometric relation

tan 60 = XZ / YZ = 43.3 / YZ

YZ = 43.3 / tan 60 = 25 meters

To know more about Trigonometry follow

brainly.com/question/24349828

#SPJ1

5 0

2 years ago

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