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VLD [36.1K]
3 years ago
11

If a>b, c<b, and d>a, then which of the following is the correct relationship?

Mathematics
1 answer:
tatuchka [14]3 years ago
7 0

x > y is the same like y < x

If a > b and b > c then a > b > c

If d > a then d > a > b > c

<h3>Answer: (D) c < b < a < d</h3>
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Which expressions are polynomials? Select each correct answer.
lilavasa [31]
Here's an example

<span>Simplify by collecting like terms: 4(<span>x2y  </span>+ 7y) – 5y(3<span>x2 </span>– y) – 10y</span>

 

<span>A) −11x2y + 5<span>y2 </span>+ 18y</span>

 

<span>B) 4<span>x2y + </span>11y – 8<span>yx2 – </span>16y</span>

 

<span>C) 4<span>x2y + </span>18y – 15<span>yx2 + </span>5<span>y2</span></span>

 

<span>D) 4<span>x2y </span>– 5y(3<span>x2 </span>– y) – 3y</span>

 

<span>A) −11x2y + 5<span>y2 </span>+ 18y </span>

<span>Correct. 4(<span>x2y  </span>+ 7y) – 5<span>y(3x2 </span>– y) – 10y = 4<span>x2y + </span>28y – 15<span>yx2</span> + 5<span>y2 – </span>10y = 4<span>x2y + </span>18y – 15<span>yx2</span> + 5<span>y2 </span>= −11x2y + 5<span>y2 </span>+ 18y.</span>

 

<span>B) 4<span>x2y + </span>11y – 8<span>yx2 – </span>16y </span>

<span>Incorrect. The 4 is distributed to both terms in the parentheses by multiplying each term by 4 resulting in 28y, not 11y. The −5y is similarly distributed to each term in the parentheses, resulting in −15<span>yx2 + </span>5<span>y2. </span>The correct answer is −11x2y + 5<span>y2 </span>+ 18y.</span>

 

<span>C) 4<span>x2y + </span>18y – 15<span>yx2 + </span>5<span>y2</span></span>

<span>Incorrect. This polynomial can be further simplified by combining the like terms 4<span>x2y </span>and −15<span>yx2. </span>The order of the variables in a term does not matter. The correct answer is </span>

−<span>11x2y + 5<span>y2 </span>+ 18y.</span>

 

<span>D) 4<span>x2y </span>– 5y(3<span>x2 </span>– y) – 3y </span>

<span>Incorrect. Before combining like terms, you must distribute to clear the parentheses. The correct answer is −11x2y + 5<span>y2 </span>+ 18y.</span>

 

 


8 0
4 years ago
Drag the point A to the location indicated in each scenario to complete each statement.
Art [367]

The graph from which the position of the point <em>A</em> can determined following

the multiplication with a scalar is attached.

Responses:

  • If <em>A</em> is in quadrant I and is multiplied by a negative scalar, <em>c</em>, then c·A is in <u>quadrant III</u>
  • If A is in quadrant II and is multiplied by a positive scalar, <em>c</em>, then c·A is in <u>quadrant II</u>
  • If <em>A</em> is in quadrant II and is multiplied by a negative scalar, <em>c</em>, then c·A is in <u>quadrant IV</u>
  • If <em>A</em> is in quadrant III and is multiplied by a negative scalar, <em>c</em>, then c·A is in <u>quadrant I</u>

<h3>Methods by which the above responses are obtained</h3>

Background information;

The question relates to the coordinate system with the abscissa represent the real number and the ordinate representing the imaginary number.

Solution:

If A is in quadrant I; A = a + b·i

When multiplied by a negative scalar, <em>c</em>, we get;

c·A = c·a + c·b·i

Therefore;

c·a is negative

c·b is negative

  • c·A = c·a + c·b·i is in the <u>quadrant III</u> (third quadrant)

If A is quadrant II, we have;

A = -a + b·i

When multiplied by a positive scalar <em>c</em>, we have;

c·A = c·(-a) + c·b·i = -c·a + c·b·i

-c·a is negative

c·b·i is positive

Therefore;

  • c·A = -c·a + c·b·i is in <u>quadrant II</u>

Multiplying <em>A</em> by negative scalar if <em>A</em> is in quadrant II, we have;

c·A = -c·a + c·b·i

-c·a is positive

c·b·i is negative

Therefore;

c·A = -c·a + c·b·i is in <u>quadrant IV</u>

If A is in quadrant III, we have;

A = a + b·i

a is negative

b is negative

Multiplying <em>A</em> with a negative scalar <em>c</em> gives;

c·A = c·a + c·b·i

c·a is positive

c·b  is positive

Therefore;

  • c·A = c·a + c·b·i is in<u> quadrant I</u>

Learn more about real and imaginary numbers here;

brainly.com/question/5082885

brainly.com/question/13573157

4 0
3 years ago
What is the answer to 5y=2x
Andrews [41]

Answer:

the answer is Y =2/5x

Step-by-step explanation:

if it help u can follow me

6 0
3 years ago
Read 2 more answers
Y=4x. Y=40. What does x equal
egoroff_w [7]

Answer:

Your answer is 10!

Y = 4x

(40) = 4x

40 = 4(10)

4 0
3 years ago
Read 2 more answers
Find the solution to the initial value problem
Karo-lina-s [1.5K]

Answer:

y = (11x + 13)e^(-4x-4)

Step-by-step explanation:

Given y'' + 8y' + 16 = 0

The auxiliary equation to the differential equation is:

m² + 8m + 16 = 0

Factorizing this, we have

(m + 4)² = 0

m = -4 twice

The complimentary solution is

y_c = (C1 + C2x)e^(-4x)

Using the initial conditions

y(-1) = 2

2 = (C1 -C2) e^4

C1 - C2 = 2e^(-4).................................(1)

y'(-1) = 3

y'_c = -4(C1 + C2x)e^(-4x) + C2e^(-4x)

3 = -4(C1 - C2)e^4 + C2e^4

-4C1 + 5C2 = 3e^(-4)..............................(2)

Solving (1) and (2) simultaneously, we have

From (1)

C1 = 2e^(-4) + C2

Using this in (2)

-4[2e^(-4) + C2] + 5C2 = 3e^(-4)

C2 = 11e^(-4)

C1 = 2e^(-4) + 11e^(-4)

= 13e^(-4)

The general solution is now

y = [13e^(-4) + 11xe^(-4)]e^(-4x)

= (11x + 13)e^(-4x-4)

3 0
3 years ago
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