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mariarad [96]
3 years ago
14

Select all the expressions that are equivalent to the polynomial below.

Mathematics
2 answers:
PIT_PIT [208]3 years ago
5 0
(4x+5)(-3x-1)=
-12x²-19x-5
So, you answers would be:
A. (-16x²+10x-3) + (4x²-29x-2)
D. (2x²-11x-9) - (14x²+8x-4)
Have a nice day! ♪
amm18123 years ago
5 0

Answer:

A. C. And D. Are same

Step-by-step explanation:

(4x+5)(-3x-1)

Applying distributive law

4x(-3x-1)+5(-3x-1)

Again applying distributive law in both the brackets

-12x^2-4x-15x-5

Adding the terms having same power of x

-12x^2-19x-5

1. -16x^2+10x-3+(4x^2-29x-2 )

Simplifying the terms containing the same power of x

-16x^2+10x-3+4x^2-29x-2

-12x^2-19x-5

Hence it is true

2. 3(x-5)-2(6x^2+9x+5)

   3x-15-12x^2-18x-10

  -12x^2-15x-25

  False

3. 2(x-1)-3(4x^2+7x+1)

   2x-2-12x^2-21x-3

  -12x^2-19x-5

True

4. 2x^2-11x-9-(14x^2+8x-4)

   2x^2-11x-9-14x^2-8x+4

  -12x^2-19x-5

True

5. -15x^2+9x-10+3x^2-10x-5

-12x^2+19x-15

False

6. 4x^2-13x-7-(16x^2+9x-5)

   4x^2-13x-7-16x^2-9x+5

  -12x^2-22x-2

False

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What is the simplified expression for Negative 3 (2 x minus y) + 2 y + 2 (x + y)?
Lana71 [14]

Answer:

7y - 4x

Step-by-step explanation:

-3 (2x-y)+2y+2(x+y)\\

Step 1:  Open the brackets

= -6x+3y +2y +2(x+y)\\=-6x +3y + 2y +2x + 2y

Step 2: Bring the similar variables together

=-6x +2x \ +3y + 2y + 2y

Step 3: Simplify by adding/subtracting the coefficients of the similar variables

= - 4x +7y

Step 4: Rearrange as required.

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3 years ago
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The ratio of the measures of the sides of a triangle is 8: 15: 17, and its perimeter is 480 in.. Find the measure of the shortes
lubasha [3.4K]

Answer:

96

Step-by-step explanation:

The ratio of the measure of sides of a triangle are 8:15:17

The perimeter is 480

8x + 15x + 17x= 480

40x= 480

x = 480/40

x = 12

Therefore the measure of the shortest side of the triangle can be calculated as follows

8x

8×12

= 96

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3 years ago
Whats the answer <br> (2*3)•4
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7 0
2 years ago
The line 5x – 5y = 2 intersects the curve x2y – 5x + y + 2 = 0 at
inna [77]

Answer:

(a) The coordinates of the points of intersection are (-2, -12/5), (2/5, 0), and (2, 8/5)

(b) The gradient of the curve at each point of intersection are;

Gradient at (-2, -12/5) = -0.92

Gradient at (2/5, 0) = 4.3

Gradient at (2, 8/5) = -0.28

Step-by-step explanation:

The equations of the lines are;

5·x - 5·y = 2......(1)

x²·y - 5·x + y + 2 = 0.......(2)

Making y the subject of equation (1) gives;

5·y = 5·x - 2

y = (5·x - 2)/5

Making y the subject of equation (2) gives;

y·(x² + 1) - 5·x + 2 = 0

y = (5·x - 2)/(x² + 1)

Therefore, at the point the two lines intersect their coordinates are equal thus we have;

y = (5·x - 2)/5 = y = (5·x - 2)/(x² + 1)

Which gives;

\dfrac{5 \cdot x - 2}{5} = \dfrac{5 \cdot x - 2}{x^2 + 1}

Therefore, 5 = x² + 1

x² = 5 - 1 = 4

x = √4 = 2

Which is an indication that the x-coordinate is equal to 2

The y-coordinate is therefore;

y = (5·x - 2)/5 = (5 × 2 - 2)/5 = 8/5

The coordinates of the points of intersection = (2, 8/5}

Cross multiplying the following equation

Substituting the value for y in equation (2) with (5·x - 2)/5 gives;

\dfrac{5 \cdot x^3 - 2 \cdot x^2 - 20 \cdot x + 8}{5} = 0

Therefore;

5·x³ - 2·x² - 20·x + 8 = 0

(x - 2)×(5·x² - b·x + c) = 5·x³ - 2·x² - 20·x + 8

Therefore, we have;

x²·b - 2·x·b -x·c + 2·c -5·x³ + 10·x²

5·x³ - 10·x² - x²·b + 2·x·b + x·c - 2·c = 5·x³ - 2·x² - 20·x + 8

∴ c = 8/(-2) = -4

2·b + c = - 20

b = -16/2 = -8

Therefore;

(x - 2)×(5·x² - b·x + c) = (x - 2)×(5·x² + 8·x - 4)

(x - 2)×(5·x² + 8·x - 4) = 0

5·x² + 8·x - 4 = 0

x² + 8/5·x - 4/5  = 0

(x + 4/5)² - (4/5)² - 4/5 = 0

(x + 4/5)² = 36/25

x + 4/5 = ±6/5

x = 6/5 - 4/5 = 2/5 or -6/5 - 4/5 = -2

Hence the three x-coordinates are

x = 2, x = - 2, and x = 2/5

The y-coordinates are derived from y = (5·x - 2)/5 as y = 8/5, y = -12/5, and y = y = 0

The coordinates of the points of intersection are (-2, -12/5), (2/5, 0), and (2, 8/5)

(b) The gradient of the curve, \dfrac{\mathrm{d} y}{\mathrm{d} x}, is given by the differentiation of the equation of the curve, x²·y - 5·x + y + 2 = 0 which is the same as y = (5·x - 2)/(x² + 1)

Therefore, we have;

\dfrac{\mathrm{d} y}{\mathrm{d} x}= \dfrac{\mathrm{d} \left (\dfrac{5 \cdot x - 2}{x^2 + 1}  \right )}{\mathrm{d} x} = \dfrac{5\cdot \left ( x^{2} +1\right )-\left ( 5\cdot x-2 \right )\cdot 2\cdot x}{\left (x^2 + 1 ^{2} \right )}.......(3)

Which gives by plugging in the value of x in the slope equation;

At x = -2, \dfrac{\mathrm{d} y}{\mathrm{d} x} = -0.92

At x = 2/5, \dfrac{\mathrm{d} y}{\mathrm{d} x} = 4.3

At x = 2, \dfrac{\mathrm{d} y}{\mathrm{d} x} = -0.28

Therefore;

Gradient at (-2, -12/5) = -0.92

Gradient at (2/5, 0) = 4.3

Gradient at (2, 8/5) = -0.28.

7 0
3 years ago
Question 10
olga_2 [115]

Answer:

1

Step-by-step explanation:

I just need points lol

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