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3 years ago

Factor the polynomial: –3x^4 – 9x^3 + 15x^2

Mathematics

2 answers:

spayn [35]3 years ago

7 0

Answer:-3x^2(9x^5-5)

Step-by-step explanation:

lara31 [8.8K]3 years ago

4 0

Step-by-step explanation:

-3x2(x2)-9x3+25x2

-3x2(x2+3x-5)

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Solve for the value of X. <br> -4x + 3 = 35

Stella [2.4K]

Answer:

-8

Step-by-step explanation:

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3 years ago

Y = -5x + 15 2x + y = 12

Nana76 [90]

Answer:

x = 1, y = 10

Step-by-step explanation:

y = -5x + 15 --- Equation 1

2x + y = 12 --- Equation 2

Substitute y = -5x + 15 into Equation 2:

2x + y = 12

2x - 5x + 15 = 12

Evaluate like terms.

15 - 3x = 12

Isolate -3x.

-3x = 12 - 15

Evaluate like terms.

-3x = -3

Find x.

x = -3 ÷ -3

x = 1

Substitute x = 1 into Equation 2:

2x + y = 12

2(1) + y = 12

2 + y = 12

Isolate y.

y = 12 - 2

y = 10

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3 years ago

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Solve for m.<br> 8m+95<−87m+5

tiny-mole [99]

Answer:

m < - $\frac{18}{19}$

Step-by-step explanation:

8m + 95 < -87m + 5

8m + 87m < 5 - 95

95m < -90

m < -<u> $\frac{90}{95}$ </u>

Simplifying gives;

m < - $\frac{18}{19}$

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3 years ago

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6 football teams enter a five a side tournament how many players are in the tournament

irakobra [83]

There are 30 players in the tournament

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3 years ago

A rectangle has a length 6 more than it's width if the width is decreased by 2 and the length decreased by 4 the resulting has a

Rashid [163]

Answer:

Length of original rectangle: 11 units.

$\frac{\text{Area of original rectangle}}{\text{Area of new rectangle}}=\frac{55}{21}$

$\text{Perimeter of new rectangle}=20$

Step-by-step explanation:

Let x represent width of the original rectangle.

We have been given that a rectangle has a length 6 more than it's width. S the length of the original rectangle would be $x+6$ .

We have been given that when the width is decreased by 2 and the length decreased by 4 the resulting has an area of 21 square units.

The width of new rectangle would be $x-2$ .

The length of new rectangle would be $x+6-4=x+2$ .

The area of new rectangle would be $(x+2)(x-2)$ .

Now we will equate area of new rectangle with 21 and solve for x as:

$(x+2)(x-2)=21$

Applying difference of squares, we will get:

$x^2-2^2=21$

$x^2-4=21$

$x^2-4+4=21+4$

$x^2=25$

Since width cannot be negative, so we will take positive square root of both sides.

$\sqrt{x^2}=\sqrt{25}$

$x=5$

Therefore, the width of original rectangle is 5 units.

Length of the original rectangle would be $x+6\Rightarrow x+5=11$ .

Therefore, the length of original rectangle is 11 units.

$\text{Area of original rectangle}=5\times 11$

$\text{Area of original rectangle}=55$

Therefore, area of the original rectangle is 55 square units.

Now we will find ratio of the original rectangle area to the new rectangle area as:

$\frac{\text{Area of original rectangle}}{\text{Area of new rectangle}}=\frac{55}{21}$

We know that perimeter of rectangle is two times the sum of length and width.

$\text{Perimeter of new rectangle}=2((x+2)+(x-2))$

$\text{Perimeter of new rectangle}=2((5+2)+(5-2))$

$\text{Perimeter of new rectangle}=2(7+3)$

$\text{Perimeter of new rectangle}=2(10)$

$\text{Perimeter of new rectangle}=20$

Therefore, the perimeter of the new rectangle is 20 units.

7 0

4 years ago