A cylindrical water tank is being filled with a hose. The depth of the water increases by 1 1/4 ft per hour.

How would you factor a quadratic expression like 9x^2-4

Ganezh [65]

Hello, we need to see it as a difference of two squares as we know that for all a and b real numbers

$a^2-b^2=(a-b)(a+b)$

So, here

$9x^2-4=(3x)^2-2^2=(3x-2)(3x+2)$

Voila !

Thank you

7 0

3 years ago

Use the graph to answer the question.

Lelu [443]

Answer:

Coordinate Q is (0.8, 0.7)

Step-by-step explanation:

We are told that the coordinates of point Pare (0.6,0.1).

This means that along the x-axis, x = 0.6 and along the y-axis, y = 0.1.

Now, by inspection of the graph, we can see that when we count boxes from the origin to the point P, we have 6 boxes. Thus, each box corresponds to 0.1. So, for point Q, from the origin to that point, on the x-axis, we have 8 boxes. Since one box = 0.1, then the x - value of Coordinate Q is 0.8.

On the y - axis, we see that we have one box from the origin up for the corresponding y-value of coordinate P.

This means that one box is 0.1.

For coordinate Q, we will count 7 boxes. Thus, y-value of coordinate Q is 0.7.

Thus,coordinate Q is (0.8, 0.7)

7 0

2 years ago

Type the correct answer in each box. Use numerals instead of words.

lubasha [3.4K]

Answer:

System A has 4 real solutions.

System B has 0 real solutions.

System C has 2 real solutions

Step-by-step explanation:

System A:

x^2 + y^2 = 17 eq(1)

y = -1/2x eq(2)

Putting value of y in eq(1)

x^2 +(-1/2x)^2 = 17

x^2 + 1/4x^2 = 17

5x^2/4 -17 =0

Using quadratic formula:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

a = 5/4, b =0 and c = -17

$x=\frac{-(0)\pm\sqrt{(0)^2-4(5/4)(-17)}}{2(5/4)}\\x=\frac{0\pm\sqrt{85}}{5/2}\\x=\frac{\pm\sqrt{85}}{5/2}\\x=\frac{\pm2\sqrt{85}}{5}$

Finding value of y:

y = -1/2x

$y=-1/2(\frac{\pm2\sqrt{85}}{5})$

$y=\frac{\pm\sqrt{85}}{5}$

System A has 4 real solutions.

System B

y = x^2 -7x + 10 eq(1)

y = -6x + 5 eq(2)

Putting value of y of eq(2) in eq(1)

-6x + 5 = x^2 -7x + 10

=> x^2 -7x +6x +10 -5 = 0

x^2 -x +5 = 0

Using quadratic formula:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

a= 1, b =-1 and c =5

$x=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(5)}}{2(1)}\\x=\frac{1\pm\sqrt{1-20}}{2}\\x=\frac{1\pm\sqrt{-19}}{2}\\x=\frac{1\pm\sqrt{19}i}{2}$

Finding value of y:

y = -6x + 5

y = -6(\frac{1\pm\sqrt{19}i}{2})+5

Since terms containing i are complex numbers, so System B has no real solutions.

System B has 0 real solutions.

System C

y = -2x^2 + 9 eq(1)

8x - y = -17 eq(2)

Putting value of y in eq(2)

8x - (-2x^2+9) = -17

8x +2x^2-9 +17 = 0

2x^2 + 8x + 8 = 0

2x^2 +4x + 4x + 8 = 0

2x (x+2) +4 (x+2) = 0

(x+2)(2x+4) =0

x+2 = 0 and 2x + 4 =0

x = -2 and 2x = -4

x =-2 and x = -2

So, x = -2

Now, finding value of y:

8x - y = -17

8(-2) - y = -17

-16 -y = -17

-y = -17 + 16

-y = -1

y = 1

So, x= -2 and y = 1

System C has 2 real solutions

4 0

3 years ago

Read 2 more answers

Can someone please help me asap

attashe74 [19]

Answer:

y=3x-4

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Elle is buying new flooring for her kitchen

const2013 [10]

Answer:

The total area of the two rooms is 188.25 square feet

Step-by-step explanation:

Given:

Area of the kitchen = 132 square feet.

Dimension of the Laundry room = 8 1/3 feet by 6 3/4 feet.

To Find:

The total area of the two rooms = ?

Solution:

Area of the two rooms = Area of the kitchen + Area of the laundry room

Area of the two rooms = 132 + Area of the laundry room

Area of the laundry room = $(8 \frac{1}{3}) \times (6 \frac{3}{4})$

Area of the laundry room = $\frac{25}{3} \times \frac{27}{4}$

Area of the laundry = $\frac{675}{12}$

Area of the two rooms = $132 + \frac{675}{12}$

Area of the two rooms = 132 + 56.25

Area of the two rooms = 188.25 square feet

3 0

2 years ago