Solve the equation check for extraneous solutions:

What is 3(2x-6)-11=4(x-3)+6

navik [9.2K]

Answer:

$x=11.5$

Step-by-step explanation:

4 0

3 years ago

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PLEASE HELP!! Solve for x. x = [?] 4x – 16 2x + 16

Annette [7]

X=16 i graphed it !!!!!!!!!!!!

6 0

3 years ago

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What is the slope in the equations: y = 4x + 3

Anestetic [448]

Using the slope-intercept form, the slope is 4

4 0

3 years ago

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Which of the following statements have the same result? Explain each step in solving each one.. . I f(3) when f(x) = 2x + 2. II

zhuklara [117]

F(x) = 2x + 2.....find f(3)
f(3) = 2(3) + 2
f(3) = 6 + 2
f(3) = 8

f(x) = (3x - 4)/5....find f^-1(4)
y = (3x - 4)/5...swap the variables
x = (3y - 4)/5 ...solve for y
5x = 3y - 4
5x + 4 = 3y
(5x + 4)/3 = y
f^-1(x) = (5x + 4)/3.....for f(4)
f(4) = (5(4) + 4)/3
f(4) = (20 + 4)/3
f(4) = 24/3
f(4) = 8

y + 10 = 2y + 1
y - 2y = 1 - 10
-y = -9
y = 9

so 1 and 2 have the same result...8

8 0

4 years ago

The lateral surface area of cone A is exactly 1/2 the lateral surface area of cylinder B. Cone A radius is r and height h - Cyli

Mrac [35]

The ratio of the lateral surface area of cone A to the lateral surface area of cylinder B is equal to $r = \frac{1}{2}\cdot \frac{\sqrt{r^{2}+h^{2}}}{h}$ . (Correct choice: False)

<h3>What is the ratio of lateral area of cone to the lateral area of the cylinder?</h3>

In accordance with space geometry, the lateral areas of the cone and cylinder are described by the following equations:

Cone

$A_{l} = \pi \cdot r \cdot \sqrt{r^{2}+h^{2}}$ (1)

Cylinder

$A_{l} = 2\pi\cdot r\cdot h$ (2)

If we divide (2) by (1), then we have the following ratio:

$r = \frac{1}{2}\cdot \frac{\sqrt{r^{2}+h^{2}}}{h}$

The ratio of the lateral surface area of cone A to the lateral surface area of cylinder B is equal to $r = \frac{1}{2}\cdot \frac{\sqrt{r^{2}+h^{2}}}{h}$ . (Correct choice: False)

To learn more on surface areas: brainly.com/question/2835293

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3 0

2 years ago