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3 years ago

Steps of 17/3 - 3/4x = 1/2x + 5

Mathematics

1 answer:

STatiana [176]3 years ago

5 0

Answer:

Step-by-step explanation:

12(17/3 - 3/4x = 1/2x + 5)

68 - 9x = 6x + 60

-15x = -8

x = 8/15

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I need the answers for 2 and 3

Korolek [52]

2- x= -2 y= -1 so the slope is -1/-2
3- x= 2 y= 3 so the slope is 1 1/2
i’m not sure if this is right but i’m learning this as well

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3 years ago

If f(x) and g(x)=x, determine the value(s) of x that satisfy the f(x)=g(x).

lions [1.4K]

Answer: All values make a true statement.

Step-by-step explanation: Since f(x) and g(x)=x, f(x)=g(x) can also be written as x=x. That is a true statement so all values of x works.

Hope this helps!

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3 years ago

What the slope of(7,3) (6,-2)

antoniya [11.8K]

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$slope = \frac{ - 2 - 3}{6 - 7} \\$

$slope = \frac{ - 5}{ - 1} \\$

$slope = 5$

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3 years ago

Neeeeeed heeelp plz!!!

Blababa [14]

|
|__x_- 4___________ 2x³/2x² =x
2x² + 2x +3 | 2x³ - 6x² +7x +3
- (2x³ +2x² +3x)
 -8x² +4x +3 -8x²/2x² = -4
-(-8x² -8x -12)
 12x +15

(x-4) + (12x +15)/(2x² + 2x +3)

6 0

4 years ago

From a piece of tin in the shape of a square 6 inches on a side, the largest possible circle is cut out. What is the ratio of th

wel

Answer:

$\sf \dfrac{1}{4} \pi \quad or \quad \dfrac{7}{9}$

Step-by-step explanation:

The width of a square is its side length.

The width of a circle is its diameter.

Therefore, the largest possible circle that can be cut out from a square is a circle whose diameter is equal in length to the side length of the square.

Formulas

$\sf \textsf{Area of a square}=s^2 \quad \textsf{(where s is the side length)}$

$\sf \textsf{Area of a circle}=\pi r^2 \quad \textsf{(where r is the radius)}$

$\sf \textsf{Radius of a circle}=\dfrac{1}{2}d \quad \textsf{(where d is the diameter)}$

If the diameter is equal to the side length of the square, then:
$\implies \sf r=\dfrac{1}{2}s$

Therefore:

$\begin{aligned}\implies \sf Area\:of\:circle & = \sf \pi \left(\dfrac{s}{2}\right)^2\\& = \sf \pi \left(\dfrac{s^2}{4}\right)\\& = \sf \dfrac{1}{4}\pi s^2 \end{aligned}$