Please help 7x-y=23 y=4x-11

What rational number represents a drop of 4 1/4 in? Enter your answer in the box.

aalyn [17]

Answer:

The fraction 1/4 is a rational number.

It stands for the ratio between the integers 1 and 4.

6 0

3 years ago

Read 2 more answers

LEAST TO GREATEST PLSSS

Cerrena [4.2K]

Answer:

-0.8, 7/9, √3

Step-by-step explanation:

5 0

3 years ago

Please solve my difficult question fast

kicyunya [14]

The 3 angles form the straight line AB. A straight line equals 180 degrees.

The 3 angles when added together need to equal 180:

2x + 65 + (x + 65) = 180

Simplify by combining like terms:

3x + 130 = 180

Subtract 130 from both sides

3x = 50

Divide both sides by 3

X = 50/3

X = 16 2/3 (16.66667 as a repeating decimal)

Now you have x if you need to solve all the angles replace x with its value and sole:

2x = 2(16 2/3) = 33 1/3

X + 65 = 16 2/3 + 65 = 81 2/3

7 0

3 years ago

Read 2 more answers

Solve the following inequality: 3-5x-x^2>=0

Dennis_Churaev [7]

Answer:

-5/2+-1/2√37≤x≤-5/2+1/2√37

Step-by-step explanation:

Step 1: Find the critical points

-x^2-5x+3=0

For this equation: a=-1, b=-5, c=3

−1x^2+−5x+3=0

x=−b±√b2−4ac/2a

x=−(−5)±√(−5)2−4(−1)(3)/2(-1)

x=5±√37 /−2

x=-5/2+1/2√37

Step 2: Check intervals in between critical points

x≤-5/2+1/2 √37 (Doesn't work in original inequality)

-5/2+-1/2√37≤x≤-5/2+1/2√37 (Works in original inequality)

x≥-5/2+1/2 √37 (Doesn't work in original inequality)

5 0

3 years ago

The point P(1,1/2) lies on the curve y=x/(1+x). (a) If Q is the point (x,x/(1+x)), find the slope of the secant line PQ correct

lukranit [14]

Answer:

See explanation

Step-by-step explanation:

You are given the equation of the curve

$y=\dfrac{x}{1+x}$

Point $P\left(1,\dfrac{1}{2}\right)$ lies on the curve.

Point $Q\left(x,\dfrac{x}{1+x}\right)$ is an arbitrary point on the curve.

The slope of the secant line PQ is

$\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{\frac{x}{1+x}-\frac{1}{2}}{x-1}=\dfrac{\frac{2x-(1+x)}{2(x+1)}}{x-1}=\dfrac{\frac{2x-1-x}{2(x+1)}}{x-1}=\\ \\=\dfrac{\frac{x-1}{2(x+1)}}{x-1}=\dfrac{x-1}{2(x+1)}\cdot \dfrac{1}{x-1}=\dfrac{1}{2(x+1)}\ [\text{When}\ x\neq 1]$