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3 years ago

OB is the bisector of AD and EC. . EO = 2x – 20 and OC = 70 – x. Solve for x.

1 answer:

4 0

We know that OB bisects EC where they intersect would be at point O,
so the two lengths EO and OC must be equal.

2x - 20 = 70 - x

[adding x + 20 to each side]

2x - 20 + x + 20 = 70 - x + x + 20

[simplifying]

3x = 90

[dividing each side by 3]

x = 30 would be the answer

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4 years ago

Find f(a), f(a+h), and 71. f(x) = 7x - 3 f(a+h)-f(a) h if h = 0. 72. f(x) = 5x²

Leni [432]

Answer:

$71. \ \ \ f(a) \ = \ 7a \ - \ 3; \ f(a+h) \ = \ 7a \ + \ 7h \ - \ 3; \ \displaystyle\frac{f(a+h) \ - \ f(a)}{h} \ = \ 7$

$72. \ \ \ f(a) \ = \ 5a^{2}; \ f(a+h) \ = \ {5a}^{2} \ + \ 10ah \ + \ {5h}^{2}; \ \displaystyle\frac{f(a+h) \ - \ f(a)}{h} \ = \ 10a \ + \ 5h$

Step-by-step explanation:

In single-variable calculus, the difference quotient is the expression

$\displaystyle\frac{f(x+h) \ - \ f(x)}{h}$ ,

which its name comes from the fact that it is the quotient of the difference of the evaluated values of the function by the difference of its corresponding input values (as shown in the figure below).

This expression looks similar to the method of evaluating the slope of a line. Indeed, the difference quotient provides the slope of a secant line (in blue) that passes through two coordinate points on a curve.

$m \ \ = \ \ \displaystyle\frac{\Delta y}{\Delta x} \ \ = \ \ \displaystyle\frac{rise}{run}$ .

Similarly, the difference quotient is a measure of the average rate of change of the function over an interval. When the limit of the difference quotient is taken as h approaches 0 gives the instantaneous rate of change (rate of change in an instant) or the derivative of the function.

Therefore,

$71. \ \ \ \ \ \displaystyle\frac{f(a \ + \ h) \ - \ f(a)}{h} \ \ = \ \ \displaystyle\frac{(7a \ + \ 7h \ - \ 3) \ - \ (7a \ - \ 3)}{h} \\ \\ \-\hspace{4.25cm} = \ \ \displaystyle\frac{7h}{h} \\ \\ \-\hspace{4.25cm} = \ \ 7$

$72. \ \ \ \ \ \displaystyle\frac{f(a \ + \ h) \ - \ f(a)}{h} \ \ = \ \ \displaystyle\frac{{5(a \ + \ h)}^{2} \ - \ {5(a)}^{2}}{h} \\ \\ \-\hspace{4.25cm} = \ \ \displaystyle\frac{{5a}^{2} \ + \ 10ah \ + \ {5h}^{2} \ - \ {5a}^{2}}{h} \\ \\ \-\hspace{4.25cm} = \ \ \displaystyle\frac{h(10a \ + \ 5h)}{h} \\ \\ \-\hspace{4.25cm} = \ \ 10a \ + \ 5h$

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2 years ago

Unitary method

tangare [24]

Answer:

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420 kilometers will be covered

Step-by-step explanation:

speed = distance ÷ time.

so 60kmh = 240km divided by 4 Hours

60km = 1 Hour of distance

240km = 4 Hours of distance

Therefore 60 x 3 hour is 180km added to 240 is 420km. So in 7 hours traveling 60 kilometers per hour the will cover 420kms.

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3 years ago