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UNO [17]
3 years ago
12

3/2 x - 5/4 y =15 rate of change​

Mathematics
1 answer:
Alenkinab [10]3 years ago
3 0

Answer:

6/5

Step-by-step explanation:

m is your rate of change you need to get y by itself.

3/2x - 15 = 5/4y

3x/2 - 30/2 = 5/4y

3x-30/2 = 5/4y

3x - 30 = 2 x 5/4y

3x - 30 = 2 x 5/4x2y (cross out 2 and 4)

3x - 30 = 5/2y

2 x 3x - 2 x 30 = 2 x 5/4y

6x - 60 = 5y

6x/5 - 60/5 = 5y/5

y = 6/5 x x - 12

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Morgages is spelled mortgages

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What is the value of the expression 5x2 + 2x when x = 5?
masha68 [24]

Answer:

135

Step-by-step explanation:

Given:

x = 5

To determine a numerical value for the expression, simply substitute the value of "x" into the expression and simplify the expression, if necessary, to determine a specific number for the expression provided.

  • \implies 5x^{2}  + 2x
  • \implies 5(5)^{2}  + 2(5) \longrightarrow \longrightarrow \longrightarrow (x = 5)
  • \implies 5(5)(5)  + 2(5)
  • \implies 125  + 10
  • \implies 135

Therefore, the value of the expression 5x^{2}  + 2x when x = 5 is 135.

Learn more about this topic: brainly.com/question/27675691

7 0
1 year ago
What is 82,119 rounded to the nearest ten thousand?
Stels [109]
80,000.
To round up to the nearest ten thousand, we would see if the next lower place value has a amount either 5 or higher or 4 or lower.
If it’s five or higher, we round up.
If it’s four or lower, we round down.
We can see that the number in the ten thousands place is 8. The number in the place value after 8 is 2.
2<5 so we round down.
The answer is 80,000.
Hope this helps!
8 0
3 years ago
The graph of f ′ (x), the derivative of f(x), is continuous for all x and consists of five line segments as shown below. Given f
Nataliya [291]

The maxima of f(x) occur at its critical points, where f '(x) is zero or undefined. We're given f '(x) is continuous, so we only care about the first case. Looking at the plot, we see that f '(x) = 0 when x = -4, x = 0, and x = 5.

Notice that f '(x) ≥ 0 for all x in the interval [0, 5]. This means f(x) is strictly increasing, and so the absolute maximum of f(x) over [0, 5] occurs at x = 5.

By the fundamental theorem of calculus,

\displaystyle f(5) = f(0) + \int_0^5 f'(x) \, dx

The definite integral corresponds to the area of a trapezoid with height 2 and "bases" of length 5 and 2, so

\displaystyle \int_0^5 f'(x) \, dx = \frac{5+2}2 \times 2 = 7

\implies \max\{f(x) \mid 0\le x \le5\} = f(5) = f(0) + 7 = \boxed{13}

8 0
2 years ago
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