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irga5000 [103]
3 years ago
8

What is the slope of the line that is perpendicular to the line v=3/4x-6

Mathematics
1 answer:
emmasim [6.3K]3 years ago
6 0
The answer is D -4/3 = slope
You might be interested in
which inequality represents the following situation 3/5 times 5 less than a number is no more than 27
andrey2020 [161]

The answer is \frac{3}{5}(x-5) \leq 27


Let's build the expression piece by piece, translating sentences into formulas:


"3/5 times..." \to \frac{3}{5} \times \ldots


"5 less than a number..." \to x-5 \ldots


So, the first block becomes


"3/5 times 5 less than a number..." \to \frac{3}{5}(x-5) \ldots


"...is no more than 27" \ldots \leq 27


So, putting the pieces together, you have the answer



6 0
3 years ago
Una fábrica mando a hacer recipientes que están formando por un cilindro y un cono unidos por la base, el volumen es de 41.86cm3
SOVA2 [1]

Answer: Suponiendo que la altura del cono y la del cilindro son iguales.

\pi r^2\times h+\frac{1}{3} \pi r^2\times h=41.86\\\pi (2)^2\times h+\frac{1}{3} \pi (2)^2\times h=41.86\\\\Solving\;for\;h\\h=2.49833

5 0
3 years ago
How do you put 7/8 into a decimal
Bingel [31]

Answer:

7.8

Step-by-step explanation:

5 0
4 years ago
Read 2 more answers
Let R be the region in the first quadrant bounded by the graphs of y =x^2 and y=2x, as shown in the figure above. The region R i
Kobotan [32]

Answer:

The area between the two functions is approximately 1.333 units.

Step-by-step explanation:

If I understand your question correctly, you're looking for the area surrounded by the the line y = 2x and the parabola y = x², (as shown in the attached image).

To do this, we just need to take the integral of y = x², and subtract that from the area under y = 2x, within that range.

First we need to find where they intersect:

2x = x²

2 = x

So they intersect at (2, 4) and (0, 0)

Now we simply need to take the integrals of each, subtracting the parabola from the line (as the parabola will have lower values in that range):

a = \int\limits^2_0 {2x} \, dx - \int\limits^2_0 {x^2} \, dx\\\\a = x^2\left \{ {{x=2} \atop {x=0}} \right - \frac{x^3}{3}\left \{ {{x=2} \atop {x=0}} \right\\\\a = (2^2 - 0^2) - (\frac{2^3}{3} - \frac{0^3}{3})\\\\a = 2^2 - \frac{2^3}{3}\\a = 4 - 8/3\\a \approx 1.333

So the correct answer is C, the area between the two functions is 4/3 units.

3 0
3 years ago
Can someone explain the differences between a rational and irrational number, and a interger
vodomira [7]
Interfere are numbers that can be neagative
Or positive. Rational numbers are numbers that can be put into fractions. Irrational numbers cannot be put into a fraction
6 0
4 years ago
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