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3 years ago

what is the x-intercept of the line with the equation 4x 2y = 12? a. (0, 6) b. (0, 3) c. (6, 0) d. (3, 0)

1 answer:

7 0

If you would like to find the x-intercept of the line with the equation 4x + 2y = 12, you can do this using the following steps:

4x + 2y = 12
y = 0
4x + 2 * 0 = 12
4x = 12
x = 12/4 = 3

(x, y) = (3, 0)

The correct result would be <span>d. (3, 0).</span>

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Answer:

Domain: $-2\leq x\leq 2$

Range: $0\leq y\leq 4$

Step-by-step explanation:

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3 years ago

If Whitney wrote the decimal representations for the first 300 positive integer multiples of 5 and did not write any other numbe

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Answer:

201 times

Step-by-step explanation:

Since it's the first 300 positive integer multiples of 5, then the last integer will be: 300 × 5 = 1500

While the first is 5.

The sequence is like this;

5, 10, 15, 20, 25, 30, 35, .. 1500

Since for every 2 integers that are multiples of 5, 1 will include the digit 5, then it means that, number of times the unit place will have 5 is; 300/5 = 150 times

Now, between 5 and 100, the only value that has 5 in it's tense place is 50 & 55.

So for every 100 numbers, we have 2 times to write 5 in the tens place. Thus, for 1500 numbers, we will write 5 in the tens place: 1500/100 × 2 = 30 times

Now, for the hundreds place, from 500 and 600, we have 21 multiples of 5 inclusive of 500 and 600 but since we want the one that has 5 in the hundreds place, then it is 20 as they all start with 5 excluding 600.

Also, from 1000 to 1500, the only number that has 5 in its hundreds place is 1500.

Thus,total times 5 is written in the hundreds place = 21 times

Total number of times 5 is written = 150 + 30 + 21 = 201 times

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3 years ago

1. (a) Solve the differential equation (x + 1)Dy/dx= xy, = given that y = 2 when x = 0. (b) Find the area between the two curves

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(a) The differential equation is separable, so we separate the variables and integrate:

$(x+1)\dfrac{dy}{dx} = xy \implies \dfrac{dy}y = \dfrac x{x+1} \, dx = \left(1-\dfrac1{x+1}\right) \, dx$

$\displaystyle \frac{dy}y = \int \left(1-\frac1{x+1}\right) \, dx$

$\ln|y| = x - \ln|x+1| + C$

When x = 0, we have y = 2, so we solve for the constant C :

$\ln|2| = 0 - \ln|0 + 1| + C \implies C = \ln(2)$

Then the particular solution to the DE is

$\ln|y| = x - \ln|x+1| + \ln(2)$

We can go on to solve explicitly for y in terms of x :

$e^{\ln|y|} = e^{x - \ln|x+1| + \ln(2)} \implies \boxed{y = \dfrac{2e^x}{x+1}}$

(b) The curves y = x² and y = 2x - x² intersect for

$x^2 = 2x - x^2 \implies 2x^2 - 2x = 2x (x - 1) = 0 \implies x = 0 \text{ or } x = 1$

and the bounded region is the set

$\left\{(x,y) ~:~ 0 \le x \le 1 \text{ and } x^2 \le y \le 2x - x^2\right\}$

The area of this region is

$\displaystyle \int_0^1 ((2x-x^2)-x^2) \, dx = 2 \int_0^1 (x-x^2) \, dx = 2 \left(\frac{x^2}2 - \frac{x^3}3\right)\bigg|_0^1 = 2\left(\frac12 - \frac13\right) = \boxed{\frac13}$

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cupoosta [38]

Please give me a couple of minutes, I’ll answer in the comment section

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