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2 years ago

What are the two solutions to x( 2x - 5) = 0

1 answer:

7 0

When you multiply anything by zero, you will get zero
Therefore you know that:
x = 0 and 2x - 5 = 0
2x - 5 = 0, solve for x
2x = 5, x = 5/2

Solution: x = 5/2, x = 0

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Joe thinks the area to be pantries on wall 2is 84 square feet.

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Please put a picture next time :)

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1 year ago

Marshall evaluates the expression 15.3 - 7.9 and gets an answer of 7.4.

creativ13 [48]

Answer:

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Step-by-step explanation:

15.3-7.9=7.4

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3 years ago

How many cars did he put in each box

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3 years ago

Read 2 more answers

In the 1996 presidential election, Bill Clinton received 2,495,683 votes in Texas. This was 177,868 more votes than he had recei

lilavasa [31]

Answer:

Votes did Bill Clinton get in Texas in 1992 = 2317815

Step-by-step explanation:

Total votes received by Bill Clinton in 1996 = 2,495,683

Given that this was 177,868 more votes than he had received in 1992.

So the vote received in 1992 by Bill Clinton in 1992 is = ( Total votes received by Bill Clinton in 1996 ) - ( 177,868 )

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Therefore votes did Bill Clinton get in Texas in 1992 = 2317815

5 0

3 years ago

Calculus 2

FinnZ [79.3K]

Answer:

See Below.

Step-by-step explanation:

We want to estimate the definite integral:

$\displaystyle \int_1^47\sqrt{\ln(x)}\, dx$

Using the Trapezoidal Rule, Midpoint Rule, and Simpson's Rule with six equal subdivisions.

The trapezoidal rule is given by:

$\displaystyle \int_{a}^bf(x)\, dx\approx\frac{\Delta x}{2}\Big(f(x_0)+2f(x_1)+...+2f(x_{n-1})+f(x_n)\Big)$

Our limits of integration are from x = 1 to x = 4. With six equal subdivisions, each subdivision will measure:

$\displaystyle \Delta x=\frac{4-1}{6}=\frac{1}{2}$

Therefore, the trapezoidal approximation is:

$\displaystyle =\frac{1/2}{2}\Big(f(1)+2f(1.5)+2f(2)+2f(2.5)+2f(3)+2f(3.5)+2f(4)\Big)$

Evaluate:

$\displaystyle =\frac{1}{4}(7)(\sqrt{\ln(1)}+2\sqrt{\ln(1.5)}+...+2\sqrt{\ln(3.5)}+\sqrt{\ln(4)})\\\\\approx18.139337$

The midpoint rule is given by:

$\displaystyle \int_a^bf(x)\, dx\approx\sum_{i=1}^nf\Big(\frac{x_{i-1}+x_i}{2}\Big)\Delta x$

Thus:

$\displaystyle =\frac{1}{2}\Big(f\Big(\frac{1+1.5}{2}\Big)+f\Big(\frac{1.5+2}{2}\Big)+...+f\Big(\frac{3+3.5}{2}\Big)+f\Big(\frac{3.5+4}{2}\Big)\Big)$

Simplify:

$\displaystyle =\frac{1}{2}(7)\Big(f(1.25)+f(1.75)+...+f(3.25)+f(3.75)\Big)\\\\ =\frac{1}{2}(7) (\sqrt{\ln(1.25)}+\sqrt{\ln(1.75)}+...+\sqrt{\ln(3.25)}+\sqrt{\ln(3.75)})\\\\\approx 18.767319$

Simpson's Rule is given by:

$\displaystyle \int_a^b f(x)\, dx\approx\frac{\Delta x}{3}\Big(f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+...+4f(x_{n-1})+f(x_n)\Big)$

So:

$\displaystyle =\frac{1/2}{3}\Big((f(1)+4f(1.5)+2f(2)+4f(2.5)+...+4f(3.5)+f(4)\Big)$

Simplify:

$\displaystyle =\frac{1}{6}(7)(\sqrt{\ln(1)}+4\sqrt{\ln(1.5)}+2\sqrt{\ln(2)}+4\sqrt{\ln(2.5)}+...+4\sqrt{\ln(3.5)}+\sqrt{\ln(4)})\\\\\approx 18.423834$

6 0

2 years ago