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3 years ago

How many solutions will the following system of linear equations have? –2x + 4y = –7 y= -1/2x + 5 A. 0 B. 1 C. 2 D. infinite

Mathematics

2 answers:

FinnZ [79.3K]3 years ago

7 0

Answer:

B. 1

Step-by-step explanation:

they only cross at one point... look below

dsp733 years ago

6 0

ANSWER

B. 1

EXPLANATION

The given system of equations is;

–2x + 4y = –7

y= -1/2x + 5

We rewrite the first equation is slope intercept form:

4y = 2x–7

$y = \frac{1}{2}x - \frac{7}{4}$

The two equations have different slopes.

This means that they will intersect at one point in the XY plane.

Hence the system has a unique solution.

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Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

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:

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

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3 years ago

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Anika [276]

Answer:

Step-by-step explanation:

59 085

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