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2 years ago

HELP PLEASE ANSWER QUICK VERY EASY!!!

Mathematics

1 answer:

sleet_krkn [62]2 years ago

5 0

Answer:

id say c. because its the closest number that corresponds with the graph.

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Solve these linear equations by Cramer's Rules Xj=det Bj / det A:

timurjin [86]

Answer:

(a) $x_1=-2,x_2=1$

(b) $x_1=\frac{3}{4} ,x_2=-\frac{1}{2} ,x_3=\frac{1}{4}$

Step-by-step explanation:

(a) For using Cramer's rule you need to find matrix $A$ and the matrix $B_j$ for each variable. The matrix $A$ is formed with the coefficients of the variables in the system. The first step is to accommodate the equations, one under the other, to get $A$ more easily.

$2x_1+5x_2=1\\x_1+4x_2=2$

$\therefore A=\left[\begin{array}{cc}2&5\\1&4\end{array}\right]$

To get $B_1$ , replace in the matrix A the 1st column with the results of the equations:

$B_1=\left[\begin{array}{cc}1&5\\2&4\end{array}\right]$

To get $B_2$ , replace in the matrix A the 2nd column with the results of the equations:

$B_2=\left[\begin{array}{cc}2&1\\1&2\end{array}\right]$

Apply the rule to solve $x_1$ :

$x_1=\frac{det\left(\begin{array}{cc}1&5\\2&4\end{array}\right)}{det\left(\begin{array}{cc}2&5\\1&4\end{array}\right)} =\frac{(1)(4)-(2)(5)}{(2)(4)-(1)(5)} =\frac{4-10}{8-5}=\frac{-6}{3}=-2\\x_1=-2$

In the case of B2, the determinant is going to be zero. Instead of using the rule, substitute the values of the variable $x_1$ in one of the equations and solve for $x_2$ :

$2x_1+5x_2=1\\2(-2)+5x_2=1\\-4+5x_2=1\\5x_2=1+4\\ 5x_2=5\\x_2=1$

(b) In this system, follow the same steps,ust remember $B_3$ is formed by replacing the 3rd column of A with the results of the equations:

$2x_1+x_2 =1\\x_1+2x_2+x_3=0\\x_2+2x_3=0$

$\therefore A=\left[\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right]$

$B_1=\left[\begin{array}{ccc}1&1&0\\0&2&1\\0&1&2\end{array}\right]$

$B_2=\left[\begin{array}{ccc}2&1&0\\1&0&1\\0&0&2\end{array}\right]$

$B_3=\left[\begin{array}{ccc}2&1&1\\1&2&0\\0&1&0\end{array}\right]$

$x_1=\frac{det\left(\begin{array}{ccc}1&1&0\\0&2&1\\0&1&2\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)} =\frac{1(2)(2)+(0)(1)(0)+(0)(1)(1)-(1)(1)(1)-(0)(1)(2)-(0)(2)(0)}{(2)(2)(2)+(1)(1)(0)+(0)(1)(1)-(2)(1)(1)-(1)(1)(2)-(0)(2)(0)}\\ x_1=\frac{4+0+0-1-0-0}{8+0+0-2-2-0} =\frac{3}{4} \\x_1=\frac{3}{4}$

$x_2=\frac{det\left(\begin{array}{ccc}2&1&0\\1&0&1\\0&0&2\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)} =\frac{(2)(0)(2)+(1)(0)(0)+(0)(1)(1)-(2)(0)(1)-(1)(1)(2)-(0)(0)(0)}{4} \\x_2=\frac{0+0+0-0-2-0}{4}=\frac{-2}{4}=-\frac{1}{2}\\x_2=-\frac{1}{2}$

$x_3=\frac{det\left(\begin{array}{ccc}2&1&1\\1&2&0\\0&1&0\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)}=\frac{(2)(2)(0)+(1)(1)(1)+(0)(1)(0)-(2)(1)(0)-(1)(1)(0)-(0)(2)(1)}{4} \\x_3=\frac{0+1+0-0-0-0}{4}=\frac{1}{4}\\x_3=\frac{1}{4}$

6 0

3 years ago

If the dimensions of a pentagonal prism are quadrupled, then the surface area of the prism is multiplied by eight.

Lubov Fominskaja [6]

Answer:

false

Step-by-step explanation:

the relationship between lengths/dimensions and areas is that areas are created by multiplying 2 dimensions.

when you quadruple (×4) the dimensions, then the areas are growing with the square of the factor (×4×4 = ×16), because the factor goes twice into the multiplication : one time for every dimension involved.

so, quadrupling the dimensions would multiply the areas by 16.

8 0

2 years ago

The base of a 15 foot long guy wire is located 5 ft from the telephone pole it is anchoring. how high up the pole does the guy w

Andrei [34K]

${5}^{2} + {h}^{2} = {15}^{2}$
h = sqrt
$sqrt{200}$

6 0

2 years ago

Use the data in LAWSCH85 for this exercise. (i) Using the same model as in Problem 4 in Chapter 3, state and test the null hypot

natka813 [3]

Answer:

Step-by-step explanation:

In the model

Log (salary) = B0 + B1LSAT +B2GPA +B3log(libvol) +B4log(cost)+B5 rank+u

The hypothesis that rank has no effect on log (salary) is H0:B5 = 0. The estimated equation (now with standard errors) is

Log (salary) = 8.34 + .0047 LSAT + .248 GPA + .095 log(libvol)

(0.53) (.0040) (.090) (.033)

+ .038 log(cost) – .0033 rank

(.032) (.0003)

n = 136, R2 = .842.

The t statistic on rank is –11(i.e. 0.0033/0.0003), which is very significant. If rank decreases by 10 (which is a move up for a law school), median starting salary is predicted to increase by about 3.3%.

(ii) LSAT is not statistically significant (t statistic ≈1.18) but GPA is very significance (t statistic ≈2.76). The test for joint significance is moot given that GPA is so significant, but for completeness the F statistic is about 9.95 (with 2 and 130 df) and p-value ≈.0001.

6 0

3 years ago

Can somebody help me on this question

Marrrta [24]

Consent is 5 coefficient is 3x hope this helps

8 0

2 years ago

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