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2 years ago

I need help with 4=2/3(2x-9

Mathematics

2 answers:

sweet [91]2 years ago

4 0

Answer:

x=15/2

Hope it helps :)

Gala2k [10]2 years ago

4 0

Answer:

x = 7.5

Step-by-step explanation:

4 = 2/3 (2x - 9)

4 = 4/3x - 6

4 + 6 = 4/3x - 6 + 6

10 = 4/3x

10 x 3/4 = x

x = 7.5

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Determine whether triangle TJD is congruent to triangle SEK given T (-4,-2), J (0,5), D (1,-1), S (-1,3), E (3,10), K (4,4) and

nataly862011 [7]

Yes, it is.

The three points of triangle SEK are the points of TJD shifted 3 units right and 5 units upward.

So they are the same triangle, just translated in the plane.

7 0

2 years ago

Find the minimum and maximum of f(x,y,z)=x^2+y^2+z^2 subject to two constraints, x+2y+z=4 and x-y=8.

Alika [10]

The Lagrangian for this function and the given constraints is

$L(x,y,z,\lambda_1,\lambda_2)=x^2+y^2+z^2+\lambda_1(x+2y+z-4)+\lambda_2(x-y-8)$

which has partial derivatives (set equal to 0) satisfying

$\begin{cases}L_x=2x+\lambda_1+\lambda_2=0\\L_y=2y+2\lambda_1-\lambda_2=0\\L_z=2z+\lambda_1=0\\L_{\lambda_1}=x+2y+z-4=0\\L_{\lambda_2}=x-y-8=0\end{cases}$

This is a fairly standard linear system. Solving yields Lagrange multipliers of $\lambda_1=-\dfrac{32}{11}$ and $\lambda_2=-\dfrac{104}{11}$ , and at the same time we find only one critical point at $(x,y,z)=\left(\dfrac{68}{11},-\dfrac{20}{11},\dfrac{16}{11}\right)$ .

Check the Hessian for $f(x,y,z)$ , given by

$\mathbf H(x,y,z)=\begin{bmatrix}f_{xx}&f_{xy}&f_{xz}\\f_{yx}&f_{yy}&f_{yz}\\f_{zx}&f_{zy}&f_{zz}\end{bmatrix}=\begin{bmatrix}=\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}$

$\mathbf H$ is positive definite, since $\mathbf v^\top\mathbf{Hv}>0$ for any vector $\mathbf v=\begin{bmatrix}x&y&z\end{bmatrix}^\top$ , which means $f(x,y,z)=x^2+y^2+z^2$ attains a minimum value of $\dfrac{480}{11}$ at $\left(\dfrac{68}{11},-\dfrac{20}{11},\dfrac{16}{11}\right)$ . There is no maximum over the given constraints.

7 0

3 years ago

Please please help me!!!!!!!

Jlenok [28]

Answer:

Step-by-step explanation:

Hope this helps! Good luck!

7 0

2 years ago

A recent survey showed that in a sample of 100 elementary school teachers, 15 were single. In a sample of 180 high school teache

trapecia [35]

Answer:

By using hypothesis test at α = 0.01, we cannot conclude that the proportion of high school teachers who were single greater than the proportion of elementary teachers who were single

Step-by-step explanation:

let p1 be the proportion of elementary teachers who were single

let p2 be the proportion of high school teachers who were single

Then, the null and alternative hypotheses are:

$H_{0}$ : p2=p1

$H_{a}$ : p2>p1

We need to calculate the test statistic of the sample proportion for elementary teachers who were single.

It can be calculated as follows:

$\frac{p(s)-p}{\sqrt{\frac{p*(1-p)}{N} } }$ where

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p is the proportion of elementary teachers who were single ( $\frac{15}{100} =0.15$ )