On a coordinate plane, triangle G H J has points (1, 1), (4, 1), (4, 5).

Mathematics

1 answer:

CaHeK987 [17]2 years ago

7 0

Answer:

the answer is 6 sq units

Step-by-step explanation:

if it is correct pls give me the brainlest

You might be interested in

Avery’s piggy bank has 300 nickels, 450 pennies, and 150 dimes. She randomly picks three coins. Each time she picks a coin, she

Mashutka [201]

Answer:

(1) .20 (2) .40 (3) .12 (4) Less than

Step-by-step explanation:

You have to look at the table. There are 5 columns with 10 rows. 5x10=50

Then simply count the boxes that have the correct number of currency for instance, if they are asking for EXACTLY 1 dime then you rule out the ones that have 2 or 3 dimes and only the count the ones that have a single dime. So you count PDN but you would not count PDD. There are 20 boxes that have a single dime in them. 20 out of the 50 boxes. 20/50=.40 (answer 2)

The estimated probability that exactly two of the three coins Avery randomly picked are nickels is . 20

The estimated probability that exactly one of the three coins Avery randomly picked is a dime is . 40

The estimated probability that all three coins Avery randomly picked are pennies is . 12

The answer to #1 is .20 or 20% and the answer to #2 is .40 or 40%. 20% is less than 40% so...

The estimated probability that exactly two of the three coins Avery randomly picked are nickels is LESS THAN the estimated probability that exactly one of the three coins Avery randomly picked is a dime.

4 0

3 years ago

Find all points on the portion of the plane x+y+z=5 in the first octant at which f(x, y, z) = xy2z2 has a maximum value.

irina1246 [14]

Lagrange multipliers:

$L(x,y,z,\lambda)=xy^2z^2+\lambda(x+y+z-5)$

$L_x=y^2z^2+\lambda=0$
$L_y=2xyz^2+\lambda=0$
$L_z=2xy^2z+\lambda=0$
$L_\lambda=x+y+z-5=0$

$\lambda=-y^2z^2=-2xyz^2=-2xy^2z$

$-y^2z^2=-2xyz^2\implies y=2x$ (if $y,z\neq0$ )

$-y^2z^2=-2xy^2z\implies z=2x$ (if $y,z\neq0$ )

$-2xyz^2=-2xy^2z\implies z=y$ (if $x,y,z\neq0$ )

In the first octant, we assume $x,y,z>0$ , so we can ignore the caveats above. Now,

$x+y+z=5\iff x+2x+2x=5x=5\implies x=1\implies y=z=2$

so that the only critical point in the region of interest is (1, 2, 2), for which we get a maximum value of $f(1,2,2)=16$ .

We also need to check the boundary of the region, i.e. the intersection of $x+y+z=5$ with the three coordinate axes. But in each case, we would end up setting at least one of the variables to 0, which would force $f(x,y,z)=0$ , so the point we found is the only extremum.