All categories

3 years ago

What is the surface area of this rectangle prism

Mathematics

2 answers:

JulijaS [17]3 years ago

8 0

356 cm square I guess

asambeis [7]3 years ago

4 0

356 is the surface area in centimeters

You might be interested in

The general formula for this parabola is y2 = 4px. Therefore, the value of p is

olga55 [171]

Simple

compare
y²=4px vs
y²=-4x

y²=y²
4px=-4x
divide both sides by x
4p=-4
divide both sides by 4
p=-1

p=-1
If correct please give brainliest!! Hope this helps

6 0

3 years ago

PLEASE HELP!! LMNP is rotated 90° clockwise around the origin. What are the coordinates of M’

lubasha [3.4K]

Answer:

The answer is "(4, -5)".

Step-by-step explanation:

Please find the complete question in the attached file.

In the given image file when we rotate it to 90 degrees clockwise so the value as the rule is $(x, y) \longrightarrow (y, -x)$

Putting the value in x = 5 and y = 4 then:

$(5, 4) \longrightarrow (4, -5)$

5 0

3 years ago

For a track meet, Tara ran the 1,600 m event at constant speed in 5 minutes and 20 seconds. What was her speed?

9966 [12]

Answer:

300 m/min

Step-by-step explanation:

Using the general formula of d = rt, where d = distance, r = rate and t = time, we can solve for the missing variable (r, or speed). Given Tara's distance of 1600 m and her total time of 5 minutes and 20 seconds, first convert her time into just minutes:

5 $\frac{20}{60}=\frac{1}{3}$

5 $\frac{1}{3}=\frac{16}{3}$

1600 = r(16/3)

Multiply both sides by reciprocal of 16/3: 3/16(1600) = r(16/3)(3/16)

Solve for 'r': 300 meters per minute

4 0

3 years ago

EXAMPLE 5 Find the maximum value of the function f(x, y, z) = x + 2y + 9z on the curve of intersection of the plane x − y + z =

geniusboy [140]

The Lagrangian,

$L(x,y,z,\lambda,\mu)=x+2y+9z-\lambda(x-y+z-1)-\mu(x^2+y^2-1)$

has critical points where its partial derivatives vanish:

$L_x=1-\lambda-2\mu x=0$

$L_y=2+\lambda-2\mu y=0$

$L_z=9-\lambda=0$

$L_\lambda=x-y+z-1=0$

$L_\mu=x^2+y^2-1=0$

$L_z=0$ tells us $\lambda=9$ , so that

$L_x=0\implies-8-2\mu x=0\implies x=-\dfrac4\mu$

$L_y=0\implies11-2\mu y=0\implies y=\dfrac{11}{2\mu}$

Then with $L_\mu=0$ , we get

$x^2+y^2=\dfrac{16}{\mu^2}+\dfrac{121}{4\mu^2}=1\implies\mu=\pm\dfrac{\sqrt{185}}2$

and $L_\lambda=0$ tells us

$x-y+z=-\dfrac4\mu-\dfrac{11}{2\mu}+z=1\implies z=1+\dfrac{19}{2\mu}$

Then there are two critical points, $\left(\pm\frac8{\sqrt{185}},\mp\frac{11}{\sqrt{185}},1\pm\frac{19}{\sqrt{185}}\right)$ . The critical point with the negative $x$ -coordinates gives the maximum value, $9+\sqrt{185}$ .

8 0

4 years ago

345+8 1/2<br><br> /= fraction

Natali5045456 [20]

153 1/2 I thinkOr you have to convert it Good luck

8 0

4 years ago

Read 2 more answers