Really need help on the problem shown in the picture above and please don’t answer if you don’t know ヽ(ಠ

X (27-x) = 182<br>Please help me by solving this!

pentagon [3]

I hope this helps you..!

3 0

2 years ago

Plsss help fast thanksss correct answer pls

maks197457 [2]

Answer:

here is the answer... for it

8 0

2 years ago

HELP! ASAP!

Novosadov [1.4K]

Step-by-step explanation:

Part A:

So the height is going to be x when you fold the sides up. So that's one part of the volume but for the width it was going to be 4 but since two corners were cut out with the length x the new width is going to be (4-2x). The same thing applies for the length which should be 8 inches but since two corners were removed with the length x it's now (8-2x)

v = x(4-2x)(8-2x)

Part B:

The volume can be graphed although there must be a domain restriction since the height, width, or length cannot be negative. So let's look at each part of the equation

so for the x in front it must be greater than 0 to make sense

for the (4-2x), the x must be less than 2 or else the width is negative.

for the (8-2x) the x must be less than 4 or else the length is negative

so the domain is going to be restricted to 0 < x < 2 so all the dimensions are greater than 0

By using a graphing calculator you can see the maximum of the given equation with the domain restrictions is 0.845 which gives a volume of 12.317

5 0

2 years ago

If <img src="https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20x%20%3D%20log_%7Ba%7D%28bc%29" id="TexFormula1" title="\rm \: x = log_{a}(

timama [110]

Use the change-of-basis identity,

$\log_x(y) = \dfrac{\ln(y)}{\ln(x)}$

to write

$xyz = \log_a(bc) \log_b(ac) \log_c(ab) = \dfrac{\ln(bc) \ln(ac) \ln(ab)}{\ln(a) \ln(b) \ln(c)}$

Use the product-to-sum identity,

$\log_x(yz) = \log_x(y) + \log_x(z)$

to write

$xyz = \dfrac{(\ln(b) + \ln(c)) (\ln(a) + \ln(c)) (\ln(a) + \ln(b))}{\ln(a) \ln(b) \ln(c)}$

Redistribute the factors on the left side as

$xyz = \dfrac{\ln(b) + \ln(c)}{\ln(b)} \times \dfrac{\ln(a) + \ln(c)}{\ln(c)} \times \dfrac{\ln(a) + \ln(b)}{\ln(a)}$

and simplify to

$xyz = \left(1 + \dfrac{\ln(c)}{\ln(b)}\right) \left(1 + \dfrac{\ln(a)}{\ln(c)}\right) \left(1 + \dfrac{\ln(b)}{\ln(a)}\right)$

Now expand the right side:

$xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} \\\\ ~~~~~~~~~~~~+ \dfrac{\ln(c)\ln(a)}{\ln(b)\ln(c)} + \dfrac{\ln(c)\ln(b)}{\ln(b)\ln(a)} + \dfrac{\ln(a)\ln(b)}{\ln(c)\ln(a)} \\\\ ~~~~~~~~~~~~ + \dfrac{\ln(c)\ln(a)\ln(b)}{\ln(b)\ln(c)\ln(a)}$

Simplify and rewrite using the logarithm properties mentioned earlier.

$xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} + \dfrac{\ln(a)}{\ln(b)} + \dfrac{\ln(c)}{\ln(a)} + \dfrac{\ln(b)}{\ln(c)} + 1$

$xyz = 2 + \dfrac{\ln(c)+\ln(a)}{\ln(b)} + \dfrac{\ln(a)+\ln(b)}{\ln(c)} + \dfrac{\ln(b)+\ln(c)}{\ln(a)}$

$xyz = 2 + \dfrac{\ln(ac)}{\ln(b)} + \dfrac{\ln(ab)}{\ln(c)} + \dfrac{\ln(bc)}{\ln(a)}$

$xyz = 2 + \log_b(ac) + \log_c(ab) + \log_a(bc)$

$\implies \boxed{xyz = x + y + z + 2}$

(C)

6 0

1 year ago

N his road trip, Leon stops to refuel and get some snacks. He has the following purchases:

tresset_1 [31]

Answer:

C. Paid $2.60 too much

Step-by-step explanation:

Gas 12 x 2.89 = 34.68

granola 2 x 1.59 = 3.18 x 1.075 = 3.4185

apple .89 x 1.075 = 0.95675

vitamin water 2 x 1.39 = 2.78 x 1.075 = 2.9885

34.68
3.4185
0.95675
+ 2.9885
---------------------
$42.04

$44.64
- $42.04
---------------------
$2.60

3 0

2 years ago

Really need help on the problem shown in the picture above and please don’t answer if you don’t know ヽ(ಠ_ಠ)ノ