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3 years ago

7

Jason unfolded a cardboard box. The figure of the unfolded box is shown below: Which calculation will give the total surface are

a of the cardboard, in square inches, that was used to make the box?
5 × 6 × 6
6 × 5 × 5
10 × 6 × 6
6 × 10 × 10

1 answer:

Luda [366]3 years ago

4 0

D.) 6 x 10 x 10 hope it helped

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The leader of the team wants to do everything him or herself. They have a very big ego and are difficult to work with. What is t

Aleks [24]

Answer:

This is a hardbut I think the awnser is C

Step-by-step explanation:

This usually happens when a person think that the others are incapable of actullay figuring out he problems .

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3 years ago

Help The area of a square

DIA [1.3K]

Answer:

$2551\dfrac{1}{2}$

Step-by-step explanation:

$\bf l = 20\dfrac{\sf 1}{4} \ in = \dfrac{81}{4} \ in$

$\bf w = 10\dfrac{1}{2} = \dfrac{21}{2} \ in\\\\\\h = 12 \ in$

Volume of rectangular prism = l*w*h

$= \dfrac{81}{4}*\dfrac{21}{2}*12\\\\\\= 81 *\dfrac{21}{2}*3\\\\\\= \dfrac{5103}{2}\\\\\\= 2551 \dfrac{1}{2} \ cubic \ inches$

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2 years ago

Read 2 more answers

A 100 gallon tank initially contains 100 gallons of sugar water at a concentration of 0.25 pounds of sugar per gallon suppose th

Vsevolod [243]

At the start, the tank contains

(0.25 lb/gal) * (100 gal) = 25 lb

of sugar. Let $S(t)$ be the amount of sugar in the tank at time $t$ . Then $S(0)=25$ .

Sugar is added to the tank at a rate of <em>P</em> lb/min, and removed at a rate of

$\left(1\frac{\rm gal}{\rm min}\right)\left(\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm gal}\right)=\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm min}$

and so the amount of sugar in the tank changes at a net rate according to the separable differential equation,

$\dfrac{\mathrm dS}{\mathrm dt}=P-\dfrac S{100}$

Separate variables, integrate, and solve for <em>S</em>.

$\dfrac{\mathrm dS}{P-\frac S{100}}=\mathrm dt$

$\displaystyle\int\dfrac{\mathrm dS}{P-\frac S{100}}=\int\mathrm dt$

$-100\ln\left|P-\dfrac S{100}\right|=t+C$

$\ln\left|P-\dfrac S{100}\right|=-100t-100C=C-100t$

$P-\dfrac S{100}=e^{C-100t}=e^Ce^{-100t}=Ce^{-100t}$

$\dfrac S{100}=P-Ce^{-100t}$

$S(t)=100P-100Ce^{-100t}=100P-Ce^{-100t}$

Use the initial value to solve for <em>C</em> :

$S(0)=25\implies 25=100P-C\implies C=100P-25$

$\implies S(t)=100P-(100P-25)e^{-100t}$

The solution is being drained at a constant rate of 1 gal/min; there will be 5 gal of solution remaining after time

$1000\,\mathrm{gal}+\left(-1\dfrac{\rm gal}{\rm min}\right)t=5\,\mathrm{gal}\implies t=995\,\mathrm{min}$

has passed. At this time, we want the tank to contain

(0.5 lb/gal) * (5 gal) = 2.5 lb

of sugar, so we pick <em>P</em> such that

$S(995)=100P-(100P-25)e^{-99,500}=2.5\implies\boxed{P\approx0.025}$

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3 years ago

Plsss helppp I’m dying!! —> Complete the square to determine if the equation is a circle, quadratic equation, or other

sashaice [31]

a) The equation represents an ellipse.

b) The equation represents a circle.

c) The equation represents a parabola.

<h3>How to infer the graphical form of a general equation</h3>

In this problem we have <em>general</em> equations of the form A · x² + B · y² + C · x + D · y + E = 0, which have to be modified into <em>standard</em> form to infer its <em>graphical</em> form. This procedure can be done by <em>algebra</em> properties:

16 · x² + 4 · y² + 96 · x - 8 · y + 84 = 0

[(4 · x)² + 2 · 12 · (4 · x)] + [(2 · y)² - 2 · 2 · (2 · y)] = - 84

[(4 · x)² + 2 · 12 · (4 · x) + 144] + [(2 · y)² - 2 · 2 · (2 · y) + 4] = 64

(4 · x + 12)² + (2 · y + 2)² = 64

4² · (x + 3)² + 2² · (y + 2)² = 64

(x + 3)² / 4 + (y + 2)² / 16 = 1 : Ellipse

x² + y² + 8 · x - 6 · y - 15 = 0

(x² + 8 · x) + (y² - 6 · y) = 15

(x² + 2 · 4 · x + 16) + (y² - 2 · 3 · y + 9) = 40

(x + 4)² + (y - 3)² = 40 : Circle

x² + 6 · x + 4 · y + 5 = 0

x² + 6 · x + 5 = - 4 · y

y = - (1 / 4) · x² - (3 / 2) · x - (5 / 4) : Parabola

To learn more on ellipses: brainly.com/question/19507943

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4 0

1 year ago

If point Q(3,-4) is rotated +270° about the origin, what are the coordinates of Q?

erica [24]

Answer:

$Q' =(-4,-3)$

Step-by-step explanation:

Given

$Q = (3,-4)$

Rotation = +270 degrees

Required

Determine the new coordinates of Q (i.e. Q')

When a point (x,y) is rotated +270° about the origin, the new coordinates is: (y,−x)

So:

$Q = (3,-4)$

Where:

$x = 3$ and $y = -4$

Q' will be:

$Q' =(-4,-3)$

8 0

3 years ago