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2 years ago

The volume of a box is given by V=3x^3 + 33x^2 + 84x. Determine expressions for a possible set of dimensions for this box.

Mathematics

1 answer:

Leto [7]2 years ago

7 0

Answer:

(-7,0) (-4,0)(0,0)

Step-by-step explanation:

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A rectangular piece of sheet metal has a length that is 10 10 in. less than twice the width. A square piece 5 5 in. on a side is

Hatshy [7]

Answer:

length: 32 in
width: 21 in

Step-by-step explanation:

If the width of the original piece of sheet metal is x, then the length is 2x-10. Subtracting a 5" square from each corner makes the bottom of the box have dimensions (x-10) and (2x-20). The volume of the box is then ...

5(x -10)(2x -20) = 1210 . . . . . . volume is the product of the dimensions

10(x -10)² = 1210 = 10(11²) . . . . factor the equation

(x -10)² = 11² . . . . . . . . divide by 10

x = 11 + 10 = 21 . . . . . .take the square root, add 10

The length and width of the original piece of sheet metal were 32 inches and 21 inches, respectively.

4 0

2 years ago

What is 18÷9-3+5+6-6+6-5

CaHeK987 [17]

It is 11 because 1 18÷9 = 2 2+3 = 5 5+5 = 10 10+6 = 16-6=10 10+6=16 16-5 equal 11

4 0

2 years ago

Read 2 more answers

De un numero se sabe que si a su cuadrado le restamos su mitad, se obtiene el mismo numero, Cuál es?

Elena-2011 [213]

Answer:

creo que es e numero 2

Step-by-step explanation:

el numero 2 al cuadrado es 4 si le restas su mitad(2) te da como resultado 2.

3 0

2 years ago

Is this correct I think it is incenter it is asking what point P in the triangle is called

KatRina [158]

Answer:

Yes. Incenter.

Step-by-step explanation:

It is constructed by taking the 3 angles bisectors of the triangle.

5 0

2 years ago

Suppose that triangle ABC is a right triangle with the right angle at C. Let line segment CD be the perpendicular from point C t

neonofarm [45]

Step-by-step explanation:

Given:

Let triangle ACD is aright angle triangle with right angle at C. A line perpendicular to AB join C.

Therefore we can say that line segment CD divides angle at C into two equal angles.

So in ΔACD and ΔCDB

∠ ACD = ∠DCB

and ∠ADC = ∠BDC = 90°

and CD =CD

∴ we can say that ΔACD and ΔCDB are similar triangles.

∴ Area of ΔACD = $\frac{1}{2}\times base\times height$

= $\frac{1}{2}\times AD\times CD$

Area of ΔCDB = $\frac{1}{2}\times base\times height$

= $\frac{1}{2}\times DB\times CD$

Therefore ratio of the areas of ΔACD and ΔCDB is

i.e. $\frac{\Delta ACD}{\Delta CDB}$ = $\frac{\frac{1}{2}\times AD\times CD}{\frac{1}{2}\times DB\times CD}$

= $\frac{AD}{DB}$

∴ Area of the ratio of $\frac{\Delta ACD}{\Delta CDB}$ = $\frac{AD}{DB}$

Hence proved

7 0

3 years ago