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3 years ago

Evaluate all three please i don't know any of those :(

Mathematics

1 answer:

gayaneshka [121]3 years ago

3 0

1. 35 - 2 x 4^2 = 3.
2. 12 divided by (1 + 3^3 - 2^4) = 1.
3. 8^2 divided by [(11-3) x 2] =4.
Have a gud day :)

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The length l, width w, and height h of a box change with time. At a certain instant the dimensions are l = 3 m and w = h = 6 m,

Gemiola [76]

Answer:

a) The rate of change associated with the volume of the box is 54 cubic meters per second, b) The rate of change associated with the surface area of the box is 18 square meters per second, c) The rate of change of the length of the diagonal is -1 meters per second.

Step-by-step explanation:

a) Given that box is a parallelepiped, the volume of the parallelepiped, measured in cubic meters, is represented by this formula:

$V = w \cdot h \cdot l$

Where:

$w$ - Width, measured in meters.

$h$ - Height, measured in meters.

$l$ - Length, measured in meters.

The rate of change in the volume of the box, measured in cubic meters per second, is deducted by deriving the volume function in terms of time:

$\dot V = h\cdot l \cdot \dot w + w\cdot l \cdot \dot h + w\cdot h \cdot \dot l$

Where $\dot w$ , $\dot h$ and $\dot l$ are the rates of change related to the width, height and length, measured in meters per second.

Given that $w = 6\,m$ , $h = 6\,m$ , $l = 3\,m$ , $\dot w =3\,\frac{m}{s}$ , $\dot h = -6\,\frac{m}{s}$ and $\dot l = 3\,\frac{m}{s}$ , the rate of change in the volume of the box is:

$\dot V = (6\,m)\cdot (3\,m)\cdot \left(3\,\frac{m}{s} \right)+(6\,m)\cdot (3\,m)\cdot \left(-6\,\frac{m}{s} \right)+(6\,m)\cdot (6\,m)\cdot \left(3\,\frac{m}{s}\right)$

$\dot V = 54\,\frac{m^{3}}{s}$

The rate of change associated with the volume of the box is 54 cubic meters per second.

b) The surface area of the parallelepiped, measured in square meters, is represented by this model:

$A_{s} = 2\cdot (w\cdot l + l\cdot h + w\cdot h)$

The rate of change in the surface area of the box, measured in square meters per second, is deducted by deriving the surface area function in terms of time:

$\dot A_{s} = 2\cdot (l+h)\cdot \dot w + 2\cdot (w+h)\cdot \dot l + 2\cdot (w+l)\cdot \dot h$

Given that $w = 6\,m$ , $h = 6\,m$ , $l = 3\,m$ , $\dot w =3\,\frac{m}{s}$ , $\dot h = -6\,\frac{m}{s}$ and $\dot l = 3\,\frac{m}{s}$ , the rate of change in the surface area of the box is:

$\dot A_{s} = 2\cdot (6\,m + 3\,m)\cdot \left(3\,\frac{m}{s} \right) + 2\cdot (6\,m+6\,m)\cdot \left(3\,\frac{m}{s} \right) + 2\cdot (6\,m + 3\,m)\cdot \left(-6\,\frac{m}{s} \right)$

$\dot A_{s} = 18\,\frac{m^{2}}{s}$

The rate of change associated with the surface area of the box is 18 square meters per second.

c) The length of the diagonal, measured in meters, is represented by the following Pythagorean identity:

$r^{2} = w^{2}+h^{2}+l^{2}$

The rate of change in the surface area of the box, measured in square meters per second, is deducted by deriving the surface area function in terms of time before simplification:

$2\cdot r \cdot \dot r = 2\cdot w \cdot \dot w + 2\cdot h \cdot \dot h + 2\cdot l \cdot \dot l$

$r\cdot \dot r = w\cdot \dot w + h\cdot \dot h + l\cdot \dot l$

$\dot r = \frac{w\cdot \dot w + h \cdot \dot h + l \cdot \dot l}{\sqrt{w^{2}+h^{2}+l^{2}}}$

Given that $w = 6\,m$ , $h = 6\,m$ , $l = 3\,m$ , $\dot w =3\,\frac{m}{s}$ , $\dot h = -6\,\frac{m}{s}$ and $\dot l = 3\,\frac{m}{s}$ , the rate of change in the length of the diagonal of the box is:

$\dot r = \frac{(6\,m)\cdot \left(3\,\frac{m}{s} \right)+(6\,m)\cdot \left(-6\,\frac{m}{s} \right)+(3\,m)\cdot \left(3\,\frac{m}{s} \right)}{\sqrt{(6\,m)^{2}+(6\,m)^{2}+(3\,m)^{2}}}$

$\dot r = -1\,\frac{m}{s}$

The rate of change of the length of the diagonal is -1 meters per second.

6 0

3 years ago

81000 in standard form

BabaBlast [244]

I believe the answer is 10^4

4 0

3 years ago

Read 2 more answers

The answer to this.

Readme [11.4K]

Answer:

The two column proof is presented as follows;

Step ${}$ Statement Reason

1 ${}$ $\overline {AC}$ ≅ $\overline {BD}$ Given

${}$ ∠CAB ≅ ∠DBA

2 ${}$ $\overline {AB}$ ≅ $\overline {AB}$ Reflexive property

3 ${}$ ΔABC ≅ ΔBAD SAS rule of congruency

Step-by-step explanation:

Given that we have;

Segment $\overline {AC}$ of ΔABC being congruent to (≅) segment $\overline {BD}$ on ΔBAD and angle ∠CAB on ΔABC is congruent to angle ∠DBA on ΔBAD, and also that the two triangles share a common side, which is segment $\overline {AB}$ , we have;

Segment $\overline {AB}$ is congruent to itself by reflexive property, therefore;

Two sides and an included angle on ΔABC are congruent to the corresponding two sides and an included angle on ΔBAD, which by Side-Angle-Side, SAS, rule of congruency, ΔABC is congruent to ΔBAD

7 0

3 years ago

Translate the following: the sum of twice a number 7

Morgarella [4.7K]

Answer:

2n + 7

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

If f(1) = 7 and f(n) = -3f(n − 1) then find the value of f(6).

ludmilkaskok [199]

Answer:

-1701

Step-by-step explanation:

The recursion relation tells you each term of the sequence is -3 times the previous term. Then the first 6 terms are ...

7, -21, 63, -189, 567, -1701

The value of f(6) is -1701.

_____

<em>Additional comment</em>

The explicit formula for a geometric sequence with first term 7 and common ratio -3 is ...

f(n) = 7×(-3)^(n-1)

Then f(6) is ...

f(6) = 7×(-3)^5 = 7×(-243) = -1701

8 0

2 years ago