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10 days ago

6

Determine whether the following lines represented by the vector equations below intersect, are parallel, are skew, or are identi

cal.
r(t)=⟨1−t,3+2t,−3t⟩
s(t)=⟨2t,−3−4t,3+6t⟩

1 answer:

zzz [11.8K]10 days ago

6 0

The vectors r(t) and s(t) are indeed parallel.

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Find the value of y if EF = EG.

AnnZ [11884]

Answer:

55°

Step-by-step explanation:

The problem does not have a suitable diagram. Please check the attachment for the diagram.

Observing the diagram, it is evident that the triangle is isosceles, as it has two sides that are congruent (i.e., the same length). Since these sides are identical, the base angles are consequently also equal.

From the diagram, <EFG = <EGF

Since <EFG = 55°, it follows that <EGF = y = 55°

Thus, the value of y is established as 55°

5 0

14 days ago

The photo shows a small coin. The scale from the actual coin to the photo is

PIT_PIT [11868]

8mm correlates to 2cm just as 8mm aligns with 20mm

The ratio of 8: 20 simplifies down to 2: 5

3.25cm ÷ 5 equals 0.65cm

Multiplying 0.65 by 2 results in 1.3cm

Your answer amounts to 1.3cm

6 0

1 month ago

How much profit was generated by the sales of gold label and black label combined?

lawyer [12081]

The profit produced by combining sales of gold label and black label equals the total of their individual profits. From the table $\begin{array}{ccc}&\text{Number of bottles sold}&\text{Average profit (per bottle)}\\\text{Gold label}&1,000&\$2.75\end{array}$ you find the gold label's profit $1,000\cdot \$2.75=\$2,750.$ and similarly from another table $\begin{array}{ccc}&\text{Number of bottles sold}&\text{Average profit (per bottle)}\\\text{Black label}&2,000&\$1.50\end{array}$ find the black label's profit $2,000\cdot \$1.50=\$3,000.$ . Adding these yields $\$2,750+\$3,000=\$5,750.$ . Thus, option 5 is the correct choice.

8 0

1 month ago

Read 2 more answers

A rectangular schoolyard is to be fenced in using the wall of the school for one side and 150

lawyer [12081]

Answer:

Part 1) The equation is $A(x)=150x-2x^2$

Part 2) When x=40 m, the area of the schoolyard is A=2,800 m^2

Part 3) The valid domain consists of all real numbers exceeding zero and below 75 meters

Step-by-step explanation:

Part 1) Formulate an expression for A(x)

Let

x -----> the length of the rectangular school yard

y ---> the width of the rectangular school yard

It is known that

The perimeter for the fencing (taking the school wall as one side) is

$P=2x+y$

$P=150\ m$

thus

$150=2x+y$

$y=150-2x$ -----> this is equation A

The area of the rectangular school yard is

$A=xy$ ----> this is equation B

Substituting equation A into equation B yields

$A=x(150-2x)$

$A=150x-2x^2$

Change to function notation

$A(x)=150x-2x^2$

Part 2) What is the area when x=40?

With x equal to 40 m

substitute the value into the expression from Part 1 to determine A

$A(40)=150(40)-2(40^2)$

$A(40)=2,800\ m^2$

Part 3) What would be a suitable domain for A(x) in this scenario?

We understand that

A signifies the area of the rectangular school yard

x characterizes the length of the rectangular school yard

It follows that

$A(x)=150x-2x^2$

This forms a vertical parabola opening downwards

The vertex indicates a maximum point

The x-coordinate of the vertex corresponds to the length that maximizes the area

The y-coordinate of the vertex denotes the maximum area

The vertex corresponds to (37.5, 2812.5)

Refer to the accompanying figure

Consequently,

The peak area achieved is 2,812.5 m^2

The x-intercepts are located at x=0 m and x=75 m

The domain for A is the range -----> (0, 75)

All real numbers greater than zero and less than 75 meters

5 0

1 month ago

1 kilogram is equivalent to about 2.2 pounds. An orangutan weighs 142 pounds. What is its mass in kilograms? Round your answer t

AnnZ [11884]

The weight of an orangutan is 64.5 kilograms

Solution:

It is stated that the orangutan's weight is 142 pounds

Additionally,

1 kilogram is approximately 2.2 pounds

We will now convert 142 pounds into kilograms

1 kilogram = 2.2 pounds

$\text{1 pound } = \frac{1}{2.2} \text{ kilograms }$

As a result, 142 pounds translates to

$142 \text{ pound } = \frac{1}{2.2} \times 142 \text{ kilogrmas }$

$\rightarrow \frac{1}{2.2} \times 142 = 64.545 \text{ kilograms }$

When rounding to the nearest tenth, we obtain 64.5 kilograms

Therefore the mass of the orangutan is 64.5 kilograms

3 0

18 days ago