All categories

3 years ago

A conversion factor can be written _________ different ways. How do you select the most appropriate conversion factor?

Mathematics

2 answers:

Komok [63]3 years ago

7 0

The blank answer is: In 2

DochEvi [55]3 years ago

5 0

Answer:

can be written two ways

Step-by-step explanation:

A conversion factor is written based on the equality between the two units

You might be interested in

10. You are buying meat for a cookout. You need to buy 8 packages of meat. A

AlladinOne [14]

Step 1. $5.19 + $1.89=____
Step 2. ___x 2 =___
Step 3. Keep multiplying until you get 31.62.

7 0

3 years ago

LA plane is trying to travel 250 miles at a bearing of 20° E of S, however, it ends 230 miles away from the

Semmy [17]

Answer:

The wind pushed the plane $65.01$ miles in the direction of $45.56 ^{\circ}$ East of North with respect to the destination point.

Step-by-step explanation:

Let origin, O, br the starting point and point D be the destination at 250 miles at a bearing of 20° E of S, but due to wind let D' be the actual position of the plane at 230 miles away from the starting point in the direction of 35° E of South as shown in the figure.

So, we have |OD|=250 miles and |OD'|=230 miles.

Vector $\overrightarrow{DD'}$ is the displacement vector of the plane pushed by the wind.

From figure, the magnitude of the required displacement vector is

$|DD'|=\sqrt{|AB|^2+|PQ|^2}\;\cdots(i)$

and the direction is $\alpha$ east of north as shown in the figure,

$\tan \alpha=\frac{|PQ|}{|AB|}\;\cdots(ii)$

From the figure,

$|AB|=|OA-OB|$

$\Rightarrow |AB|=|OD\cos 20 ^{\circ}-OD'\cos 35 ^{\circ}|$

$\Rightarrow |AB|=|250\cos 20 ^{\circ}-230\cos 35 ^{\circ}|$

$\Rightarrow |AB|=45.52$ miles

Again, $|PQ|=|OP-OQ|$

$\Rightarrow |PQ|=|OD\sin 20 ^{\circ}-OD'\sin 35 ^{\circ}|$

$\Rightarrow |PQ|=|250\sin 20 ^{\circ}-230\sin 35 ^{\circ}|$

$\Rightarrow |PQ|=46.42$ miles

Now, from equations (i) and (ii), we have

$|DD'|=\sqrt{|45.52|^2+|46.42|^2}=65.01$ miles, and

$\tan \alpha=\frac{|46.42|}{|45.52|}$

$\alpha=\tan^{-1}\left(\frac{|46.42|}{|45.52|}\right)=45.56 ^{\circ}$

Hence, the wind pushed the plane $65.01$ miles in the direction of $45.56 ^{\circ}$ E astof North with respect to the destination point.

5 0

2 years ago

a cone has a radius of 6cm and height 9 calculate a)the volume and curved surface area b) total surface area

Orlov [11]

Answer:

V = 339.3 cm³; L = 203.9 cm²; A = 317.0 cm²

Step-by-step explanation:

a) Volume

The formula for the volume (V) of a cone is

V = ⅓πr²h

V = ⅓π × 6² × 9

V = ⅓π × 36 × 9

V = 108π cm³

V ≈ 339.3 cm³

=====

Curved surface area

The formula for the lateral surface area (L) of a cone is

L = πr√(r² + h²)

L = π×6√(36 + 81)

L = 6π√[9(4 + 9)]

L = 6π√(9 × 13)

L = 18π√13 cm²

L ≈ 203.9 cm²

===============

b) Base surface area

The base is a circle, so the formula for base surface area (B) is

B = πr²

B = π×6²

B = 36π cm²

B ≈ 113.1 cm²

=====

Total surface area

A = L + B

A = 18π√(13 +36π)

A = 18π(2 + √13) cm²

A ≈ 203.9 + 113.1

A ≈ 317.0 cm²

3 0

3 years ago

A student solves a quadratic and gets solutions of x=-7 and x=8. Could the discriminant of this equation equal -188?

MrMuchimi

Answer:

Step-by-step explanation:

8 0

3 years ago

Three students, Alicia, Benjamin, and Caleb, are constructing a square inscribed in a circle with center at point C. Alicia draw

True [87]

Benjamin is correct about the diameter being perpendicular to each other and the points connected around the circle.

<h3>Inscribing a square</h3>

The steps involved in inscribing a square in a circle include;

A diameter of the circle is drawn.

A perpendicular bisector of the diameter is drawn using the method described as the perpendicular of the line sector. Also known as the diameter of the circle.

The resulting four points on the circle are the vertices of the inscribed square.

Alicia deductions were;

Draws two diameters and connects the points where the diameters intersect the circle, in order, around the circle

Benjamin's deductions;

The diameters must be perpendicular to each other. Then connect the points, in order, around the circle

Caleb's deduction;

No need to draw the second diameter. A triangle when inscribed in a semicircle is a right triangle, forms semicircles, one in each semicircle. Together the two triangles will make a square.

It can be concluded from their different postulations that Benjamin is correct because the diameter must be perpendicular to each other and the points connected around the circle to form a square.

Thus, Benjamin is correct about the diameter being perpendicular to each other and the points connected around the circle.

Learn more about an inscribed square here:

brainly.com/question/2458205

#SPJ1

6 0

2 years ago