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3 years ago

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Televisions and computer screens are usually advertised based on the length of their diagonals. If the height of a computer scre

en is 7 in. and the width is 9 in., what is the length of the diagonal? Round to the nearest inch.

1 answer:

iren2701 [21]3 years ago

7 0

Would it not be 63 square inches?

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One batch of cookies makes 18. How many

Viktor [21]

Answer:

6

Step-by-step explanation:

100/18=5 1/2

18*6=108

6 0

2 years ago

Read 2 more answers

What is the volume of a brick that is 20.3 cm long, 8.9 cm wide, and 5.7 cm high?

Anvisha [2.4K]

Answer:

1029.819

Step-by-step explanation:

20.3 * 8.9 * 5.7 = 1029.819

4 0

3 years ago

Read 2 more answers

A tetrahedron is a triangular pyramid with 4 equilateral faces. Han has a box of tetrahedrons with each of the 4 vertices marked

Daniel [21]

Theoretically, the number of the vertex comes at the top must be equal for infinitely many numbers of trials, but it's a tedious job to roll the dice for a large number of times. Rolling the tetrahedron only 10 times will not predict the correct result.

Han suspects that this tetrahedron is weighted in some way to make 1 appear more often than the other numbers.

So, check his suspect experimentally with the help of the concept of center of mass. A fair tetrahedron must have the center of mass at the center of the body.

The given tetrahedron has 4 equilateral triangular faces, so at first, measure all the sides of triangular faces and make sure that all are having the same length.

Then, conduct an experiment to check the location of the center of mass.

Hang the tetrahedron by attaching a thin string at the vertex 1 as shown in the figure and observe the base surface ( triangle 234) whether it is parallel to the ground or not.

Now, repeat the same by hanging it with vertices 2, 3, and 4 and observe the base surface.

For the center of mass to be located at the center of the tetrahedron, the base must be parallel to the ground (flat ground).i.e the hanging string must be perpendicular to the base for all the four cases.

If this is so, then the tetrahedron is a fair tetrahedron otherwise not.

8 0

3 years ago

Quadrilateral KLMN is a rectangle. The coordinates of L are L(1,-4), and the coordinates of Mare M(3,-2). Find the slopes of sid

lozanna [386]

L(1, -4)=(xL, yL)→xL=1, yL=-4
M(3, -2)=(xM, yM)→xM=3, yM=-2

Slope of side LM: m LM = (yM-yL) / (xM-xL)
m LM = ( -2 - (-4) ) / (3-1)
m LM = ( -2+4) / (2)
m LM = (2) / (2)
m LM = 1

The quadrilateral is the rectangle KLMN
The oposite sides are: LM with NK, and KL with NK
In a rectangle the opposite sides are parallel, and parallel lines have the same slope, then:
Slope of side LM = m LM = 1 = m NK = Slope of side NK
Slope of side NK = m NK = 1

Slope of side KL = m KL = m MN = Slope of side MN

The sides KL and LM (consecutive sides) are perpendicular (form an angle of 90°), then the product of their slopes is equal to -1:
(m KL) (m LM) = -1
Replacing m LM = 1
(m KL) (1) = -1
m KL = -1 = m MN

Answer:
Slope of side LM =1
Slope of side NK =1
Slope of side KL = -1
Slope of side MN = -1

3 0

3 years ago

Both the La Plata river dolphin (Pontoporia blainvillei) and

Citrus2011 [14]

Answer:

<em>1. A sperm whale is </em><em>3</em><em> </em><em>orders of magnitude</em><em> heavier than a La Plata river dolphin; 2. The radius of the larger ball is </em><em>one (1) order of magnitude</em><em> bigger than the radius of the smaller ball; 3. The volume of the larger ball is </em><em>3 orders of magnitude</em><em> bigger than the volume of the smaller ball</em>.

Step-by-step explanation:

If we expressed a number as:

$\\ N = a * 10^{b}$ (1)

Where

$\\ \frac{1}{\sqrt{10}} \leq a < \sqrt{10}$ (2)

or

$\\ 1 \leq a < 10$ (3)

Then, <em>b</em> represents the <em>order of magnitude </em>of such a number (<em>Order of magnitude (2020), </em>in Wikipedia).

The order of magnitude can be defined as "...the smallest power of ten needed to represent a quantity" (Weisstein, Eric W. "Order of Magnitude". From MathWorld--A Wolfram Web Resource).

Having gathered all this information, we can proceed as follows:

<h3>First case</h3>

The<em> La Plata river dolphin</em> weighs between 30 and 50kg and the sperm whale weighs between 35,000 and 40,000kg.

Then, considering (1) and (3) to express the dolphin and whale's weight (since in this way the order of magnitude is the same as the exponent part in the <em>scientific notation</em>):

$\\ 30kg \leq Dolphin_{weight} \leq 50kg$

$\\ 3*10^{1}kg \leq Dolphin_{weight} \leq 5*10^{1}kg$

$\\ 35000kg \leq Whale_{weight} \leq 40000kg$

$\\ 3.5*10^{4}kg \leq Whale_{weight} \leq 4.0*10^{4}kg$

Since the range for the weights are in the same order of magnitude for both dolphin and whale (considering the definition above):

$\\ Dolphin_{weight} = 10^{1}\;(order\;of\;magnitude=1)$

$\\ Whale_{weight} = 10^{4}\;(order\;of\;magnitude=4)$

Then

$\\ \frac{Whale_{weight} = 10^{4}}{Dolphin_{weight} = 10^{1}}$

$\\ \frac{Whale_{weight}}{Dolphin_{weight}} = \frac{10^{4}}{10^{1}}$

$\\ \frac{Whale_{weight}}{Dolphin_{weight}} = 10^{4-1}$

$\\ \frac{Whale_{weight}}{Dolphin_{weight}} = 10^{3}$

Thus

<em>A sperm whale is </em><em>3</em><em> </em><em>orders of magnitude</em><em> heavier than a La Plata river dolphin.</em>

<h3>Second case</h3>

Following the same reasoning, we can conclude that <em>the radius of the larger ball is </em><em>one (1) order of magnitude</em><em> bigger than the radius of the smaller ball:</em>

$\\ \frac{Larger\;ball_{radius}}{Smaller\;ball_{radius}} = \frac{10^{1}}{10^{0}}$

$\\ \frac{Larger\;ball_{radius}}{Smaller\;ball_{radius}} = 10^{1-0} = 10^{1}$

<h3>Third case</h3>

For this case, we need to calculate <em>the volume of a sphere</em> for both radii (1cm and 10cm).

The volume of a sphere is

$\\ V_{sphere} = \frac{4}{3}*\pi*R^{3}$

Then, the volume of the <em>ball of radius 1cm</em> is:

$\\ V_{radius=1} = \frac{4}{3}*\pi*(1cm)^{3}$

$\\ V_{radius=1} \approx 4.19*10^{0}cm^{3}$

And, the volume of the <em>ball of radius 10cm</em> is:

$\\ V_{radius=10} = \frac{4}{3}*\pi*(10cm)^{3}$

$\\ V_{radius=10} \approx 4.19*10^{3}cm^{3}$

Thus

$\\ \frac{10^{3}}{10^{0}} = 10^{3}$

As a result, <em>the volume of the larger ball is </em><em>3 orders of magnitude</em><em> bigger than the volume of the smaller ball</em>.

4 0

3 years ago