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3 years ago

14

Together, teammates Pedro and Ricky 2681 base hits last season. Pedro had 285 more hits than Ricky. How many hits did each playe

r have

1 answer:

lara31 [8.8K]3 years ago

6 0

Pedro did 1,626 hits and ricky did 1,056. Can you please help me with my question?

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How does an earthquake of magnitude 7.9 compare in intensity with an earthquake of magnitude of 3.7 on the Richter scale? The l

Andre45 [30]

We are required to compare the magnitude for two earthquake intensity:
the larger intensity is 7.9
the smaller intensity is 3.7
thus the ratio of the intensity will be:
(larger intensity)/(smaller intensity)
=7.9/3.7
=2.135
~2
Thus we conclude that:
The larger earthquake's intensity was 2 times as great as the smaller earthquake's intensity.

5 0

3 years ago

A rectangular school banner has a length of 36 inches and a width of 52 inches. A sign is made that is similar to the school ban

zalisa [80]

Answer:

<u>4212 : 1573</u> is the ratio.

Step-by-step explanation:

Given:

Length of rectangular school banner is 36 inches and width is 52 inches.

And, the sign made is similar to banner.

The sign's length is 22 inches.

Now, we have to find the ratio of the area of the banner to the area of the sign.

So, we have dimensions of rectangular school banner:

Length = 36 inches.

Width = 52 inches.

But, we have only length of sign:

Length = 22 inches.

Now, we have to find the width of sign by using cross multiplication method:

Let the width of sign be $x.$

<em>As, sign is similar to banner.</em>

So, if 36 inches is equivalent to 52 inches.

Then, 22 inches is equivalent to $x.$

$\frac{36}{52} =\frac{22}{x}$

<em>By cross multiplying we get:</em>

<em /> $36x=1144$ <em />

<em>Dividing both sides by 36 we get:</em>

$x=31\frac{7}{9} .$

<em>Hence, the width of sign is </em> $31\frac{7}{9}$ <em> inches.</em>

Now, we find the area of banner and the area of sign by putting formula:

Area of banner = length × width.

$Area\ of\ banner=36\times 52$

$Area\ of\ banner=1872\ square\ inches.$

Now, area of sign:

$Area\ of\ sign=length\times width\\\\Area\ of\ sign=22\times 31\frac{7}{9} \\\\Area\ of\ sign=22\times \frac{286}{9} \\\\Area\ of\ sign=\frac{6292}{9} \ square\ inches.$

Now, to get the ratio of the area of the school banner to the area of the sign:

$1872:\frac{6292}{9}$

$=\frac{1872}{\frac{6292}{9} } \\\\=\frac{16848}{6292}$

<em>On simplifying we get:</em>

$=\frac{4212}{1573}$

$=4212:1573.$

Therefore, the ratio of the area of the school banner to the area of the sign is 4212:1573.

8 0

3 years ago

N divided Into 8 is what

Goryan [66]

Answer:

$\frac{N}{8}$

Dividing is the same as putting numbers into a fraction. This is N/8. We don't get the answer until we figure out what N is equal to.

7 0

3 years ago

If JK and LM are congruent and parallel, then JKL and MLK are congruent.

viktelen [127]

Answer:

Given statement is TRUE.

Step-by-step explanation:

Given that line segment JK and LM are parallel. From picture we see that LK is transversal line.

We know that corresponding angles formed by transversal line are congruent.

Hence ∠JKL = ∠ MLK ...(i)

Now consider triangles JKL and MLK

JK = LM {Given}

∠JKL = ∠ MLK { Using (i) }

KL = KL {common sides}

Hence by SAS property of congruency of triangles, ΔJKL and ΔMLK are congruent.

Hence given statement is TRUE.

3 0

3 years ago

What is the area of the regular trapezoid below?<br> - 6 cm<br> T<br> 8 cm<br> 10 cm

Elenna [48]

Answer:

Area = 64 cm²

Step-by-step explanation:

To find the area, we have to follow some steps:

=> <u>Step 1: Make the bases similar:</u>

Since we have 2 bases given 6 and 10 we have to come up with a third base by finding their average:

(6 + 10) / 2

(16) / 2

8

So now that we know our base,

=><u> Our second step is to multiply the base we found by the height of the trapezoid:</u>

Area = common base * height

Area = 8 * 8

Area = 64 cm²

Hope this helps!

5 0

3 years ago