Ask question

Ask question

All categories

3 years ago

13

the results of a swim meet are shown. who swam the fastest. tawny 50m. 40.8 seconds. pepita. 100m 60.2 seconds or Susana 200m 11

2.4 seconds

1 answer:

Tema [17]3 years ago

3 0

Susana swam the fastest at 112.4 seconds, if peptic swam the 200m her time would be 120.4

You might be interested in

SHOW YOUR WORK<br><br> Can someone help?<br> (no files or links)<br> (Will give brainliest)

DanielleElmas [232]

2 1/3
You would do 6 3/4- 1 3/4=5
5-1 1/3= 2 2/3

7 0

3 years ago

You have two circles, one with radius r and the other with radius R. You wish for the difference in the areas of these two circl

gtnhenbr [62]

Answer:

maximum difference of radii = $(r-R)=\frac{1}{2\pi ^{2}}$

Step-by-step explanation:

We know that area of circle is given by

$A=\pi \times (radius)^{2}$

For circle with radius 'r' we have

$A_{1}=\pi \times (r)^{2}$

For circle with radius 'R' we have

$A_{2}=\pi \times (R)^{2}$

Now according to given condition we have

$A_{1}-A_{2}\leq \frac{5}{\pi }$

$\Rightarrow \pi r^{2}-\pi R^{2}\leq \frac{5}{\pi }\\\\\Rightarrow (r^{2}-R^{2})\leq \frac{5}{\pi ^{2}}\\\\(r+R)(r-R)\leq \frac{5}{\pi ^{2}}\\\\\because (a^{2}-b^{2})=(a+b)(a-b)\\\\(r+R)=10(Given)\\\\\Rightarrow(r-R)\leq \frac{5}{10\pi ^{2}}\\\\\therefore (r-R)\leq\frac{1}{2\pi ^{2}}$

Thus maximum difference of radii = $(r-R)=\frac{1}{2\pi ^{2}}$

6 0

4 years ago

Pls help I just need the answer

inn [45]

That’s mainey me too

8 0

3 years ago

What is -(-E) in the picture shown below?

KIM [24]

Answer:

-(-E)=0.6 , I am not sure of my answer

6 0

3 years ago

20. 2x + 4(x - 1) + 2

RideAnS [48]

Answer:

6x - 2

Step-by-step explanation:

2x + 4(x - 1) + 2

=2x + 4x - 4 + 2

=6x - 2

8 0

3 years ago

Read 2 more answers