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DaniilM [7]
4 years ago
7

Marc can run 2 miles in 16 minutes. At this rate, how long will it take Marc to run 5 miles?

Mathematics
2 answers:
tatyana61 [14]4 years ago
6 0
C 40 mins ................
skad [1K]4 years ago
3 0
8= 1 minute so you do 8x5 because he runs 5 miles and your answer will be C=40 minutes. Hope this help :)
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If Tucsos average rainfall is 12 1/4 inches and yumas is 3 3/5 inches, how much more rain, on the average, does Tucson get than
Taya2010 [7]
12 1/4 - 3 3/5 =
49/4 - 18/5 =
245/20 - 72/20 =
173/20 = 
8 13/20
5 0
3 years ago
I AM CALLING ALL ACES, EXPERTS, SMARTY PANTS, ETC!!! :)) (MATH 50 POINTS)
8_murik_8 [283]

10. The volume of the triangular prism is V=A_{base}\cdot H=\dfrac{1}{2} height \cdot base\cdot H, then

V=\dfrac{1}{2} \cdot 4\cdot 3.5\cdot 3.3=23.1.

11. The volume of the triangular pyramid is V=\dfrac{1}{3}A_{base}\cdot H=\dfrac{1}{3}\cdot \dfrac{1}{2} height \cdot base\cdot H, then

V=\dfrac{1}{3}\cdot \dfrac{1}{2} \cdot 4\cdot 3.5\cdot 3.3=7.7.

12. They have common base (common triangle). y=7.7, r=23.1, then r:y=23.1:7.7=3.

13. See 10 and 11 for notations in symbols.

6 0
3 years ago
A garden store has the following miscellaneous bulbs in a basket:
ArbitrLikvidat [17]
Okay. Since we know that the customer before them bought one of each type of bulb, we can subtract one bulb from each type so now, here is what the store has:

5 amaryllis

6 daffodils

3 lilies

2 tulips

Next, let’s find the probability of the customer picking an amaryllis.

We know there are 5 amaryllis, so we can put that as the numerator.

Next, we have to add up all the flowers to get the denominator.

5+6+3+2

I got 16.

So now we have 5/16

So, that is our answer! 5/16, or as a decimal 0.3125, or as a percentage 31.25%.

Hope this helps! Comment if you have any questions! :)
4 0
3 years ago
When h has value 4 calculate 5h -3
Lapatulllka [165]
5x4-3
5x4=20
so 20-3 = 17
your answer is 17.
You replace h with 4, then multiply 4 by 5 and then deduct 3 from the product.
5 0
3 years ago
How would you verify the trigonometric identity 2x = 2sinx cosx
dem82 [27]

Step-by-step explanation:

<em>Hi</em><em>,</em><em> </em><em>there</em><em>!</em><em>!</em>

<em>I</em><em> </em><em>hope</em><em> </em><em>you</em><em> </em><em>mean</em><em> </em><em>sin2x</em><em>=</em><em>2</em><em>sinx</em><em> </em><em>.</em><em> </em><em>cosx</em>

<em>so</em><em>,</em><em> </em><em>let's</em><em> </em><em>begin</em><em> </em><em>in</em><em> </em><em>a</em><em> </em><em>simple</em><em> </em><em>way</em><em>;</em><em> </em><em>by</em><em> </em><em>adding</em><em>,</em><em> </em><em>alright</em><em>:</em>

<em>sin2x</em><em>=</em><em> </em><em>sin</em><em>(</em><em>x</em><em>+</em><em>x</em><em>)</em><em> </em><em> </em><em> </em><em> </em><em> </em><em>(</em><em>as</em><em> </em><em>2</em><em>x</em><em>=</em><em>x</em><em>+</em><em>x</em><em>)</em><em>.</em>

<em>now</em><em>,</em><em> </em><em>let's</em><em> </em><em>use</em><em> </em><em>compound</em><em> </em><em>formula</em><em> </em><em>for</em><em> </em><em>sin</em><em>,</em>

<em>so</em><em>,</em><em> </em><em>we</em><em> </em><em>get</em><em>:</em>

<em>sin</em><em> </em><em>(</em><em>x</em><em>+</em><em>x</em><em>)</em><em>=</em><em> </em><em>sinx.cosx</em><em> </em><em>+</em><em> </em><em>cosx</em><em>.</em><em> </em><em>sinx</em><em> </em><em> </em><em> </em><em>(</em><em>as</em><em> </em><em>sin</em><em>(</em><em>A</em><em>+</em><em>B</em><em>)</em><em>=</em><em>sin</em><em> </em><em>A</em><em>.</em><em> </em><em>cosB</em><em> </em><em>+</em><em> </em><em>cosA</em><em> </em><em>.</em><em> </em><em>sinB</em><em>)</em>

<em>or</em><em>,</em><em> </em><em>sin</em><em> </em><em>(</em><em>x</em><em>+</em><em>x</em><em>)</em><em>=</em><em>2</em><em>sinx</em><em>.</em><em>cosx</em><em> </em><em> </em><em>(</em><em>adding</em><em> </em><em>both</em><em>)</em><em>.</em>

<em>Therefore</em><em>, </em><em> </em><em>sin</em><em> </em><em>2</em><em>x</em><em> </em><em>=</em><em> </em><em>2sinx</em><em>.</em><em> </em><em>cosx</em><em>.</em>

<em><u>Hope</u></em><em><u> </u></em><em><u>it helps</u></em><em><u>.</u></em><em><u>.</u></em>

5 0
3 years ago
Read 2 more answers
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