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2 years ago

8

Javier is a graphic designer. He is asked to add a triangular shaped piece of art to a magazine cover. The requirement for the t

riangle includes sides with lengths 3 cm, 5 cm, and 9 cm. Can Javier draw a triangle with these dimensions? Explain how you know.

2 answers:

Studentka2010 [4]2 years ago

6 0

Answer:

11

Step-by-step explanation:

5x5+1=11

yarga [219]2 years ago

6 0

Answer:

Step-by-step explanation: the ghetto

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Application of Geometric Series

Gnoma [55]

Answer:

7.2224

Step-by-step explanation:

The value of the summation is given by the formula ...

Sn = (a1)(1 -r^n)/(1 -r) . . . . . where a1 is the first of n terms, and r is the common ratio.

Your sum has first term and ratio ...

a1 = 2(0.6) . . . . . summation term for n=1

r = 0.6

So the sum is ...

$\displaystyle \sum_{n=1}^4{2(0.6^n)}=2(0.6)\dfrac{1-0.6^4}{1-0.6}=\dfrac{1.2}{0.4}(0.8704) = 2.6112$

Then the value of the entire given expression is ...

$\displaystyle 2+2\sum_{n=1}^4{}2(0.6^n)=2+2(2.6112)=\boxed{7.2224}$

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A calculator can help you find the value.

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1 year ago

Verilenlerden<br>hangisi birim<br>kesirdir?<br>2/5<br>B<br>5/8<br>2/3<br>D<br>

o-na [289]

Answer:

2/3 ..................

4 0

2 years ago

The U.S. Energy Department states that 60% of all U.S. households have ceiling fans. In addition, 29% of all U.S. households hav

hram777 [196]

Answer:

Very sorry if this is no help

Step-by-step explanation:

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1 year ago

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis. Verify y

mariarad [96]

Answer:

the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis is;

$\frac{\pi }{2} [e^2 - 1 ]$ or 10.036

Step-by-step explanation:

Given the data in the question;

y = $y = e^{(x - 1 )$ , y = 0, x = 1, x = 2.

Now, using the integration capabilities of a graphing utility

y = $y = e_2}^{(x - 1 )_$ , y = 0

Volume = $\pi \int\limits^2_1 ( e^{x-1)^2} - (0)^2 dx$

Volume = $\pi \int\limits^2_1 ( e^{x-1)^2 dx$

Volume = $\pi \int\limits^2_1 e^{2x-2}dx$

Volume = $\frac{\pi }{e^2} \int\limits^2_1 e^{2x}dx$

Volume = $\frac{\pi }{e^2} [\frac{e^{2x}}{2}]^2_1$

Volume = $\frac{\pi }{2e^2} [e^4 - e^2 ]$

Volume = $\frac{\pi }{2} [e^2 - 1 ]$ or 10.036

Therefore, the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis is;

$\frac{\pi }{2} [e^2 - 1 ]$ or 10.036

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2 years ago

A 13 L solution that was 12% vinegar was mixed with a 9 L solution that was 70% vinegar. What is the concentration of the new mi

kolbaska11 [484]

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Step-by-step explanation:

BIG BRAIN...

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2 years ago