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____ [38]
3 years ago
11

How do i do this, I'm really really really really really really really really really really really really really really really c

onfused

Mathematics
2 answers:
loris [4]3 years ago
8 0
I told this to someone else also...

The best thing ever is Photomath! Go to the app store on your phone, tablet, etc. and download Photomath. It literally does all your math for you and even shows the steps!! I use it all the time. :)
zimovet [89]3 years ago
5 0
Here you go hope it helps

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How do you solve the problem 94 - (5 x 6) ÷ 6 x 11
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3 years ago
Write a definite integral that represents the area of the region. (Do not evaluate the integral.) y1 = x2 + 2x + 3 y2 = 2x + 12F
Svet_ta [14]

Answer:

A = \int\limits^3__-3}{9}-{x^{2}} \, dx = 36

Step-by-step explanation:

The equations are:

y = x^{2} + 2x + 3

y = 2x + 12

The two graphs intersect when:

x^{2} + 2x + 3 = 2x + 12

x^{2} = 0

x_{1}  = 3\\x_{2}  = -3

To find the area under the curve for the first equation:

A_{1} = \int\limits^3__-3}{x^{2} + 2x + 3} \, dx

To find the area under the curve for the second equation:

A_{2} = \int\limits^3__-3}{2x + 12} \, dx

To find the total area:

A = A_{2} -A_{1} = \int\limits^3__-3}{2x + 12} \, dx -\int\limits^3__-3}{x^{2} + 2x + 3} \, dx

Simplifying the equation:

A = \int\limits^3__-3}{2x + 12}-({x^{2} + 2x + 3}) \, dx = \int\limits^3__-3}{9}-{x^{2}} \, dx

Note: The reason the area is equal to the area two minus area one is that the line, area 2, is above the region of interest (see image).  

3 0
3 years ago
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