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andreyandreev [35.5K]
3 years ago
14

Brainliest and 30 points please help

Mathematics
1 answer:
kolbaska11 [484]3 years ago
4 0

Answer:

22. Perimeter = 52 units

     Area = 160 Square Units

23. Notation form :  2 \times 10^{5}

     Standard Form : 200000

Step-by-step explanation:

22.

The formula for perimeter is

P = 2 (length + Width )

  = 2 ( 2x+3x+1)

  =  2(5x+1)

The Formula for Area of a rectangle is

A= length x width

A= 2x \times (3x+1)

A=6x^2+2x

Now we have to find the Perimeter and Area for x = 5

P = 2(5 x 5 +1)

P= 2(26)

P=52 units

A= 6 x 5 x 5 + 2 x 5

A= 150 + 10

A= 160 square units

23.

A. By using rule of exponents , we can determine that both the results will be same.

B.

Miriam's calculation

\frac{2.8 \times 10^{9}}{1.4 \times 10^{4}}

\frac{2.8}{1.4} \times 10^{9-4}

2 \times 10^5

Priya's calculation

\frac{2.8 \times 10^{2}}{1.4 \times 10^{-3}}

\frac{2.8}{1.4} \times 10^{2-(-3)}

2 \times 10^5

Standard notation

2 \times 10^5 = 200000

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3 years ago
Use the fundamental theorem of calculus to find the area of the region between the graph of the function x^5 + 8x^4 + 2x^2 + 5x
BaLLatris [955]

Answer:

The area of the region is 25,351 units^2.

Step-by-step explanation:

The Fundamental Theorem of Calculus:<em> if </em>f<em> is a continuous function on </em>[a,b]<em>, then</em>

                                   \int_{a}^{b} f(x)dx = F(b) - F(a) = F(x) |  {_a^b}

where F is an antiderivative of f.

A function F is an antiderivative of the function f if

                                                    F^{'}(x)=f(x)

The theorem relates differential and integral calculus, and tells us how we can find the area under a curve using antidifferentiation.

To find the area of the region between the graph of the function x^5 + 8x^4 + 2x^2 + 5x + 15 and the x-axis on the interval [-6, 6] you must:

Apply the Fundamental Theorem of Calculus

\int _{-6}^6(x^5+8x^4+2x^2+5x+15)dx

\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\\\\int _{-6}^6x^5dx+\int _{-6}^68x^4dx+\int _{-6}^62x^2dx+\int _{-6}^65xdx+\int _{-6}^615dx

\int _{-6}^6x^5dx=0\\\\\int _{-6}^68x^4dx=\frac{124416}{5}\\\\\int _{-6}^62x^2dx=288\\\\\int _{-6}^65xdx=0\\\\\int _{-6}^615dx=180\\\\0+\frac{124416}{5}+288+0+18\\\\\frac{126756}{5}\approx 25351.2

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