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Alex
3 years ago
12

What is the length of AA' , rounded to the nearest hundredth? Type a numerical answer is the space provided. Do not type spaces

in your answer

Mathematics
1 answer:
inna [77]3 years ago
3 0
Coordinate of A are (0,0)
Coordinates of A' are (5,2)

We can find the distance from A to A' using the distance formula:

AA'= \sqrt{(5-0)^{2}+ (2-0)^{2}  } \\  \\ 
AA'= \sqrt{25+4} \\  \\ 
AA'= \sqrt{29} \\  \\ 
AA'=5.39

Thus, rounded to nearest hundredth, AA' is equal to 5.39
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If you add​ natalie's age and​ fred's age, the result is 43. if you add​ fred's age to 44 times​ natalie's age, the result is 88
Pavel [41]
Let Natalie be x, and Fred, y.
So based on 1st condition, we get
x + y =43         - Equation 1   
â´ x = 43-y - Equation 2
Now based on 2nd condition, we get
44x+y=88
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3 years ago
Please help with this Calculus questions
Triss [41]

Answer:

\int_{0}^{1}\left( - \frac{3 u^{9}}{2} + \frac{2 u^{4}}{5} + 2 \right)du=\frac{193}{100}=1.93.

Step-by-step explanation:

To find \int_{0}^{1}\left( - \frac{3 u^{9}}{2} + \frac{2 u^{4}}{5} + 2 \right)du.

First, calculate the corresponding indefinite integral:

Integrate term by term:

\int{\left(- \frac{3 u^{9}}{2} + \frac{2 u^{4}}{5} + 2\right)d u}} =\int{2 d u} + \int{\frac{2 u^{4}}{5} d u} - \int{\frac{3 u^{9}}{2} d u}

Apply the constant rule \int c\, du = c u

\int{2 d u}} + \int{\frac{2 u^{4}}{5} d u} - \int{\frac{3 u^{9}}{2} d u} = {\left(2 u\right)} + \int{\frac{2 u^{4}}{5} d u} - \int{\frac{3 u^{9}}{2} d u}

Apply the constant multiple rule \int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du

2 u - {\int{\frac{3 u^{9}}{2} d u}} + \int{\frac{2 u^{4}}{5} d u} = 2 u - {\left(\frac{3}{2} \int{u^{9} d u}\right)} + \left(\frac{2}{5} \int{u^{4} d u}\right)

Apply the power rule \int u^{n}\, du = \frac{u^{n + 1}}{n + 1}

2 u - \frac{3}{2} {\int{u^{9} d u}} + \frac{2}{5} {\int{u^{4} d u}}=2 u - \frac{3}{2} {\frac{u^{1 + 9}}{1 + 9}}+ \frac{2}{5}{\frac{u^{1 + 4}}{1 + 4}}

Therefore,

\int{\left(- \frac{3 u^{9}}{2} + \frac{2 u^{4}}{5} + 2\right)d u} = - \frac{3 u^{10}}{20} + \frac{2 u^{5}}{25} + 2 u = \frac{u}{100} \left(- 15 u^{9} + 8 u^{4} + 200\right)

According to the Fundamental Theorem of Calculus, \int_a^b F(x) dx=f(b)-f(a), so just evaluate the integral at the endpoints, and that's the answer.

\left(\frac{u}{100} \left(- 15 u^{9} + 8 u^{4} + 200\right)\right)|_{\left(u=1\right)}=\frac{193}{100}

\left(\frac{u}{100} \left(- 15 u^{9} + 8 u^{4} + 200\right)\right)|_{\left(u=0\right)}=0

\int_{0}^{1}\left( - \frac{3 u^{9}}{2} + \frac{2 u^{4}}{5} + 2 \right)du=\left(\frac{u}{100} \left(- 15 u^{9} + 8 u^{4} + 200\right)\right)|_{\left(u=1\right)}-\left(\frac{u}{100} \left(- 15 u^{9} + 8 u^{4} + 200\right)\right)|_{\left(u=0\right)}=\frac{193}{100}

6 0
3 years ago
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