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Alisiya [41]
2 years ago
8

I need helppp pleasee

Mathematics
1 answer:
gogolik [260]2 years ago
8 0
50p=c
So $50 x the number of people = the price in dollars
So for 11 people it would cost $550
I hope this helped :)
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Please answer correctly !!!!!! Will mark brainliest !!!!!!!!!!!!
djverab [1.8K]

Answer:

x+5

Step-by-step explanation:

Side is equal to sqrt area

= sqrt x^2 +10x + 25

= x+5

So one side of the square is x+5 metres

3 0
3 years ago
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HELP PLEASE 18 POINTS
ICE Princess25 [194]

Answer:

Step-by-step explanation:

line 1  m=y2-y1 / x2-x1  [formula for slope ]

m= 0 - (-3) / 8-(-4)

m= 3 /12      

m= 1/4  slope of line 1

line2  

m = 2-6 / 0-(-1)

m = -4 / 1

m = -4  slope of line 2

line 3

m = -4 - (-7)  / -3 -  6

m = 3 / -9

m = - 1/3  slope of line 3

lines 1 & 2   perpendicular b/c they are reciprocals with a sign change :)

lines 1 & 3     neither.. differet slopes and not reciprocals

lines 2 & 3  perpendicular  b/c they are reciprocals with a sign change :)

6 0
3 years ago
Factor the GCF: 12a3b + 8a2b2 − 20ab3
Harrizon [31]
<span>12a^3b + 8a^2b^2 − 20ab^3

</span>12a^3b = 4ab(3a^2)
8a^2b^2 = 4ab(2ab)
20ab^3 = 4ab(5b^2)

GCF = 4ab

12a3b + 8a2b2 − 20ab3 = 4ab(3a^2 + 2ab - 5b^2)
6 0
3 years ago
Read 2 more answers
Exam
MrRissso [65]

Answer:

B. 1/9 cuz it's just saying y is 1/9 of x

3 0
3 years ago
Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n.
Otrada [13]

I guess the "5" is supposed to represent the integral sign?

I=\displaystyle\int_1^4\ln t\,\mathrm dt

With n=10 subintervals, we split up the domain of integration as

[1, 13/10], [13/10, 8/5], [8/5, 19/10], ... , [37/10, 4]

For each rule, it will help to have a sequence that determines the end points of each subinterval. This is easily, since they form arithmetic sequences. Left endpoints are generated according to

\ell_i=1+\dfrac{3(i-1)}{10}

and right endpoints are given by

r_i=1+\dfrac{3i}{10}

where 1\le i\le10.

a. For the trapezoidal rule, we approximate the area under the curve over each subinterval with the area of a trapezoid with "height" equal to the length of each subinterval, \dfrac{4-1}{10}=\dfrac3{10}, and "bases" equal to the values of \ln t at both endpoints of each subinterval. The area of the trapezoid over the i-th subinterval is

\dfrac{\ln\ell_i+\ln r_i}2\dfrac3{10}=\dfrac3{20}\ln(ell_ir_i)

Then the integral is approximately

I\approx\displaystyle\sum_{i=1}^{10}\frac3{20}\ln(\ell_ir_i)\approx\boxed{2.540}

b. For the midpoint rule, we take the rectangle over each subinterval with base length equal to the length of each subinterval and height equal to the value of \ln t at the average of the subinterval's endpoints, \dfrac{\ell_i+r_i}2. The area of the rectangle over the i-th subinterval is then

\ln\left(\dfrac{\ell_i+r_i}2\right)\dfrac3{10}

so the integral is approximately

I\approx\displaystyle\sum_{i=1}^{10}\frac3{10}\ln\left(\dfrac{\ell_i+r_i}2\right)\approx\boxed{2.548}

c. For Simpson's rule, we find a quadratic interpolation of \ln t over each subinterval given by

P(t_i)=\ln\ell_i\dfrac{(t-m_i)(t-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+\ln m_i\dfrac{(t-\ell_i)(t-r_i)}{(m_i-\ell_i)(m_i-r_i)}+\ln r_i\dfrac{(t-\ell_i)(t-m_i)}{(r_i-\ell_i)(r_i-m_i)}

where m_i is the midpoint of the i-th subinterval,

m_i=\dfrac{\ell_i+r_i}2

Then the integral I is equal to the sum of the integrals of each interpolation over the corresponding i-th subinterval.

I\approx\displaystyle\sum_{i=1}^{10}\int_{\ell_i}^{r_i}P(t_i)\,\mathrm dt

It's easy to show that

\displaystyle\int_{\ell_i}^{r_i}P(t_i)\,\mathrm dt=\frac{r_i-\ell_i}6(\ln\ell_i+4\ln m_i+\ln r_i)

so that the value of the overall integral is approximately

I\approx\displaystyle\sum_{i=1}^{10}\frac{r_i-\ell_i}6(\ln\ell_i+4\ln m_i+\ln r_i)\approx\boxed{2.545}

4 0
3 years ago
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