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2 years ago

6

Please I need help I don’t understand this!

1 answer:

Alex_Xolod [135]2 years ago

8 0

When reflected across the y axis, the x coordinate becomes the opposite but the y coordinate stays the same, so N is (2, -5)
when reflected across the x axis, the y coordinate becomes the opposite but the x coordinate stays the same, so R is (2, 5)

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The sum of three consecutive integers is 84. What are the three numbers?

Mashcka [7]

Answer:

they are 27, 28, and 29.

Step-by-step explanation:

7 0

2 years ago

Simple Algebra problem. Includes the x and y axis. Also need help finding the perimeter.

Leona [35]

Answer:

31.5cm²

Step-by-step explanation:

area of a triangle =1/2 b h

b= 7cm

h=9cm

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= 1/2×63

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c²=a²+b²

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3 0

3 years ago

10. The number of people, N, employed in a chain of cafes is related to the number of cafes, n, by the equation:

il63 [147K]

Answer:

b.

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= 10*14+120

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Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

1. -8(-2x+6)+(-10)<br> 2.9(x-8)-17x<br> show work please

Damm [24]

1)-8(-2x+6)+(-10)
16x-48-10
16x-50
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3 0

2 years ago

Read 2 more answers

Use the Midpoint Rule with n = 5 to estimate the volume V obtained by rotating about the y-axis the region under the curve y = 1

velikii [3]

Using the shell method, the volume is given exactly by the definite integral,

$2\pi\displaystyle\int_0^1x(1+9x^3)\,\mathrm dx$

Splitting up the interval [0, 1] into 5 subintervals gives the partition,

[0, 1/5], [1/5, 2/5], [2/5, 3/5], [3/5, 4/5], [4/5, 5]

with left and right endpoints, respectively, for the $i$ -th subinterval

$\ell_i=\dfrac{i-1}5$

$r_i=\dfrac i5$

where $1\le i\le5$ . The midpoint of each subinterval is

$m_i=\dfrac{\ell_i+r_i}2=\dfrac{2i-1}{10}$

Then the Riemann sum approximating the integral above is

$2\pi\displaystyle\sum_{i=1}^5m_i(1+9{m_i}^3)\frac{1-0}5$

$\dfrac{2\pi}5\displaystyle\sum_{i=1}^5\left(\frac{2i-1}{10}+9\left(\frac{2i-1}{10}\right)^4\right)$

$\dfrac{2\pi}{5\cdot10^4}\displaystyle\sum_{i=1}^5\left(16i^4-32i^3+24i^2+1992i-999\right)=\frac{112,021\pi}{25,000}\approx\boxed{14.08}$

(compare to the actual value of the integral of about 14.45)

3 0

3 years ago