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GREYUIT [131]
3 years ago
6

Muriel says she has written a system of two linear equations that has an infinite number of solutions. One of the equations of t

he system is 3y = 2x – 9. Which could be the other equation?
Mathematics
1 answer:
Assoli18 [71]3 years ago
3 0

Answer: B) y= 2/3x - 3

Step-by-step explanation:

On edge

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Kim has small wooden cube shaped blocks. To make the next size cube, she needs 8 blocks. Write a sequence to show the number of
ipn [44]
The situation describes a geometric sequence with first term as 1, and the common ratio of 8.

1st term = 1
2nd term = 1 x 8 = 8
3rd term = 8 x 8 = 64
4th term = 64 x 8 = 512
5th term = 512 x 8 = 4,096
6th term = 4,096 x 8 = 32,768
5 0
3 years ago
A) how many 15-year old students took part in the survey?
trapecia [35]

Answer: 60 or something near that sorry if wrong

Step-by-step explanation:

8 0
3 years ago
What value of k makes the trinomial x^2- 10x + k a perfect square?
Advocard [28]

Answer: 25

Step-by-step explanation:     (x-a)² = x² - 2ax + a²

  From x² - 10x + k  we deduce that 10x = 2·5x  and a = 5

Then k = 5²2

3 0
3 years ago
Find the height of a parallelogram with base 6.75m and an area of 218.72 square meters
oksian1 [2.3K]

Answer:

about 32.4 meters

Step-by-step explanation:

The area of a parallelogram is the equation A=Base times Height.

Put in the known values 6.75m and 218.72meters squared into the equation.

It becomes the total area 218.72 = 6.75(h)

To get h or height by itself, you have to divide both sides by 6.75. You will get the answer 32.402, or you can round it to 32.4 meters.

7 0
3 years ago
Evaluate the definite integral. <br> 1 x4(1 + 2x5)5 dx.
Nata [24]

Answer:

Step-by-step explanation:

Given the definite integral \int\limits {\dfrac{x^4}{(1-2x^5)^5} } \, dx, we to evaluate it. Using integration by substitution method.

Let u = 1-2x⁵ ...1

du/dx = -10x⁴

dx = du/-10x⁴.... 2

Substitute equation 1 and 2 into the integral function and evaluate the resulting integral as shown;

= \int\limits {\dfrac{x^4}{u^5} } \, \dfrac{du}{-10x^4}

= \dfrac{-1}{10} \int\limits {\dfrac{du}{u^5} }  \\\\= \dfrac{-1}{10} \int\limits {{u^{-5}du }  \\= \dfrac{-1}{10} [{\frac{u^{-5+1}}{-5+1}]  \\\\= \dfrac{-1}{10} ({\frac{u^{-4}}{-4})\\\\

= \dfrac{u^{-4}}{40} \\\\\\= \dfrac{1}{40u^4} +C

substitute u = 1-2x⁵ into the result

= \dfrac{1}{40(1-2x^5)^4} +C

Hence\int\limits {\dfrac{x^4}{(1-2x^5)^5} } \, dx = \dfrac{1}{40(1-2x^5)^4} +C

5 0
3 years ago
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