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3 years ago

The last answer I got for this was not help full. Please help. Will give brainliest.

Mathematics

1 answer:

dimulka [17.4K]3 years ago

7 0

Suppose JJ = Jhalil , J = Joman

1 - JJ 860 J620

2 - JJ 870 J 670

3- JJ880 J 720

4- JJ890 J 770

5- JJ900 J 820

6- JJ910 J 870

7- JJ920 J 920 - Catched up

8- JJ 930 J 970

9- JJ940 J 1020

10- JJ 950 J 1070

11- JJ960 J 1120

12- JJ 970 J 1170

so a - week 7

b- Joman

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3 years ago

What is the least common multiple of the number 64, 16, 2, and 8?

Lina20 [59]

Answer:

Step-by-step explanation:

Since 64 is a multiple of itself and of all the other numbers, the answer is 64.

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3 years ago

Find the intersection of the line and the circle given below <br> y=-x-3<br> x^2+y^2=17

Salsk061 [2.6K]

Answer:

There are two points of intersection

(-4,1) and (1,-4)

Step-by-step explanation:

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2 years ago

Read 2 more answers

I need help with this please!!

vovangra [49]

Answer:

${6x}^{3} + {9x}^{2} - 18x + 3$

Step-by-step explanation:

1. Expand by distributing sum groups. $3x( {2x}^{2} + 5x - 1) - 3( {2x}^{2} + 5x - 1)$

2. Expand by distributing terms.

${6x}^{3} + {15x}^{2} - 3x - 3( {2x}^{2} + 5x - 1)$

3.Expand by distributing terms.

${6x}^{3} + {15x}^{2} - 3x - ( {6x}^{2} + 15x - 3)$

4. Remove parentheses.

${6x}^{3} + {15x}^{2} - 3x - {6x}^{2} - 15x + 3$

5. Collect like terms.

${6x}^{3} + {15x}^{2} - {6x}^{2} ) + ( - 3x - 15x) + 3$

6. Simplify.

${6x}^{3} + {9x}^{2} - 18x + 3$

Thus. the answer is,

${6x}^{3} + {9x}^{2} - 18x + 3$

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2 years ago

Find the domain of the function. (Enter your answer using interval notation.)

Andreas93 [3]

Answer:

$(-\infty,-9)\cup(-9,\infty)$

Step-by-step explanation:

The domain of a rational function is all real numbers <em>except </em>for when the denominator equals 0.

So, to find the domain restrictions, set the denominator to 0 and solve for x.

We have the rational function:

$s(y)=\frac{7y}{y+9}$

Set the denominator to 0:

$y+9=0$

Subtract 9:

$y\neq-9$

So, the domain is all real numbers except for -9.

In other words, our domain is all values to the left of negative 9 and to the right of negative 9.

In interval notation, this is:

$(-\infty,-9)\cup(-9,\infty)$

And we're done :)

4 0

3 years ago