What is the algebraic expression for the quotient of j and 8

Rewrite the equation by completing the square g(x) = x^2 - x - 6

borishaifa [10]

Answer:

g(x)=x2−x−6g(x)=x2-x-6

Set x2−x−6x2-x-6 equal to 00.

x2−x−6=0x2-x-6=0

Solve for xx.

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Factor x2−x−6x2-x-6 using the AC method.

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(x−3)(x+2)=0(x-3)(x+2)=0

If any individual factor on the left side of the equation is equal to 00, the entire expression will be equal to 00.

x−3=0x-3=0

x+2=0x+2=0

Set x−3x-3 equal to 00 and solve for xx.

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x=3x=3

Set x+2x+2 equal to 00 and solve for xx.

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x=−2x=-2

The final solution is all the values that make (x−3)(x+2)=0(x-3)(x+2)=0 true.

x=3,−2

5 0

3 years ago

Need help with this math question please

lesya692 [45]

Answer:

See below for answer.

Step-by-step explanation:

RP/RT =RQ/RS Given

∠R = ∠R

ΔPQR similar to Δ TSR If the measures of two sides of a triangle are proportional to the measures of two corresponding sides of another triangle and their included angles are congruent, then the triangles are similar.

3 0

3 years ago

What does it mean to solve a system of linear equation

Nana76 [90]

Answer:

to figure out the answer / an assignment of values to the variables

Step-by-step explanation:

7 0

3 years ago

What is the mode of this data set: 17, 13, 18, 20, 17, 15, 12<br><br> A. 19<br><br> B. 17

telo118 [61]

<h3>Answer: B. 17</h3>

The mode is the most frequent or most occurring value. The value 17 shows up the most times, at two times, so this is the mode. The value 19 is not even part of the given set of values, so you can immediately cross of choice A.

Side note: it is possible for a set to have more than one mode, and it's also possible to not have a mode at all.

7 0

3 years ago

A box in a supply room contains 24 compact fluorescent lightbulbs, of which 8 are rated 13-watt, 9 are rated 18-watt, and 7 are

Marrrta [24]

Answer:

a) There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

b) There is a 8.65% probability that all three of the bulbs have the same rating.

c) There is a 12.45% probability that one bulb of each type is selected.

Step-by-step explanation:

There are 24 compact fluorescent lightbulbs in the box, of which:

8 are rated 13-watt

9 are rated 18-watt

7 are rated 23-watt

(a) What is the probability that exactly two of the selected bulbs are rated 23-watt?

There are 7 rated 23-watt among 23. There are no replacements(so the denominators in the multiplication decrease). Then can be chosen in different orders, so we have to permutate.

It is a permutation of 3(bulbs selected) with 2(23-watt) and 1(13 or 18 watt) repetitions. So

$P = p^{3}_{2,1}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = \frac{3!}{2!1!}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 3*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 0.1764$