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Anika [276]
3 years ago
9

Which algebraic expression six less than half a number

Mathematics
1 answer:
densk [106]3 years ago
8 0

1/2n-6?

hope it helps!

if not, sorry...

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Six equilateral triangles are connected to create a regular hexagon. The area of the hexagon is 24a2 – 18 square units. Which is
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<span>Six equilateral triangles are connected to create a regular hexagon. The area of the hexagon is 24a2 – 18 square units. Which is an equivalent expression for the area of the hexagon based on the...

Answer: </span> 6(4a2 - 3)

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Which of the following is true of the location of the terminal side of an angle, Theta, whose sine value is One-half?
FinnZ [79.3K]

Answer:

Theta has a reference angle of 30° and is in Quadrant I or II

Step-by-step explanation:

Sin(theta) = ½

Basic angle: 30

Angles:

30,

180-30 = 150

Because sin is positive in quadrants 1 and 2

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3 years ago
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Triphasil-28 birth control tablets are taken sequentially, 1 tablet per day for 28 days, with the tablets containing the followi
bazaltina [42]

Answer:

1.925mg of levonorgestrel and 0.68mg of ethinyl estradiol are taken during the 28-day period.

Step-by-step explanation:

The total milligrams of levonorgestrel is:

L = L_{1} + L_{2} + L_{3}

In which L_{1}, L_{2} and L_{3} are the number of miligrams of levonorgestrel taken in each phase.

The total milligrams of ethinyl estradiol is:

E = E_{1} + E_{2} + E_{3}

In which E_{1},E_{2} and E_{3} are the number of miligrams of ethinyl estradiol taken in each phase.

We can solve this by phase.

Phase 1: 6 tablets, each containing 0.050 mg of levonorgestrel and 0.030 mg ethinyl estradiol

In this phase,

6*0.050mg = 0.30mg of levonorgestrel and 6*0.030mg = 0.18mg of ethinyl estradiol are taken.

So L_{1} = 0.30 and E_{1} = 0.18

Phase 2: 5 tablets, each containing 0.075 mg of levonorgestrel and 0.040 mg ethinyl estradiol

In this phase,

5*0.075mg = 0.375mg of levonorgestrel and 5*0.040mg = 0.20mg of ethinyl estradiol are taken.

So L_{2} = 0.375 and E_{2} = 0.20

Phase 3: 10 tablets, each containing 0.125 mg of levonorgestrel and 0.030 mg ethinyl estradiol

In this phase,

10*0.125mg = 1.25mg of levonorgestrel and 10*0.030mg = 0.30mg of ethinyl estradiol are taken.

So L_{3} = 1.25 and E_{3} = 0.30

The total milligrams of levonorgestrel is:

L = L_{1} + L_{2} + L_{3} = 0.30 + 0.375 + 1.25 = 1.925

The total milligrams of ethinyl estradiol is:

E = E_{1} + E_{2} + E_{3} = 0.18 + 0.20 + 0.30 = 0.68

1.925mg of levonorgestrel and 0.68mg of ethinyl estradiol are taken during the 28-day period.

6 0
2 years ago
I need help please.
GuDViN [60]

Answer:

The factors of 2q²-5pq-2q+5p are (2q-5p) (q-1)....

Step-by-step explanation:

The given expression is:

2q²-5pq-2q+5p

Make a pair of first two terms and last two terms:

(2q²-5pq) - (2q-5p)

Now factor out the common factor from each group.

Note that there is no common factor in second group. So we will take 1 as a common factor.

q(2q-5p) -1(2q-5p)

Now factor the polynomial by factoring out the G.C.F, 2q-5p

(2q-5p) (q-1)

Thus the factors of 2q²-5pq-2q+5p are (2q-5p) (q-1)....

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3 years ago
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\displaystyle\\&#10;15(2a - 2) = 5(a^2 -1) \\  \\ &#10;30a - 30 = 5a^2-5\\\\&#10;-5a^2 + 30a -30 + 5 =0\\\\&#10;-5a^2 + 30a -25 =0 ~~~~~\Big| \times (-1) \\\\&#10;5a^2 - 30a +25 = 0~~~~~\Big| : 5 \\\\&#10;a^2 - 6a +5 = 0 \\  \\ &#10;a_{12}= \frac{-b\pm \sqrt{b^2-4ac}}{2a}= \frac{6\pm \sqrt{36-20}}{2}= \frac{6\pm \sqrt{16}}{2}=\frac{6\pm 4}{2}= \boxed{3\pm 2} \\  \\ &#10;a_1 = 3+ 2 = \boxed{5}\\\\&#10;a_2 = 3-2 = \boxed{1}



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3 years ago
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