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Gnesinka [82]
3 years ago
8

Cindy woks Monday through Friday at a rate of $15.18 per hour. She takes a 1-hour lunch break. If she

Mathematics
1 answer:
Alex Ar [27]3 years ago
7 0
11 hours-1
10*15.18
151.80*5
Answer is$ 759 is her total pay
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The second statement is the<br> of the first.<br> Foundations of Geometry
Elina [12.6K]

i think its c. contrapositive

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3 years ago
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Which is the equation of a line that is perpendicular to 8x + 6y = –5?
IrinaVladis [17]

The perpendicular line will have an equation of y=3/4x-3/4

To find this, we first have to solve our equation for slope intercept form.

8x + 6y = -5 ----> subtract 8x

6y = -8x + -5 ----> divide by 6

y = -4/3x - 5/6.

So we know the slope of this equation to be -4/3. Since perpendicular lines have opposite and reciprocal slopes, we know we can simply flip the fraction and make it a negative to get the new slope of 3/4. Since B is the only option with that slope, we know it to be the correct answer.

4 0
4 years ago
ZX bisects ∠WZY. If the measure of ∠YXZ is (6m – 12)°, what is the value of m?
krok68 [10]

Answer:

The correct answer is option B.  17

Step-by-step explanation:

It is given that, ZX bisects ∠WZY. If the measure of ∠YXZ is (6m – 12)°

<u>To find the value of m</u>

From the figure we can see that, triangle WYZ is an isosceles  triangle.

ZW = ZY

Then <YXZ = <WXZ = 90°

It is given ∠YXZ  = (6m – 12)°

(6m – 12)° =  90°

6m = 90 + 12  = 102

m = 102/6 = 17

Therefore the value of m = 17

The correct answer is option B.  17

8 0
3 years ago
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Please help.........
Schach [20]

Answer: what the heck do you have to answer

Step-by-step explanation:

3 0
3 years ago
Expansion Numerically Impractical. Show that the computation of an nth-order determinant by expansion involves multiplications,
posledela

Answer:

  • number of multiplies is n!
  • n=10, 3.6 ms
  • n=15, 21.8 min
  • n=20, 77.09 yr
  • n=25, 4.9×10^8 yr

Step-by-step explanation:

Expansion of a 2×2 determinant requires 2 multiplications. Expansion of an n×n determinant multiplies each of the n elements of a row or column by its (n-1)×(n-1) cofactor determinant. Then the number of multiplies is ...

  mpy[n] = n·mp[n-1]

  mpy[2] = 2

So, ...

  mpy[n] = n! . . . n ≥ 2

__

If each multiplication takes 1 nanosecond, then a 10×10 matrix requires ...

  10! × 10^-9 s ≈ 0.0036288 s ≈ 0.004 s . . . for 10×10

Then the larger matrices take ...

  n=15, 15! × 10^-9 ≈ 1307.67 s ≈ 21.8 min

  n=20, 20! × 10^-9 ≈ 2.4329×10^9 s ≈ 77.09 years

  n=25, 25! × 10^-9 ≈ 1.55112×10^16 s ≈ 4.915×10^8 years

_____

For the shorter time periods (less than 100 years), we use 365.25 days per year.

For the longer time periods (more than 400 years), we use 365.2425 days per year.

8 0
4 years ago
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