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3 years ago

15

A lab technician mixed a 590 ml solution of water and alcohol. If 3% of the solution is alcohol, how many milliliters of alcoh

ol were used, and how many milliliters of water were used?

1 answer:

mafiozo [28]3 years ago

3 0

3 milliliters was used to the answer

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The leg of a right triangle is 3 units and the hypotenuse is 4 units. What is the length, in units, of the other leg of the tria

Free_Kalibri [48]

Answer:

5

Step-by-step explanation:

This is honestly a pure guess, but I am guessing they want you to find a perfect triangle, and 3,4,5 is the smallest perfect triangle. Considering they also said it was a right triangle, I strongly beleive the answer is 5.

6 0

3 years ago

Read 2 more answers

3. Two cars are parked, 150m apart, and on opposite sides of a building. A camera on the top of the building rotates and can vie

allsm [11]

Answer:

<em>The building is 61.5 m tall</em>

Step-by-step explanation:

The image below is a diagram where all the given distances and angles are shown. We have additionally added some variables:

h = height of the building

a, b = internal angles of each triangle

x = base of each triangle

The angles a and b can be easily found by subtracting the given angles from 90° since they are complementary angles, thus:

a = 90° - 37° = 53°

b = 90° - 42° = 48°

Now we apply the tangent ratio on both triangles separately:

$\displaystyle \tan a=\frac{\text{opposite leg}}{\text{adjacent leg}}$

$\displaystyle \tan 53^\circ=\frac{150-x}{h}$

$\displaystyle \tan 48^\circ=\frac{x}{h}$

From the last equation:

$x=h.\tan 48^\circ$

Substituting into the first equation:

$\displaystyle \tan 53^\circ=\frac{150-h.\tan 48^\circ}{h}$

Operating on the right side:

$\displaystyle \tan 53^\circ=\frac{150}{h}-\tan 48^\circ$

Rearranging:

$\displaystyle \tan 53^\circ+\tan 48^\circ=\frac{150}{h}$

Solving for h:

$\displaystyle h=\frac{150}{\tan 53^\circ+\tan 48^\circ}$

Calculating:

h = 61.5 m

The building is 61.5 m tall

5 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

3 years ago

Camilo estimates the weight of the dog . The veterinarian says that the dog weighs 84 pounds . The percent error in Camillo's es

blsea [12.9K]

Answer:

The estimated weight is in the range:

75.6 pounds $\leq$ Weight $\leq$ 92.4 pounds

Step-by-step explanation:

Since, the maximum error is 10%.

Therefore, the maximum and minimum vales will be 10% more and 10% less than 84 pounds, respectively.

<u>For Maximum Limit</u>:

$Weight_{max}$ = (1.1)(84 pounds)

$Weight_{max}$ = <u>92.4 pounds</u>

<u>For Minimum Limit</u>:

$Weight_{min}$ = (0.9)(84 pounds)

$Weight_{min}$ = <u>75.6 pounds</u>

Hence, the estimated weight is in the range:

75.6 pounds $\leq$ Weight $\leq$ 92.4 pounds

5 0

3 years ago

58 percent of what is 61.4

Nana76 [90]

Answer:

35.612

Step-by-step explanation:

3 0

2 years ago