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3 years ago

100 km per hour equals how many mph?

Mathematics

2 answers:

trasher [3.6K]3 years ago

8 0

Answer:

62.1371

To obtain an approximate result, divide the speed value by 1.609

den301095 [7]3 years ago

7 0

Answer:

62.14 mph

Step-by-step explanation:

To solve this problem we use a conversion factor.

We know that one mile equals 1.60934 km.

Then to calculate how many mph are 100 km/h we solve the following multiplication:

$100\ km/h *\frac{1\ mile}{1.60934\ km}=62.14\ mph$

Finally the answer is 62.14 mph

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DanielleElmas [232]

Answer:

Step-by-step explanation:

First, convert all the secants and cosecants to cosine and sine, respectively. Recall that $csc(x)=1/sin(x)$ and $sec(x)=1/cos(x)$ .

Thus:

$\frac{sec(x)}{cos(x)} -\frac{sin(x)}{csc(x)cos^2(x)}$

$=\frac{\frac{1}{cos(x)} }{cos(x)} -\frac{sin(x)}{\frac{1}{sin(x)}cos^2(x) }$

Let's do the first part first: (Recall how to divide fractions)

$\frac{\frac{1}{cos(x)} }{cos(x)}=\frac{1}{cos(x)} \cdot \frac{1}{cos(x)}=\frac{1}{cos^2(x)}$

For the second term:

$\frac{sin(x)}{\frac{cos^2(x)}{sin(x)} } =\frac{sin(x)}{1} \cdot\frac{sin(x)}{cos^2(x)}=\frac{sin^2(x)}{cos^2(x)}$

So, all together: (same denominator; combine terms)

$\frac{1}{cos^2(x)}-\frac{sin^2(x)}{cos^2(x)}=\frac{1-sin^2(x)}{cos^2(x)}$

Note the numerator; it can be derived from the Pythagorean Identity:

$sin^2(x)+cos^2(x)=1; cos^2(x)=1-sin^2(x)$

Thus, we can substitute the numerator:

$\frac{1-sin^2(x)}{cos^2(x)}=\frac{cos^2(x)}{cos^2(x)}=1$

Everything simplifies to 1.

7 0

3 years ago

Someone pls help with question 2

kirza4 [7]

Answer:

Step-by-step explanation:

5 0

3 years ago

In triangle PQR point c is the centroid if PQ= 30 then PX=

vova2212 [387]

Answer:

Step-by-step explanation:

since c is the centroid, therefore PX=XQ=1/2 x PQ

∴ PX=15

3 0

3 years ago

Mason answers a question correctly in a trivia game and now has 0 points. The question was worth

zhannawk [14.2K]

Answer:

Mason recieved 15 points for answering the question correctly so that means that he had -15 points before. The points would be at -15 and 0.

8 0

3 years ago

Read 2 more answers

The prior probabilities for events A1 and A2 are P(A1) = 0.35 and P(A2) = 0.50. It is also known that P(A1 ∩ A2) = 0. Suppose P(

forsale [732]

Answer:

Step-by-step explanation:

Hello!

Given the probabilities:

P(A₁)= 0.35

P(A₂)= 0.50

P(A₁∩A₂)= 0

P(BIA₁)= 0.20

P(BIA₂)= 0.05

Two events are mutually exclusive when the occurrence of one of them prevents the occurrence of the other in one repetition of the trial, this means that both events cannot occur at the same time and therefore they'll intersection is void (and its probability zero)

Considering that P(A₁∩A₂)= 0, we can assume that both events are mutually exclusive.

Considering that $P(BIA)= \frac{P(AnB)}{P(A)}$ you can clear the intersection from the formula $P(AnB)= P(B/A)*P(A)$ and apply it for the given events: