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Varvara68 [4.7K]
3 years ago
15

How many classmates does J u l i a n have?

Mathematics
1 answer:
Sav [38]3 years ago
6 0

Answer:

the answer is d

Step-by-step explanation:

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Find d:<br> 6 1/3*d= 4 2/9
Marrrta [24]
(3d x 6) + 1 = (9 x 4) + 2

-> 18d + 1 = 36 + 2
-> 18d = 38 - 1
-> 18d = 37
-> d = 37/18
-> d = 2.056
8 0
3 years ago
Correct answer get brainliest!! Determine if the following triangles are congruent if yes state what theorem proves them to be
AnnZ [28]

Answer: AAS

Step-by-step explanation:

We know that \angle TSN \cong \angle HSU because they are vertical angles, meaning the triangles are congruent by <u>AAS.</u>

3 0
1 year ago
If p and q are distinct primes. Find the number of positive divisors of<br> p^2 * q^2
kkurt [141]

Answer:

9

Step-by-step explanation:

We know that for p^mq^n  where p and q are different prime numbers the number of positive divisors are (m+1) (n+!)

We have given p^2q^2

So here m=2 and n=2

So number of positive divisors =(2+1)\times (2+1) =9

So the number of positive divisors of  p^2q^2 is 9

3 0
3 years ago
Read 2 more answers
Determine the type and number of solutions of x2+2x+3=0
erastova [34]

Solve the equation for x by finding a, b, and c of the quadratic then applying the quadratic formula.

x  =  −1  ±  i√2

There's no real solutions

7 0
3 years ago
A motorboat is capable of traveling at a speed of 14 miles per hour in still water. On a particular day, it took 15 minutes long
Anon25 [30]

By solving a system of equations we will find that the rate of current in the stream is S = 2 mi/h.

When the motorboat travels downstream, the total velocity will be the velocity of the motorboat in still water plus the velocity of the stream, while if the motorboat travels upstream, we have the velocity of the stream subtracted.

So upstream the speed is:

(14 mi/h - S)

Downstream the speed is:

(14 mi/h + S)

Where S is the rate of current in the stream.

We know that downstream it takes 15 minutes more to travel 12 miles, then we can write the system of equations:

(14 mi/h + S)*T = 12 mi

(14 mi/h - S)*(T - 15 min) = 12 mi

To solve this, we need to isolate one of the variables in one of the equations, I will isolate T in the first one:

T = (12 mi)/(14 mi/h + S)

Replacing that in the other equation we will get:

(14 mi/h - S)*((12 mi)/(14 mi/h + S) - 15 min) = 12 mi

Now we can solve this for S. Now we can multiply both sides by (14 mi/h + S).

(14 mi/h - S)*12 mi  - (14 mi/h + S)*(14 mi/h - S)*(- 15 min) = 12 mi*(14 mi/h + S)

Also notice that the speeds are in hours, so we can rewrite:

- 15 min = -0.25 h

(14 mi/h - S)*12 mi  - (14 mi/h + S)*(14 mi/h - S)*(- 0.25 h) = 12 mi*(14 mi/h + S)

168 mi^2/h - 12mi*S  + 49mi^2/h + 0.25h*S^2 = 168mi^2/h + 12mi*S

- 12mi*S  + 49mi^2/h - 0.25h*S^2 = 12mi*S

-24mi*S -  0.25h*S^2  + 49mi^2/h = 0

This is a quadratic equation, the solutions are:

S = \frac{24mi \pm \sqrt{(-24mi)^2 - 4*(49mi^2/h)*(-0.25h)}  }{2*-0.25h} \\\\S =  \frac{24mi \pm 25 mi  }{-0.5h}

We only take the positive solution, so we get:

S = (24 mi - 25 mi)/(-0.5 mi) = 2 mi/h

The rate of current in the stream is 2 mi/h.

If you want to learn more, you can read:

4 0
2 years ago
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