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3 years ago

2. During typhoon Ambo, PAGASA tracks the amount of accumulating rainfall. For

Mathematics

1 answer:

Yuki888 [10]3 years ago

4 0

Answer:

The answer is below

Step-by-step explanation:

Let t represent the time in hour and A(t) represent the amount of accumulating rainfall at time t.

In the first three hours the rain fell at a constant rate of 25mm per

hour, hence:

A(t) = 25t 0 ≤ t ≤ 3

The rain then slows down and remains constant at 75 mm [i.e. 25(3)] for an hour.

The rain increases at a rate of 20t from 4 ≤ t ≤ 6. Hence the piecewise equation is given as:

$A(t)=\left \{ {{25t\ \ \ \ \ \ \ \ \ \ for\ 0\leq\ t\ \leq\ 3} \atop {75\ \ \ \ \ \ \ \ \ \ for\ 0\leq\ t\ \leq\ 6 \atop {75+20(t-4)\ \ for\ 4\ \leq\ t\ \leq\ 6}} } \right. \\\\\left$

You might be interested in

Solve log base 6 (x-6) - log base 6 (x + 4) = 2

Ray Of Light [21]

Answer:

no solutions

Step-by-step explanation:

Since the two terms have the same base, we are able to use the rule for subtracting logarithms:

$log_{b}(x) - log_{b}(y) = log_{b}(\frac{x}{y} )$

Therefore, the equation can be written as:

$log_{6}(\frac{x-6}{x+4} )=2$

By using the definition of a logarithm we can say that:

$\frac{x-6}{x+4} = 6^{2}\\\frac{x-6}{x+4} = 36\\x -6 = 36x+144\\35x = -150\\x =-\frac{30}{7}$

When plugging this solution in, you find that the term $log_{6}(x-6)$ has x-6 evaluate to a number less than 0. This is not included in the domain of log functions, so $-\frac{30}{7}$ is not a valid solution. This means that there are no solutions.

4 0

2 years ago

There were 50 students in the chess club. The membership went up by 10%. How many more students joined the club?

Vlad [161]

10% of 50 is 5.

Therefore 50+5=55

Your answer is 5

7 0

3 years ago

Can someone plzz help me with these 2 questions. thanks!

Deffense [45]

9514 1404 393

Answer:

4. height: 5√3 cm; area: 25√3 cm²

5. 30√2 ft ≈ 42.43 ft

Step-by-step explanation:

4. The height of an equilateral triangle is (√3)/2 times the side length, so is ...

height = (√3)/2 × (10 cm) = 5√3 cm

The area is given by the formula ...

A = 1/2bh

A = 1/2(10 cm)(5√3 cm) = 25√3 cm²

5. The diagonal of a square is √2 times the side length, so the distance from 3rd to 1st base is ...

(30 ft)√2 ≈ 42.43 ft

The length of the diagonal of a square is something you should be familiar with. If you're not, you can figure the distance using the Pythagorean theorem.

d = √(30² +30²) = √1800 ≈ 42.43 . . . feet

6 0

3 years ago

Plz help me!!!!!!!!!

Over [174]

Answer: $\bold{\dfrac{4\pm \sqrt{6}}{2}}$

<u>Step-by-step explanation:</u>

$\dfrac{3}{y-2}-2=\dfrac{1}{y-1}\\\\\\\text{Multiply by the LCD (y-2)(y-1) to clear the denominator:}\\\\\dfrac{3}{y-2}(y-2)(y-1)-2(y-2)(y-1)=\dfrac{1}{y-1}(y-2)(y-1)\\\\\\3(y-1)-2(y-2)(y-1)=1(y-2)\\\\3y-3-2(y^2-3y+2)=y-2\\\\3y-3-2y^2+6y-4=y-2\\\\-2y^2+9y-7=y-2\\\\0=2y^2-8y+5\quad \rightarrow \quad a=2,\ b=-8,\ c=5\\\\\\\text{Quadratic formula is: }x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\\x=\dfrac{-(-8)\pm \sqrt{(-8)^2-4(2)(5)}}{2(2)}\\\\\\.\ =\dfrac{8\pm \sqrt{64-40}}{2(2)}$

$.\ =\dfrac{8\pm \sqrt{24}}{2(2)}\\\\\\.\ =\dfrac{8\pm 2\sqrt{6}}{2(2)}\\\\\\.\ =\dfrac{4\pm \sqrt{6}}{2}$

7 0

3 years ago

MR is the angle bisector of

Otrada [13]

Answer:

∠NMP, so m∠1 = m∠2.

Step-by-step explanation:

6 0

3 years ago