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3 years ago

PLEASE HELP! DUE TODAY!

Mathematics

2 answers:

Kitty [74]3 years ago

4 0

Answer:

but it I'd not see your question

SVEN [57.7K]3 years ago

3 0

Answer:

post question plz

Step-by-step explanation:

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I need help with this question. It is due tomorrow. #10. 35 points.

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you need to clear up the graphics its all blurry.

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3 years ago

Tim made 4 stacks of wooden blocks. The first stack was 4 blocks tall. The second stack was 6 blocks taller than the first. The

Troyanec [42]

Tim used 26 blocks of wood

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4 years ago

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Find the greatest common factor : 25p^2q^3, 15p^2q^2 and 35pq^4

olga nikolaevna [1]

Answer:

The greatest common factor is $5pq^2$ .

Step-by-step explanation:

The given expressions are $25p^2q^3$ , $15p^2q^2$ and $35pq^4$ .

Greatest common factor of two numbers is the greatest number that divides both numbers completely.

Find factors of each expression.

$25p^2q^3=5\times 5\times p\times p\times q\times q\times q$

$15p^2q^2=3\times 5\times p\times p\times q\times q$

$35pq^4=5\times 7\times p\times q\times q\times q\times q$

The common factors are 5,p,q and q. So, the greatest common factor of given expressions is

$G.C.F.=5\times p\times q\times q$

$G.C.F.=5pq^2$

Therefore the greatest common factor is $5pq^2$ .

3 0

3 years ago

A particle is moving with the given data. Find the position of the particle. a(t)=10sin(t)+3cos(t) s(0)=0 s(2pi)=12

zubka84 [21]

$a(t)=10\sin t+3\cos t$

$v(t)=\displaystyle\int a(t)\,\mathrm dt$
$v(t)=\displaystyle\int(10\sin t+3\cos t)\,\mathrm dt$
$v(t)=-10\cos t+3\sin t+C_1$

$s(t)=\displaystyle\int v(t)\,\mathrm dt$
$s(t)=\displaystyle\int(-10\cos t+3\sin t+C_1)\,\mathrm dt$
$s(t)=-10\sin t-3\cos t+C_1t+C_2$

With the boundary conditions $s(0)=0$ and $s(2\pi)=12$ , we have

$\begin{cases}0=-3+C_2&s(0)=0\\12=-3+2\pi C_1+C_2&s(2\pi)=12\end{cases}\implies C_1=\dfrac6\pi,C_2=3$

so that the position of the particle is given by

$s(t)=-10\sin t-3\cos t+\dfrac{6t}\pi+3$

3 0

4 years ago

The height of the water in an above ground pool is 3

Firdavs [7]

<em>It will take 6 minutes to drain the pool.</em>

<h2>Explanation:</h2>

As the problem states the height of the water in an above ground pool is 3ft. In other word:

3 ft is our starting point.

The pool needs to be drained at a Rate of Change of 1/2 inches per minutes. In other words:

The rate of change is constant, so we'll have a line with slope -1/2.

Therefore, we can write an equation as:

$h=3-\frac{1}{2}t \\ \\ \\ h:Height \\ \\ t:time \\ \\ \\ The \ slope \ is \ negative \ because \ the \ pool \ needs \ to \ be \ drained$

To drain the pool implies that we don't have any water in the pool, in other words this is true if and only if:

$h=0$

So:

$h=3-\frac{1}{2}t \\ \\0=3-\frac{1}{2}t \\ \\ \\ Solving \ for \ t: \\ \\ \frac{1}{2}t=3 \\ \\ t=6$

So, <em>it will take 6 minutes to drain the pool.</em>

<h2>Learn more:</h2>

Average Rate of Change: brainly.com/question/14196399#

#LearnWithBrainly

8 0

4 years ago

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