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Usimov [2.4K]
2 years ago
12

3 factor string for 840

Mathematics
2 answers:
miskamm [114]2 years ago
8 0
8 x 15 x 7

24 x 5 x 7

56 x 3 x 5

35 x 3 x 8

Hope that helped :D
elixir [45]2 years ago
5 0
Well I have more than 4 but whatever....

2 x 420 
3 x 280 
4 x 210 
5 x 168 
6 x 140 
7 x 120 
8 x 105 
10 x 84 
12 x 70 
14 x 60 
15 x 56 
20 x 24 
21 x 40 
24 x 35 
28 x 30 
2 x 2 x 210 
2 x 3 x 140 
2 x 5 x 84 
2 x 7 x 60 
3 x 5 x 56 
3 x 7 x 40 
5 x 7 x 24 
2 x 2 x 2 x 105 
2 x 2 x 3 x 70 
2 x 2 x 5 x 42 
2 x 2 x 7 x 30 
2 x 3 x 5 x 28 
2 x 3 x 7 x 20 
3 x 5 x 7 x 8 
2 x 2 x 2 x 3 x 35 
2 x 2 x 2 x 5 x 21 
2 x 2 x 2 x 7 x 15 
2 x 2 x 3 x 5 x 14 
2 x 2 x 3 x 7 x 10 
2 x 3 x 4 x 5 x 7 
<span>2 x 2 x 2 x 3 x 5 x 7</span>


Hope I Helped You!!! :-)

Have A Good Day!!!
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Naddika [18.5K]

Answer:

8.944271909999

Step-by-step explanation:

hope you happy

6 0
2 years ago
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What's 5x10 exponent 1
strojnjashka [21]
The anser is 50 becasue if u had an mini 1 above 1 that would mean u had to multiply it 5x10 2 time fro example 5x20 = 100
6 0
3 years ago
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Find an equation to the line tangent to y = 5 + |x – 2| at the coordinate (2, 5) pls answer
myrzilka [38]

Answer:

Step-by-step explanation:

Finding an equation of a tangent line to that function requires that we find the derivative of the function at that point. Since this is an absolute value function with its cusp at (2, 5), the function is not differentiable here.

3 0
3 years ago
Consider the line 5x-8y=8 what is the slope of a line perpendicular to this line what is the slope of a line parallel to this li
Sophie [7]

\text{Let}\ k:y=m_1x+b_1\ \text{and}\ l:y=m_2x+b_2,\ \text{then}\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\\\l\ \parallel\ k\iff m_1=m_2\\\\\text{Convert the equation to the slope-intercept form}\ y=mx+b.\\\\5x-8y=8\qquad\text{subtract}\ 5x\ \text{from both sides}\\\\-8y=-5x+8\qquad\text{divide both sides by (-8)}\\\\y=\dfrac{5}{8}x-1\to m_1=\dfrac{5}{8}\\\\perpendicular:\ m_2=-\dfrac{1}{\frac{5}{8}}=-\dfrac{8}{5}\\\\parallel:\ m_2=\dfrac{5}{8}

8 0
3 years ago
Write each expression as an algebraic​ (nontrigonometric) expression in​ u, u &gt; 0.
max2010maxim [7]

Answer:

\displaystyle \sin\left(2\sec^{-1}\left(\frac{u}{10}\right)\right)=\frac{20\sqrt{u^2-100}}{u^2}\text{ where } u>0

Step-by-step explanation:

We want to write the trignometric expression:

\displaystyle \sin\left(2\sec^{-1}\left(\frac{u}{10}\right)\right)\text{ where } u>0

As an algebraic equation.

First, we can focus on the inner expression. Let θ equal the expression:

\displaystyle \theta=\sec^{-1}\left(\frac{u}{10}\right)

Take the secant of both sides:

\displaystyle \sec(\theta)=\frac{u}{10}

Since secant is the ratio of the hypotenuse side to the adjacent side, this means that the opposite side is:

\displaystyle o=\sqrt{u^2-10^2}=\sqrt{u^2-100}

By substitutition:

\displaystyle= \sin(2\theta)

Using an double-angle identity:

=2\sin(\theta)\cos(\theta)

We know that the opposite side is √(u² -100), the adjacent side is 10, and the hypotenuse is u. Therefore:

\displaystyle =2\left(\frac{\sqrt{u^2-100}}{u}\right)\left(\frac{10}{u}\right)

Simplify. Therefore:

\displaystyle \sin\left(2\sec^{-1}\left(\frac{u}{10}\right)\right)=\frac{20\sqrt{u^2-100}}{u^2}\text{ where } u>0

4 0
2 years ago
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