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nadezda [96]
2 years ago
5

Write the following in expanded exponential form a. 30405

Mathematics
1 answer:
lana [24]2 years ago
7 0

Answer:

30000+0+400+0+5

6000000+800000+0+4000+100+50+2

Step-by-step explanation:

hope it's right plz mark me brainlists

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4,274 days old. Hope that helps!
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Given a right triangle with a hypotenuse of 6 cm and a leg of 4cm, what is the measure of the other leg of the triangle rounded
slega [8]

Answer:

4.5 cm=L2

Step-by-step explanation:

This problem can be solved using the pythagorian theoreme:

H=\sqrt{L1^{2}+L2^{2}  }

Where:

H=6 cm

L1=4 cm

L2=?

We isolate our incognita an we get:

\sqrt{H^{2}-L1^{2}} =L2

We supplant with the given data:

\sqrt{(6cm)^{2}-(4cm)^{2}} =L2

\sqrt{36cm^{2}-16cm^{2}} =L2

\sqrt{36cm^{2}-16cm^{2}} =L2

\sqrt{20cm^{2}} =L2

4.5 cm=L2

5 0
3 years ago
I need help on number 10 please!
hammer [34]

Answer: y = 4x/3 - 5/2

Step-by-step explanation:

The equation of a straight line can be represented in the slope intercept form as

y = mx + c

Where

c represents the y intercept

m = slope = (y2 - y1)/(x2 - x1)

The given line, L1 passes through A(6, - 7) and B(- 6, 2). The slope of line L1 is

m = (2 - - 7)/(- 6 - 6) = 9/ -12 = - 3/4

If two lines are perpendicular, it means that the slope of one line is the negative reciprocal of the slope of the other line.

Therefore, the slope of line L2 passing through the midpoint, M is 4/3

The formula determining the midpoint of a line is expressed as

[(x1 + x2)/2 , (y1 + y2)/2]

Midpoint, M = [(6 + -6)/2 , (- 7 + 2)/2]

= (0, - 5/2]

This means that the y intercept of line L2 is - 5/2

The equation of L2 becomes

y = 4x/3 - 5/2

4 0
3 years ago
Use​ l'Hôpital's Rule to find the following limit. ModifyingBelow lim With x right arrow 0StartFraction 3 sine (x )minus 3 x Ove
steposvetlana [31]

Answer:

\lim_{x \to 0} \frac{3sinx-3x}{7x^3}=-\frac{1}{14}

Step-by-step explanation:

The limit is:

\lim_{x \to 0} \frac{3sinx-3x}{7x^3}=\frac{0}{0}

so, you have an indeterminate result. By using the l'Hôpital's rule you have:

\lim_{x \to 0} \frac{a(x)}{b(x)}= \lim_{x \to 0} \frac{a'(x)}{b'(x)}

by replacing, and applying repeatedly you obtain:

\lim_{x \to 0} \frac{3sinx-3x}{7x^3}= \lim_{x \to 0}\frac{3cosx-3}{21x^2}= \lim_{x \to 0}\frac{-3sinx}{42x}= \lim_{x \to 0}\frac{-3cosx}{42}\\\\ \lim_{x \to 0} \frac{3sinx-3x}{7x^3}=\frac{-3cos0}{42}=-\frac{1}{14}

hence, the limit of the function is -1/14

8 0
3 years ago
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