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mote1985 [20]
3 years ago
13

What is the slope of the line that passed through the points (1,-7) and (10,-7)

Mathematics
1 answer:
Natali [406]3 years ago
8 0

Answer:

the slope is 0

Step-by-step explanation:

We can see by reading the coordinates of the points that they have the same y coordinate -7 so we can tell from the beginning  that the slope is 0 because the line going trough the points will NOT go up or down will STAY FLAT at -7.

OR we can actually calculate the slope

The slope of a line in general is m = (y2-y1)/(x2-x1)

The slope of the line the line that passed through the points

(x1 = 1, y1 = -7) and (x2 = 10, y2 = -7) is

m = (-7- -7) / (10-1) =

m = (-7+7)/(10-1)

m = 0/9

m =0

The slope 0 tells us that the two points are on a horizontal line.

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If x= 1/2-√3 , prove that x³-2x²-7x+5=3
prohojiy [21]
x=\frac{1}{2}-\sqrt3\\\\x^3-2x^2-7x+5=3\\\\(\frac{1}{2}-\sqrt3)^3-2(\frac{1}{2}-\sqrt3)^2-7(\frac{1}{2}-\sqrt3)+5\\\\=(\frac{1}{2})^3-3\cdot(\frac{1}{2})^2\cdto\sqrt3+3\cdot\frac{1}{2}\cdot(\sqrt3)^2-(\sqrt3)^3-...\\\\...-2[(\frac{1}{2})^2-2\cdot\frac{1}{2}\cdot\sqrt3+(\sqrt3)^2]-\frac{7}{2}+7\sqrt3+5

=\frac{1}{8}-\frac{3\sqrt3}{4}+\frac{9}{2}-3\sqrt3-2(\frac{1}{4}-\sqrt3+3)-\frac{7}{2}+7\sqrt3+5\\\\=\frac{1}{8}+\frac{9}{2}-\frac{1}{2}-6-\frac{7}{2}+5-\frac{3}{4}\sqrt3-3\sqrt3+2\sqrt3+7\sqrt3\\\\=\frac{1}{8}+\frac{1}{2}-1+5\frac{1}{4}\sqrt3=\frac{1}{8}-\frac{1}{2}+5\frac{1}{4}\sqrt3=\frac{1}{8}-\frac{4}{8}+5\frac{1}{4}\sqrt3\\\\=-\frac{3}{8}+5\frac{1}{4}\sqrt3\neq3
6 0
3 years ago
The Monthly Bank pays 3 percent interest, compounded monthly, on their savings accounts. The Daily Bank pays 3 percent interest,
Novosadov [1.4K]

Answer:

The amount that we should deposit in each bank is around $942.

Step-by-step explanation:

Case 1:

A=$1000

n = 12

t = 2

r = 3% or 0.03

p = ?

The compound interest formula is :

A=p(1+\frac{r}{n})^{nt}

Substituting values in the formula;

1000=p(1+\frac{0.03}{12})^{12\times2}

=> 1000=p(1.0025)^{24}

=> 1000=1.06175p

p=\frac{1000}{1.06175}

p = $941.84

Case 2:

A=$1000

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t = 2

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1000=p(1+\frac{0.03}{365})^{365\times2}

=> 1000=p(1.0000822)^{730}

=> 1000=1.06184p

p=\frac{1000}{1.06184}

p = $941.76

The amount that we should deposit in each bank is around $942.

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