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3 years ago

What does e represent in e/2=3

Mathematics

1 answer:

enot [183]3 years ago

5 0

e represents 6

because 6/2=3

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Complete the statement of each of these rules:

never [62]

Answer:

a) (+) × (?) = (+)

(?) = (+)

b) (-) × (+) = ?

? = (-)

c) (?) × (-) = ?

? is undefined

Step-by-step explanation:

a) (+) × (?) = (+)

From the given equation in the question, by dividing both side of the equation by (+), we have;

((+) × (?))/(+) = (+)/(+)

(?) = (+)/(+) = (+)

Therefore, we have;

(+) × (+) = (+)

(?) = (+)

b) (-) × (+) = ?

From the given equation in the question we apply the operation rules as follows;

(-) × (-) = (+)

(-) × (+) = (-)

Therefore;

(-) × (+) = ? = (-)

? = (-)

c) (?) × (-) = ?

From the rules of multiplication, we have;

(-) × (-) = (+)

(+) × (-) = (-)

Therefore, when (?) = (-), ? = +, from which we have;

(?) ≠ ?, which is an error

Similarly, when (?) = (+), ? = -, therefore, (?) and ?, always changes sign

Therefore, the equation, (?) × (-) = ?, is undefined.

3 0

3 years ago

find the value of q for which point p(q,2) is the midpoint of the line segment joining the points q(-5,0) and r(-1,4)

Elza [17]

Answer:

step by step solution:

q=(-1+(-5))/2

q=-3

6 0

2 years ago

If A and B are two angles in standard position in Quadrant I, find cos( A +B ) for the given function values. sin A = 8/17 and c

horsena [70]

Answer:

Part 1) $cos(A + B) = \frac{140}{221}$

Part 2) $cos(A - B) = \frac{153}{185}$

Part 3) $cos(A - B) = \frac{84}{85}$

Part 4) $cos(A + B) = -\frac{36}{85}$

Part 5) $cos(A - B) = \frac{63}{65}$

Part 6) $cos(A+ B) = -\frac{57}{185}$

Step-by-step explanation:

<u><em>the complete answer in the attached document</em></u>

Part 1) we have

$sin(A)=\frac{8}{17}$

$cos(B)=\frac{12}{13}$

Determine cos (A+B)

we know that

$cos(A + B) = cos(A) cos(B)-sin(A) sin(B)$

step 1

Find the value of cos(A)

Remember that

$cos^2(A)+sin^2(A)=1$

substitute the given value

$cos^2(A)+(\frac{8}{17})^2=1$

$cos^2(A)+\frac{64}{289}=1$

$cos^2(A)=1-\frac{64}{289}$

$cos^2(A)=\frac{225}{289}$

$cos(A)=\pm\frac{15}{17}$

The angle A belong to the I quadrant, the cosine is positive

$cos(A)=\frac{15}{17}$

step 2

Find the value of sin(B)

Remember that

$cos^2(B)+sin^2(B)=1$

substitute the given value

$sin^2(B)+(\frac{12}{13})^2=1$

$sin^2(B)+\frac{144}{169}=1$

$sin^2(B)=1-\frac{144}{169}$

$sin^2(B)=\frac{25}{169}$

$sin(B)=\pm\frac{25}{169}$

The angle B belong to the I quadrant, the sine is positive

$sin(B)=\frac{5}{13}$

step 3

Find cos(A+B)

substitute in the formula

$cos(A + B) = \frac{15}{17} \frac{12}{13}-\frac{8}{17}\frac{5}{13}$

$cos(A + B) = \frac{180}{221}-\frac{40}{221}$

$cos(A + B) = \frac{140}{221}$

Part 2) we have

$sin(A)=\frac{3}{5}$

$cos(B)=\frac{12}{37}$

Determine cos (A-B)

we know that

$cos(A - B) = cos(A) cos(B)+sin(A) sin(B)$

step 1

Find the value of cos(A)

Remember that

$cos^2(A)+sin^2(A)=1$

substitute the given value

$cos^2(A)+(\frac{3}{5})^2=1$

$cos^2(A)+\frac{9}{25}=1$

$cos^2(A)=1-\frac{9}{25}$

$cos^2(A)=\frac{16}{25}$

$cos(A)=\pm\frac{4}{5}$

The angle A belong to the I quadrant, the cosine is positive

$cos(A)=\frac{4}{5}$

step 2

Find the value of sin(B)

Remember that

$cos^2(B)+sin^2(B)=1$

substitute the given value

$sin^2(B)+(\frac{12}{37})^2=1$

$sin^2(B)+\frac{144}{1,369}=1$

$sin^2(B)=1-\frac{144}{1,369}$

$sin^2(B)=\frac{1,225}{1,369}$

$sin(B)=\pm\frac{35}{37}$

The angle B belong to the I quadrant, the sine is positive

$sin(B)=\frac{35}{37}$

step 3

Find cos(A-B)

substitute in the formula

$cos(A - B) = \frac{4}{5} \frac{12}{37}+\frac{3}{5} \frac{35}{37}$

$cos(A - B) = \frac{48}{185}+\frac{105}{185}$

$cos(A - B) = \frac{153}{185}$

Part 3) we have

$sin(A)=\frac{15}{17}$

$cos(B)=\frac{3}{5}$

Determine cos (A-B)

we know that

$cos(A - B) = cos(A) cos(B)+sin(A) sin(B)$

step 1

Find the value of cos(A)

Remember that

$cos^2(A)+sin^2(A)=1$

substitute the given value

$cos^2(A)+(\frac{15}{17})^2=1$

$cos^2(A)+\frac{225}{289}=1$

$cos^2(A)=1-\frac{225}{289}$

$cos^2(A)=\frac{64}{289}$

$cos(A)=\pm\frac{8}{17}$

The angle A belong to the I quadrant, the cosine is positive

$cos(A)=\frac{8}{17}$

step 2

Find the value of sin(B)

Remember that

$cos^2(B)+sin^2(B)=1$

substitute the given value

$sin^2(B)+(\frac{3}{5})^2=1$

$sin^2(B)+\frac{9}{25}=1$

$sin^2(B)=1-\frac{9}{25}$

$sin^2(B)=\frac{16}{25}$

$sin(B)=\pm\frac{4}{5}$

The angle B belong to the I quadrant, the sine is positive

$sin(B)=\frac{4}{5}$

step 3

Find cos(A-B)

substitute in the formula

$cos(A - B) = \frac{8}{17} \frac{3}{5}+\frac{15}{17} \frac{4}{5}$

$cos(A - B) = \frac{24}{85}+\frac{60}{85}$

$cos(A - B) = \frac{84}{85}$

Part 4) we have

$sin(A)=\frac{15}{17}$

$cos(B)=\frac{3}{5}$

Determine cos (A+B)

we know that

$cos(A + B) = cos(A) cos(B)-sin(A) sin(B)$

step 1

Find the value of cos(A)

Remember that

$cos^2(A)+sin^2(A)=1$

substitute the given value

$cos^2(A)+(\frac{15}{17})^2=1$

$cos^2(A)+\frac{225}{289}=1$

$cos^2(A)=1-\frac{225}{289}$

$cos^2(A)=\frac{64}{289}$

$cos(A)=\pm\frac{8}{17}$

The angle A belong to the I quadrant, the cosine is positive

$cos(A)=\frac{8}{17}$

step 2

Find the value of sin(B)

Remember that

$cos^2(B)+sin^2(B)=1$

substitute the given value

$sin^2(B)+(\frac{3}{5})^2=1$

$sin^2(B)+\frac{9}{25}=1$

$sin^2(B)=1-\frac{9}{25}$

$sin^2(B)=\frac{16}{25}$

$sin(B)=\pm\frac{4}{5}$

The angle B belong to the I quadrant, the sine is positive

$sin(B)=\frac{4}{5}$

step 3

Find cos(A+B)

substitute in the formula

$cos(A + B) = \frac{8}{17} \frac{3}{5}-\frac{15}{17} \frac{4}{5}$

$cos(A + B) = \frac{24}{85}-\frac{60}{85}$

$cos(A + B) = -\frac{36}{85}$

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4 0

3 years ago

Write the slope-intercept form of the equation of the line that passes thrqugh (-4, -3) and is

Zielflug [23.3K]

Answer:

Step-by-step explanation:

Equation of the line y = mx + c

The new line is perpendicular to 2x - 3y = -5

- 3y = - 5 - 2x

3y = 2x + 5

y = $\frac{2x}{3} + \frac{5}{3}$

Since the lines are perpendicular to each other : $m_{1} . m_{2} = -1$

where $m_{1} \ slope \ of \ the \ given \ line \ and \ m_{2} \ slope \ of \ the \ new \line\\$ .

Slope of the given line

$m_{1} = \frac{2}{3}$

Slope of the new line

$\\\frac{2}{3}.m_{2} = -1\\ m_{2} = -1 . \frac{3}{2} = \frac{-3}{2}$

The equation to the new line passing through (-4, -3) and perpendicular to 2x - 3y = -5

$(y - y_{1}) = m_{2}(x-x_{2} )\\\\(y - (-4)) = \frac{-3}{2}(x - (-3)) \\\\(y + 4) = \frac{-3}{2} (x+ 3)\\\\2(y +4) = -3(x+3)\\\\2y + 8 = -3x -9 \\\\2y = -3x - 9 -8\\\\y = \frac{-3x}{2} - \frac{17}{2}$

8 0

3 years ago

This exercise uses the population growth model. The population of a certain species of fish has a relative growth rate of 1.9% p

BlackZzzverrR [31]

Answer:

n(t) = 11e^0.019t

Step-by-step explanation:

The estimated population in 2010 = 11,000,000 = initial population

The growth rate = 1.9% per year = 1.9/100 =

The exponential growth model follows the general form :

n(t) = ae^rt

a = Initial population ; r = growth rate ; t = period

Hence, we have ;

n(t) = 11e^0.019t in millions

5 0

3 years ago