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3 years ago

1. Walmart’s stock price started at $70. It went up by $2, then went down $4. What is the final stock price?

Mathematics

1 answer:

alina1380 [7]3 years ago

7 0

Answer:

1. 68

2. 5

3.7684

4. 7

5. 37

Step-by-step explanation:

1. number 1 is equivalent to the expression 70 - 4 + 2 which is equal to 68

2. the net gain is equal to all the losses and gains added together, so 5 + (-2) + 6 + (-3) which simplifies to 5 - 2 + 6 -3 = 6

3.7677 - (-7) = 7684

4. this expression represents the scenario 12 - 3 - 2; which is equal to 7.

5. 60 - 8 - 15 represents the situation. this equals 37

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4 years ago

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Answer: 1. x= -12 2. x= 12

Step-by-step explanation:

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3 years ago

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3 years ago

Use the given information to find (a) sin(s+t), (b) tan(s+t), and (c) the quadrant of s+t. cos s = - 12/13 and sin t = 4/5, s an

Anton [14]

Answer:

Part a) $sin(s + t) =-\frac{63}{65}$

Part b) $tan(s + t) = -\frac{63}{16}$

Part c) (s+t) lie on Quadrant IV

Step-by-step explanation:

[Part a) Find sin(s+t)

we know that

$sin(s + t) = sin(s) cos(t) + sin(t)cos(s)$

step 1

Find sin(s)

$sin^{2}(s)+cos^{2}(s)=1$

we have

$cos(s)=-\frac{12}{13}$

substitute

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

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

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

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

$sin(s)=\frac{5}{13}$ ---> is positive because s lie on II Quadrant

step 2

Find cos(t)

$sin^{2}(t)+cos^{2}(t)=1$

we have

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

substitute

$(\frac{4}{5})^{2}+cos^{2}(t)=1$

$(\frac{16}{25})+cos^{2}(t)=1$

$cos^{2}(t)=1-(\frac{16}{25})$

$cos^{2}(t)=\frac{9}{25}$

$cos(t)=-\frac{3}{5}$ is negative because t lie on II Quadrant

step 3

Find sin(s+t)

$sin(s + t) = sin(s) cos(t) + sin(t)cos(s)$

we have

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

$cos(t)=-\frac{3}{5}$

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

$cos(s)=-\frac{12}{13}$

substitute the values

$sin(s + t) = (\frac{5}{13})(-\frac{3}{5}) + (\frac{4}{5})(-\frac{12}{13})$

$sin(s + t) = -(\frac{15}{65}) -(\frac{48}{65})$

$sin(s + t) =-\frac{63}{65}$

Part b) Find tan(s+t)

we know that

tex]tan(s + t) = (tan(s) + tan(t))/(1 - tan(s)tan(t))[/tex]

we have

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

$cos(t)=-\frac{3}{5}$

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

$cos(s)=-\frac{12}{13}$

step 1

Find tan(s)

$tan(s)=sin(s)/cos(s)$

substitute

$tan(s)=(\frac{5}{13})/(-\frac{12}{13})=-\frac{5}{12}$

step 2

Find tan(t)

$tan(t)=sin(t)/cos(t)$

substitute

$tan(t)=(\frac{4}{5})/(-\frac{3}{5})=-\frac{4}{3}$

step 3

Find tan(s+t)

$tan(s + t) = (tan(s) + tan(t))/(1 - tan(s)tan(t))$

substitute the values

$tan(s + t) = (-\frac{5}{12} -\frac{4}{3})/(1 - (-\frac{5}{12})(-\frac{4}{3}))$

$tan(s + t) = (-\frac{21}{12})/(1 - \frac{20}{36})$

$tan(s + t) = (-\frac{21}{12})/(\frac{16}{36})$

$tan(s + t) = -\frac{63}{16}$

Part c) Quadrant of s+t

we know that

$sin(s + t) =negative$ ----> (s+t) could be in III or IV quadrant

$tan(s + t) =negative$ ----> (s+t) could be in III or IV quadrant

Find the value of cos(s+t)

$cos(s+t) = cos(s) cos(t) -sin (s) sin(t)$

we have

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

$cos(t)=-\frac{3}{5}$

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

$cos(s)=-\frac{12}{13}$

substitute

$cos(s+t) = (-\frac{12}{13})(-\frac{3}{5})-(\frac{5}{13})(\frac{4}{5})$

$cos(s+t) = (\frac{36}{65})-(\frac{20}{65})$

$cos(s+t) =\frac{16}{65}$

we have that

$cos(s+t)=positive$ -----> (s+t) could be in I or IV quadrant

$sin(s + t) =negative$ ----> (s+t) could be in III or IV quadrant

$tan(s + t) =negative$ ----> (s+t) could be in III or IV quadrant

therefore

(s+t) lie on Quadrant IV

4 0

3 years ago

What function do you know from calculus is such that its first derivative is itself?The above function is a solution of which of

Nataliya [291]

Answer:

a. y= e raise to power y

c. y = e^ky

Step-by-step explanation:

The first derivative is obtained by making the exponent the coefficient and decreasing the exponent by 1 . In simple form the first derivative of

x³ would be 2x³-² or 2x².

But when we take the first derivative of y= e raise to power y

we get y= e raise to power y. This is because the derivative of e raise to power is equal to e raise to power y.

On simplification

y= e^y

Applying ln to both sides

lny= ln (e^y)

lny= 1

Now we can apply chain rule to solve ln of y

lny = 1

1/y y~= 1

y`= y

therefore

derivative of e^y = e^y

The chain rule states that when we have a function having one variable and one exponent then we first take the derivative w.r.t to the exponent and then with respect to the function.

Similarly when we take the first derivative of y= e raise to power ky

we get y=k multiplied with e raise to power ky. This is because the derivative of e raise to a constant and power is equal to constant multiplied with e raise to power y.

On simplification

y= k e^ky

Applying ln to both sides

lny=k ln (e^y)

lny=ln k

Now we can apply chain rule to solve ln of y ( ln of constant would give a constant)

lny = ln k

1/y y~= k

y`=k y

therefore

derivative of e^ky = ke^ky

3 0

3 years ago