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3 years ago

Given the function

Mathematics

1 answer:

LiRa [457]3 years ago

8 0

Answer:

y=6+2x

Step-by-step explanation:

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A certain country has $10 billion in paper currency in circulation, and each day $50 million comes into the country's banks. The

mash [69]

Answer:

$\bf x(t)=10(1-e^{-0.005*t})$

Step-by-step explanation:

The differential equation

$\bf \displaystyle\frac{dx}{dt}=0.005(10-x)$

can be solved by separation of variables. Write the equation as

$\bf \displaystyle\frac{dx}{10-x}=0.005dt$

Integrate on both sides

$\bf \int\displaystyle\frac{dx}{10-x}=\int0.005dt\Rightarrow -ln(10-x)=0.005t+C\Rightarrow\\\\\Rightarrow ln(10-x)^{-1}=0.005t+C\Rightarrow (10-x)^{-1}=e^{0.005t}e^C$

where C is a constant.

$\bf e^C$ is also a constant and we will keep calling it C, (there is no reason to change the letter). We have then

$\bf (10-x)^{-1}=Ce^{0.005t}\Rightarrow \displaystyle\frac{1}{10-x}=Ce^{0.005t}\Rightarrow 10-x=\displaystyle\frac{1}{Ce^{0.005t}}\Rightarrow\\\\\Rightarrow x(t)=10-(1/C)e{-0.005t}$

(1/C) is a constant, and for the same reason we will keep calling it C. So the general solution is

$\bf x(t)=10-Ce^{-0.005t}$

Now, we use the initial condition x(0)=0

$\bf x(0)=10-Ce^{-0.005*0}=0\Rightarrow C=10$

and the particular solution is

$\bf x(t)=10-10e^{-0.005*t}=10(1-e^{-0.005*t})\\\\\boxed{x(t)=10(1-e^{-0.005*t})}$

7 0

3 years ago

36% of the tarts that a baker baked were strawberry tarts. He baked 98 more chocolate tarts than strawberry tarts. How many tart

poizon [28]

Answer:

350

Step-by-step explanation:

100% - 36% = 64% (we first find out the remaining percentage)

Now we know there are 98 more chocolate tarts than strawberry tarts. We have to find out that percentage.

So, 64% - 36% = 28%

So 98 of the chocolate tarts make up 28%

x --------> 100%

98 --------> 28%

Just follow the percentage formula.

98 * 100 = 9800

9800/28 = 350

3 0

3 years ago

Can anyone do 4,5 and 6 for me plz

cricket20 [7]

For question 4, $sinA =\frac{a}{c}, cosA= \frac{b}{c}, tan B = \frac{b}{a}, sin J = \frac{j}{l}, cosK = \frac{j}{l}, tanK = \frac{k}{j}.$

Question 5. Option a and question 6. Option j

Step-by-step explanation:

Step 1:

The three basic formula needed to solve these questions are:

$sin\theta = \frac{oppositeside}{hypotenuse} , cos\theta = \frac{adjacentside}{hypotenuse}, tan\theta= \frac{opposite side}{adjacent side}.$

Step 2:

Using the above formula, we solve the following values

$sinA = \frac{oppositeside}{hypotenuse}$ $=\frac{a}{c}.$

$cosA = \frac{adjacentside}{hypotenuse}$ $= \frac{b}{c}.$

$tanB= \frac{opposite side}{adjacent side}$ $= \frac{b}{a}.$

$sinJ = \frac{oppositeside}{hypotenuse}$ $= \frac{j}{l}.$

$cosK = \frac{adjacentside}{hypotenuse}$ $= \frac{j}{l}.$

$tanK= \frac{opposite side}{adjacent side}$ $= \frac{k}{j}.$

Step 3:

For question 5, The triangle's angle = 23°, opposite side = BC inches and hypotenuse = 4 inches.

$sin\theta= \frac{opposite side}{hypotenuse}. sin 23^{\circ}= \frac{BC}{4}, sin23^{\circ} = 0.3907,BC = (0.3907)(4) = 1.5628.$

SO BC is 1.5628 inches, rounding this off to the nearest tenth, we get BC = 1.6 inches which is option a.

Step 4:

For question 6, The triangle's angle = 50°, opposite side = QR m the adjacent side = 8.1 m.

$tan\theta= \frac{opposite side}{adjacentside}. tan 50^{\circ}=\frac{QR}{8.1}, tan50^{\circ} = 1.1917,QR = (1.1917)(8.1) = 9.65277$

SO QR is 9.65277 meters, rounding this off to the nearest tenth, we get QR = 9.7 inches which is option j.

7 0

3 years ago

Given the function:

anyanavicka [17]

You are correct. The answer is choice D

The only way for g(x) to be differentiable at x = 0 is for two things to happen
(1) g(x) is continuous at x = 0
(2) g ' (x) is continuous at x = 0

To satisfy property (1) above, the value of b must be 1. This can be found by plugging x = 0 into each piece of the piecewise function and solving for b.

So the piecewise function becomes
$g(x) = \begin{cases}x+1 \ \text{ if } \ x \ < \ 0\\ \cos(x) \text{ if } \ x \ge 0\end{cases}$
after plugging in b = 1

--------------------------------

Now differentiate each piece with respect to x to get
$g'(x) = \begin{cases}1 \ \text{ if } \ x \ < \ 0\\ -\sin(x) \text{ if } \ x \ge 0\end{cases}$
The first piece of g ' (x) is always going to be equal to 1. The second piece is equal to zero when x = 0
Because -sin(x) = -sin(0) = 0

So there's this disconnect on g ' (x) meaning its not continuous

Therefore, the value b = 1 will not work.

So there are no values of b that work to satisfy property (1) and property (2) mentioned at the top.

5 0

3 years ago

Kiran and Mai are trying to figure out if this equation is an identity, what do you think

ValentinkaMS [17]

Expanding the left side of the equation, it is found that since both sides are equal, yes, it is an identity.

An equality represents an identity if both sides are equal.

In this problem:

$(a - b)^4 = a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4$