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3 years ago

7

I need help please thanks

1 answer:

maksim [4K]3 years ago

6 0

Answer:

Equation: $\frac{4}{3}x - 3$

Step-by-step explanation:

Slope = 4/3

Y-intercept = -3

Therefore;

Equation: y = mx + b

Equation: $\frac{4}{3}x - 3$

Hope this helps!

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Find the area, help please

mafiozo [28]

Answer:

Area- 22.5

Step-by-step explanation:

To find the area in a triangle you will want to multiple the base with the height then divide it by two. in this photo you have two options 9X5/2 or 7.5X6/2

remember- a= 1/2 b X h

4 0

3 years ago

A retired couple has up to $50,000 to invest. As their financial adviser, you recommend that they place at least $35,000 in Trea

patriot [66]

Answer:

See Explanation

Step-by-step explanation:

According to the Question,

Given that, A retired couple has up to $50,000 to invest. As their financial adviser, you recommend that they place at least $35,000 in Treasury bills yielding 1% and at most $10,000 in corporate bonds yielding 3%.

Therefore,

Let, x be the amount of money invested in Treasury bills

y be the amount invested in corporate bonds

Thus, the system of equations of linear inequalities is

x + y ≤ 50000

x ≥ 35000

0 ≤ y ≤ 10000

For, the graph of the system of equations of linear inequalities describes the possible amounts of each investment.

(Please find in the attachment)

3 0

3 years ago

The use of mathematical methods to study the spread of contagious diseases goes back at least to some work by Daniel Bernoulli i

harina [27]

Answer:

a

$y(t) = y_o e^{\beta t}$

b

$x(t) = x_o e^{\frac{-\alpha y_o }{\beta }[e^{-\beta t} - 1] }$

c

$\lim_{t \to \infty} x(t) = x_oe^{\frac{-\alpha y_o}{\beta } }$

Step-by-step explanation:

From the question we are told that

$\frac{dy}{y} = -\beta dt$

Now integrating both sides

$ln y = \beta t + c$

Now taking the exponent of both sides

$y(t) = e^{\beta t + c}$

=> $y(t) = e^{\beta t} e^c$

Let $e^c = C$

So

$y(t) = C e^{\beta t}$

Now from the question we are told that

$y(0) = y_o$

Hence

$y(0) = y_o = Ce^{\beta * 0}$

=> $y_o = C$

So

$y(t) = y_o e^{\beta t}$

From the question we are told that

$\frac{dx}{dt} = -\alpha xy$

substituting for y

$\frac{dx}{dt} = - \alpha x(y_o e^{-\beta t })$

=> $\frac{dx}{x} = -\alpha y_oe^{-\beta t} dt$

Now integrating both sides

$lnx = \alpha \frac{y_o}{\beta } e^{-\beta t} + c$

Now taking the exponent of both sides

$x(t) = e^{\alpha \frac{y_o}{\beta } e^{-\beta t} + c}$

=> $x(t) = e^{\alpha \frac{y_o}{\beta } e^{-\beta t} } e^c$

Let $e^c = A$

=> $x(t) =K e^{\alpha \frac{y_o}{\beta } e^{-\beta t} }$

Now from the question we are told that

$x(0) = x_o$

So

$x(0)=x_o =K e^{\alpha \frac{y_o}{\beta } e^{-\beta * 0} }$

=> $x_o = K e^{\frac {\alpha y_o }{\beta } }$

divide both side by $(K * x_o)$

=> $K = x_o e^{\frac {\alpha y_o }{\beta } }$

So

$x(t) =x_o e^{\frac {-\alpha y_o }{\beta } } * e^{\alpha \frac{y_o}{\beta } e^{-\beta t} }$

=> $x(t)= x_o e^{\frac{-\alpha * y_o }{\beta} + \frac{\alpha y_o}{\beta } e^{-\beta t} }$

=> $x(t) = x_o e^{\frac{\alpha y_o }{\beta }[e^{-\beta t} - 1] }$

Generally as t tends to infinity , $e^{- \beta t}$ tends to zero

so

$\lim_{t \to \infty} x(t) = x_oe^{\frac{-\alpha y_o}{\beta } }$

5 0

3 years ago

CAN SOMEONE HELP ME WITH NUMBER 7??<br> I WILL GIVE BRAINLIEST!

nirvana33 [79]

Answer:

tanx-(\sqrt(11))/(5)cosx>0

Step-by-step explanation:

Because to do this kind of trig you first have to know how to do basic trg

4 0

3 years ago

Which function has the largest amplitude?

frutty [35]

<h3>Answer: Choice A</h3>

The -3 out front of that function means |-3| = 3 is the amplitude, which is the largest of the four functions.

In general, the sine function is

y = A*sin(B(x-C)) + D

where |A| is the amplitude. In the first answer choice, A = -3 so |A| = |-3| = 3.

Cosine has the same format since cosine is just a phase shift of sine.

4 0

3 years ago