Function 1: y = 4x + 5

Find the quotient 7192÷9

Lelechka [254]

Answer:

799

Step-by-step explanation:

5 0

3 years ago

2 miles equal how many feet explain your answer

Verdich [7]

2 miles equal 10560 feets since 1 mile equals 5280 feets and you want 2 you just add 5280 + 5280 = 10560 feets

6 0

3 years ago

Read 2 more answers

The coordinates of quadrilateral VWXY are given below. Find the coordinates of its image after a dilation with the given scale f

Lapatulllka [165]

Answer:

The coordinates of its image after dilation by a scale factor of 2 will be:

V'(12, 4)

W'(-4, 8)

X'(-6, -10)

Y'(6, -10)

Step-by-step explanation:

Given the quadrilateral VWXY

V(6, 2)

W(-2, 4)

X(-3, -5)

Y(3, -5)

A dilation by a scale factor of 2 means every coordinate of the original quadrilateral gets multiplied by the scale factor (x2).

<u>Using the formula of dilation by a scale factor of 2</u>

Dilation with scale factor 2, multiply by 2

(x, y) → (2x, 2y)

Thus, the coordinates of its image after a dilation will be:

(x, y) → (2x, 2y)

V(6,2) → (2(6), 2(2)) = V'(12, 4)

W(-2, 4) → (2(-2), 2(4)) = W'(-4, 8)

X(-3, -5) → (2(-3), 2(-5)) = X'(-6, -10)

Y(3, -5) → (2(3), 2(-5)) = Y'(6, -10)

Therefore, the coordinates of its image after dilation by a scale factor of 2 will be:

V'(12, 4)

W'(-4, 8)

X'(-6, -10)

Y'(6, -10)

8 0

3 years ago

The value of x for the exponential function f(x) = e^x is a set of real numbers?true or false

Fofino [41]

Answer:

a) f(x) = x^2

b) f(x) = x

c) any pair of numbers

Step-by-step explanation:

5 0

3 years ago

A model for the population in a small community after t years is given by P(t)=P0e^kt.

LUCKY_DIMON [66]

$\bf \textit{Amount of Population Growth}\\\\
A=Ie^{rt}\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
I=\textit{initial amount}\\
r=rate\to r\%\to \frac{r}{100}\\
t=\textit{elapsed time}\\
\end{cases}$

a)

so, if the population doubled in 5 years, that means t = 5. So say, if we use an amount for "i" or P in your case, to be 1, then after 5 years it'd be 2, and thus i = 1 and A = 2, let's find "r" or "k" in your equation.

$\bf \textit{Amount of Population Growth}\\\\
A=Ie^{rt}\qquad 
\begin{cases}
A=\textit{accumulated amount}\to &2\\
I=\textit{initial amount}\to &1\\
r=rate\\
t=\textit{elapsed time}\to &5\\
\end{cases}
\\\\\\
2=1\cdot e^{5r}\implies 2=e^{5r}\implies ln(2)=ln(e^{5r})\implies ln(2)=5r
\\\\\\
\boxed{\cfrac{ln(2)}{5}=r}\qquad therefore\qquad \boxed{A=e^{\frac{ln(2)}{5}\cdot t}} \\\\\\
\textit{how long to tripling?}\quad 
\begin{cases}
A=3\\
I=1
\end{cases}\implies 3=1\cdot e^{\frac{ln(2)}{5}\cdot t}$

$\bf 3=e^{\frac{ln(2)}{5}\cdot t}\implies ln(3)=ln\left( e^{\frac{ln(2)}{5}\cdot t} \right)\implies ln(3)=\cfrac{ln(2)}{5} t
\\\\\\
\cfrac{5ln(3)}{ln(2)}=t\implies 7.9\approx t$

b)

A = 10,000, t = 3

$\bf \begin{cases}
A=10000\\
t=3
\end{cases}\implies 10000=Ie^{\frac{ln(2)}{5}\cdot 3}\implies \cfrac{10000}{e^{\frac{3ln(2)}{5}}}=I
\\\\\\
6597.53955 \approx I$

3 0

4 years ago