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2 years ago

I really need help with this question so can you tell me the answer please

Mathematics

1 answer:

LenaWriter [7]2 years ago

8 0

Answer:

x=1.7320508075688...... equal to 1.73

Step-by-step explanation:

equation is: y=x*x+1

so put 4 instead of y

4=x*x+1

so x*x=3

x=1.7320508075688......

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Robert owns two dogs. Each day, one dog eats

notsponge [240]

Answer:

1/3

Step-by-step explanation:

1) 1/6+1/6= 2/6

2) 2 and 6 can both be divide be 2.

2/2=1 and 6/2=3

6 0

3 years ago

One serving of a certain brand of yogurt contains 7g of fat. The low-fat version contains 38.25% less fat per serving. How man

Dafna1 [17]

Answer:

7g-100%

x-38,25

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Step-by-step explanation:

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4 years ago

A freight carrier charges $0.57 for a package weighing up to one ounce plus and additional $0.32 for each additional ounce or fr

NISA [10]

There are approximately 8 ounces (7.901) in 224 grams.

6 0

4 years ago

Read 2 more answers

(3, -3); m = -2<br> In standard form

xz_007 [3.2K]

In y=mx +b?

If so then -3=-2(3)+b
B=3

6 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