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3 years ago

Solve the systems of equations. y=x + 3 y = -x + 1 x = ??? y = ???

Mathematics

1 answer:

riadik2000 [5.3K]3 years ago

6 0

Answer:

x = -1

Step-by-step explanation:

-1 + 3 = 2

-(-1) + 1 = 2 or 1 + 1 = 2

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8.6 - 5.36 plz help me

olga nikolaevna [1]

Answer:

3.24

Step-by-step explanation:

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3 years ago

Read 2 more answers

Q1. In a fruit cocktail, for every 30 ml

lora16 [44]

Answer:

3/10

Step-by-step explanation:

$\frac{30}{30 + 20 + 50} = \frac{3}{10}$

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3 years ago

A student on a piano stool rotates freely with an angular speed of 2.85 rev/s . The student holds a 1.50 kg mass in each outstre

Vlad1618 [11]

Answer:r'=0.327 m

Step-by-step explanation:

Given

$N=2.85 rev/s$

angular velocity $\omega =2\pi N=17.90 rad/s$

mass of objects $m=1.5 kg$

distance of objects from stool $r_1=0.789 m$

Combined moment of inertia of stool and student $=5.53 kg.m^2$

Now student pull off his hands so as to increase its speed to 3.60 rev/s

$\omega _2=2\pi N_2$

$\omega _2=2\pi 3.6=22.62 rad/s$

Initial moment of inertia of two masses $I_0=2mr_^2$

$I_0=2\times 1.5\times (0.789)^2=1.867$

After Pulling off hands so that r' is the distance of masses from stool

$I_0'=2\times 1.5\times (r')^2$

Conserving angular momentum

$I_1\omega =I_2\omega _2$

$(5.53+1.867)\cdot 17.90=(5.53+I_o')\cdot 22.62$

$I_0'=1.397\times 0.791$

$I_0'=5.851$

$5.53+2\times 1.5\times (r')^2=5.851$

$2\times 1.5\times (r')^2=0.321$

$r'^2=0.107009$

$r'=0.327 m$

7 0

3 years ago

What is the answer plzzzzzz help a kid out

Komok [63]

Answer:

the answer is 8 by one of this question.

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4 years ago

Businesses deposit large sums of money into bank accountsImagine an account with $10 million dollars in it.

adoni [48]

again, let's assume daily compounding means 365 days per year.

$~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\\ P=\textit{original amount deposited}\dotfill & \$10000000\\ r=rate\to 2.12\%\to \frac{2.12}{100}\dotfill &0.0212\\ t=years\dotfill &1 \end{cases} \\\\\\ I = (10000000)(0.0212)(1)\implies \boxed{I=212000}$

$~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$10000000\\ r=rate\to 2.12\%\to \frac{2.12}{100}\dotfill &0.0212\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{daily, thus 365} \end{array}\dotfill &365\\ t=years\dotfill &1 \end{cases}$

$A=10000000\left(1+\frac{0.0212}{365}\right)^{365\cdot 1}\implies A\approx 10214256.88 \\\\\\ \underset{\textit{earned interest amount}}{10214256.88~~ - ~~10000000 ~~ \approx ~~ \boxed{214256.88}}$

what's their difference? well

$\stackrel{\textit{compounded daily}}{\approx 214256.88}~~ - ~~\stackrel{\textit{simple interest}}{212000}\implies \boxed{2256.88}$

7 0

2 years ago