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3 years ago

Solve the system of inequalities:

Mathematics

1 answer:

Flura [38]3 years ago

5 0

Answer: (here is a link to the same problem.) LINK >>>. graph system of inequality and solve for system

brainly.com/question/11934937?utm_source=android&utm_medium=share&utm_campaign=question

Step-by-step explanation: Solve the system of inequalities:

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Andru [333]

$43$

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Valentin [98]

16.5
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3 years ago

Find one of the solutions to the following absolute value equation: 26 + 4) = 48

Sindrei [870]

Answer : MAYBE 18 (?)
I’m not 100% sure
Explanation: you always solve what’s is parentheses “( )” first so then you get the equation
30 = 48
Then ( this is the part I’m not sure about)
You subtract 30 out that means 30-30=0 & 48-30 = 18 so that’s where I found my answer but idk I haven’t done absolute value since Algebra 1 !!

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3 years ago

X÷-1.2= -0.3 i need help with this asap pls and thank you

Aleks [24]

0.3×1.2

=0.36

−0.3,1.2→Negative

−0.36

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2 years ago

You deposit 2000 in account A, which pays 2.25% annual interest compounded monthly. You deposit another 2000 in account b, which

stellarik [79]

To model this situation, we are going to use the compound interest formula: $A=P(1+ \frac{r}{n} )^{nt}$
where
$A$ is the final amount after $t$ years
$P$ is the initial deposit
$r$ is the interest rate in decimal form
$n$ is the number of times the interest is compounded per year
$t$ is the time in years

For account A:
We know for our problem that $P=2000$ and $r= \frac{2.25}{100} =0.0225$ . Since the interest is compounded monthly, it is compounded 12 times per year; therefore, $n=12$ . Lets replace those values in our formula:
$A=2000(1+ \frac{0.0225}{12} )^{12t}$

For account B:
$P=2000$ , $r= \frac{3}{100} =0.03$ , $n=12$ . Lest replace those values in our formula:
$A=2000(1+ \frac{0.03}{12} )^{12t}$

Since we want to find the time, $t$ , when the sum of the balance in both accounts is at least 5000, we need to add both accounts and set that sum equal to 5000:
 $2000(1+ \frac{0.0225}{12} )^{12t}+2000(1+ \frac{0.03}{12} )^{12t}=5000$

Now that we have our equation, we just need to solve for $t$ :
$2000[(1+ \frac{0.0225}{12} )^{12t}+(1+ \frac{0.03}{12} )^{12t}]=5000$
$(1+ \frac{0.0225}{12} )^{12t}+(1+ \frac{0.03}{12} )^{12t}= \frac{5000} {2000}$
$(1.001875)^{12t}+(1.0025 )^{12t}= \frac{5}
{2}$
$ln(1.001875)^{12t}+ln(1.0025 )^{12t}=ln( \frac{5} {2})$
$12tln(1.001875)+12tln(1.0025 )=ln( \frac{5} {2})$
$t[12ln(1.001875)+12ln(1.0025 )]=ln( \frac{5} {2})$
$t= \frac{ln( \frac{5}{2} )}{12ln(1.001875)+12ln(1.0025 )}$
$17.47$

We can conclude that after 17.47 years the sum of the balance in both accounts will be at least 5000.

5 0

3 years ago