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3 years ago

Express 3.01 as a mixed number

1 answer:

8 0

The three would be the whole number while 01 would be the fraction. And since the 1 is 2 places from the decimal it would mean that it is going to be out of 100. Therefore the answer is 3 1/100

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Henry invested $59,000 in an account paying an interest rate of 2% compounded quarterly. Assuming no deposits or withdrawals are

AfilCa [17]

Answer:

Step-by-step explanation:

$A=P(1+r/4)^{(4t)}\\ \\ A=P(1+.02/4)^{(4t)}\\ \\ A=59000(1.005)^{4(20)}\\ \\ A=\$ 87929.98$

5 0

3 years ago

If ten boys T-pose in the bathroom and X of them leave then 2 come back how many boys are in the bathroom now

oksano4ka [1.4K]

Answer:

360 no-scope

Step-by-step explanation:

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4 0

3 years ago

HELPPPP will mark brainliest

ArbitrLikvidat [17]

Answer:

12x-2

Step-by-step explanation:

You need to add 5x+1 and 7x-3.

5x+1+7x-3.

We need to organize this into common terms.

5x+7x+1-3

Now we simplify.

12x-2

That is the final answer!!

Hope this helped!!!! :D

6 0

3 years ago

Which polynomial is the perfect square trinomial? And why??? Help Please!!

lutik1710 [3]

<h3>Answer:</h3>

See the attached

<h3>Step-by-step explanation:</h3>

When you square the binomial (a -b), you get ...

... (a -b)² = a² -2ab +b²

That is, both the a² and b² terms have positive signs, and the middle term is twice the product of the roots of the squared terms.

The last two selections have negative signs on the constant, so cannot be perfect square trinomials.

The first selection has a middle term that is -ab, not -2ab, so it is not a perfect square trinomial, either.

The second selection is the correct one:

... 4a² -20a +25 = (2a +5)²

4 0

3 years ago

What value of b will cause the system to have an infinite number of solutions?

irga5000 [103]

b must be equal to -6 for infinitely many solutions for system of equations $y = 6x + b$ and $-3 x+\frac{1}{2} y=-3$

<u>Solution: </u>

Need to calculate value of b so that given system of equations have an infinite number of solutions

$\begin{array}{l}{y=6 x+b} \\\\ {-3 x+\frac{1}{2} y=-3}\end{array}$

Let us bring the equations in same form for sake of simplicity in comparison

$\begin{array}{l}{y=6 x+b} \\\\ {\Rightarrow-6 x+y-b=0 \Rightarrow (1)} \\\\ {\Rightarrow-3 x+\frac{1}{2} y=-3} \\\\ {\Rightarrow -6 x+y=-6} \\\\ {\Rightarrow -6 x+y+6=0 \Rightarrow(2)}\end{array}$

Now we have two equations

$\begin{array}{l}{-6 x+y-b=0\Rightarrow(1)} \\\\ {-6 x+y+6=0\Rightarrow(2)}\end{array}$

Let us first see what is requirement for system of equations have an infinite number of solutions

If $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y+c_{2}=0$ are two equation

$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ then the given system of equation has no infinitely many solutions.

In our case,

$\begin{array}{l}{a_{1}=-6, \mathrm{b}_{1}=1 \text { and } c_{1}=-\mathrm{b}} \\\\ {a_{2}=-6, \mathrm{b}_{2}=1 \text { and } c_{2}=6} \\\\ {\frac{a_{1}}{a_{2}}=\frac{-6}{-6}=1} \\\\ {\frac{b_{1}}{b_{2}}=\frac{1}{1}=1} \\\\ {\frac{c_{1}}{c_{2}}=\frac{-b}{6}}\end{array}$

As for infinitely many solutions $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

$\begin{array}{l}{\Rightarrow 1=1=\frac{-b}{6}} \\\\ {\Rightarrow6=-b} \\\\ {\Rightarrow b=-6}\end{array}$

Hence b must be equal to -6 for infinitely many solutions for system of equations $y = 6x + b$ and $-3 x+\frac{1}{2} y=-3$

8 0

3 years ago