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2 years ago

The graph shows the solution to which system of equations?

Mathematics

2 answers:

Kruka [31]2 years ago

5 0

The green line is the graph of [ y = -2x + 1 ].

The red line is the graph of [ y = 2x - 3 ].

Choice 'C' says this.

ss7ja [257]2 years ago

3 0

The answer is C hope this helps

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2 1/2 x 3 1/5 fast please

vova2212 [387]

Step-by-step explanation:

Let's turn each mixed fraction into an improper fraction:

$2 \frac{1}{2} =\frac{2}{2} +\frac{2}{2} +\frac{1}{2} =\frac{5}{2}$

$3\frac{1}{5} =\frac{5}{5} +\frac{5}{5} +\frac{5}{5} +\frac{1}{5} =\frac{16}{5}$

Multiply:

$\frac{5}{2} \times \frac{16}{5} =\frac{80}{10}$

Simplify:

$\frac{80 \div 10}{10 \div 10} =\frac{8}{1} =8$

8 is your simplified product.

4 0

2 years ago

HELP!!!!!! I believe its B but I am not sure! :(

n200080 [17]

Answer:

Step-by-step explanation:

When solving for x as an exponent, we need to use logarithms in order to undo the operation and rearrange the terms. We use log rules to bring down the exponent and solve. Logarithms are the inverse operations to exponents and vice versa. We have one special kind of logarithm called the natural logarithm whose base is e. We write it as ln. Since our base is e here, we will use the natural logarithm to rearrange and isolate x.

$e^{4x-1} =3$

We begin by applying the natural logarithm to each side.

$ln(e^{4x-1}) =ln(3)$

Log rules allow use to rearrange the exponent as multiplication in front of the log.

$(4x-1)ln(e) =ln(3)$

ln e as an inverse simplifies to 1.

$(4x-1)(1)=ln(3)$

We now apply the inverse operations for subtraction and multiplication.

$4x-1+1=1+ln(3)\\4x=1+ln(3)\\\frac{4x}{4} =\frac{1+ln3}{4} \\x =\frac{1+ln3}{4}$

Option A is correct.

5 0

2 years ago

Situation:A researcher in North America discoversa fossile that contains 65% of its originalamount of C-14..-ktN=NoeNo inital am

zheka24 [161]

SOLUTION

We have been given the equation of the decay as

$\begin{gathered} N=N_0e^{-kt} \\ where\text{ } \\ N_0=initial\text{ amount of C-14 at time t} \\ N=amount\text{ of C-14 at time t = 65\% of N}_0=0.65N_0 \\ k=0.0001 \\ t=time\text{ in years = ?} \end{gathered}$

So we are looking for the time

Plugging the values into the equation, we have