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3 years ago

10

How many solutions does the nonlinear system of equations graphed below

2 answers:

Sunny_sXe [5.5K]3 years ago

7 0

Anser is D two hope this helps

vovikov84 [41]3 years ago

6 0

Answer:

<h2> D. Two</h2>

Step-by-step explanation:

Two common points of the graphs (two intersects) means two solutions.

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Whas does 16g+3<-17+15g equal

ipn [44]

Answer:

g∠-20

Step-by-step explanation:

4 0

3 years ago

In 1990, the population of Boston, Massachusetts was 574,283. In 2000, the population was 589,141. What is the percent of change

MrMuchimi

Answer:

Percent of change in population is $2.6$ %

Step-by-step explanation:

Given population of Boston is $574,283$ measured in year $1990$ .

And population of Boston in year $2000$ is $589,141$ .

We need to find percent of change in population.

So, percent change $=\frac{New\ population-Old\ population}{Old\ population}\times100$ %

Here old population is $574,283$ and new population is $589,141$

Plugging these value in formula we get,

Percent change $=\frac{589,141-574,283}{574,283}\times100\\\\=\frac{14858}{574,283}\times100=0.0258\times100\\\\=2.587$

Rounding to nearest tenth 2.6%

8 0

4 years ago

What is the unit rate for 3 boxes for $5

Andreas93 [3]

I think the answer to your question is
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6 0

3 years ago

Solve the following equations.<br> log2(x^2 − 16) − log^2(x − 4) = 1

Alenkasestr [34]

Answer:

$x=\frac{4*(2+e)}{e-2}$

Step-by-step explanation:

Let's rewrite the left side keeping in mind the next propierties:

$log(\frac{1}{x} )=-log(x)$

$log(x*y)=log(x)+log(y)$

Therefore:

$log(2*(x^{2} -16))+log(\frac{1}{(x-4)^{2} })=1\\ log(\frac{2*(x^{2} -16)}{(x-4)^{2}})=1$

Now, cancel logarithms by taking exp of both sides:

$e^{log(\frac{2*(x^{2} -16)}{(x-4)^{2}})} =e^{1} \\\frac{2*(x^{2} -16)}{(x-4)^{2}}=e$

Multiply both sides by $(x-4)^{2}$ and using distributive propierty:

$2x^{2} -32=16e-8ex+ex^{2}$

Substract $16e-8ex+ex^{2}$ from both sides and factoring:

$-(x-4)*(-8-4e-2x+ex)=0$

Multiply both sides by -1:

$(x-4)*(-8-4e-2x+ex)=0$

Split into two equations:

$x-4=0\hspace{3}or\hspace{3}-8-4e-2x+ex=0$

Solving for $x-4=0$

Add 4 to both sides:

$x=4$

Solving for $-8-4e-2x+ex=0$

Collect in terms of x and add $4e+8$ to both sides:

$x(e-2)=4e+8$

Divide both sides by e-2:

$x=\frac{4*(2+e)}{e-2}$

The solutions are:

$x=4\hspace{3}or\hspace{3}x=\frac{4*(2+e)}{e-2}$

If we evaluate x=4 in the original equation:

$log(0)-log(0)=1$

This is an absurd because log (x) is undefined for $x\leq 0$

If we evaluate $x=\frac{4*(2+e)}{e-2}$ in the original equation:

$log(2*((\frac{4e+8}{e-2})^2-16))-log((\frac{4e+8}{e-2}-4)^2)=1$

Which is correct, therefore the solution is:

$x=\frac{4*(2+e)}{e-2}$

6 0

4 years ago

How would u do -6-(-1)=

dsp73

The answer would be -5 because:

-6-(-1)

Keep Change Flip

-6 + 1

Different signs Subtract..so the answer would be 5 but then you add the negative to it because -6 is greater than the positive 1

5 0

3 years ago

Read 2 more answers