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3 years ago

What am I doing right

Mathematics

1 answer:

Oduvanchick [21]3 years ago

5 0

Although your -3 point is correct you need to graph the other point on (3,1) rather than (3,0)
so change that.

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SCORPION-xisa [38]

Binomial means there are two terms, so c² − 16 is the answer
mononomial means there is only one term, so -8c is the answer
written in descending form means from the highest degree (exponent) to the lowest, 4x² − 3x + 9 is the answer for question 3.
Question 4: use the same method for question 3, the third term is -7x

Question 5: not sure what this is: 2x2y. Is it 2x²y or 2x2y?
you need to add up the exponent of x and y. In this case, the highest degree is y's 4th power plus x's 1st power, which makes a 5.

7 0

3 years ago

Point C does not lie on line xy. Can point C lie in the same plane as live xy? Explain

uysha [10]

The answer is yes.

A line and a point outside the line define a unique plane. In other words, there is a single plane that contains a line and a point outside the line. If you are given line XY and point C outside the line, there is a single plane containing both line XY and point C. Therefore, line XY and point C must lie in the same plane.

4 0

3 years ago

Show work please \sqrt(x+12)-\sqrt(2x+1)=1

Nesterboy [21]

Answer:

$x=4$

Step-by-step explanation:

Given $\displaystyle\\\sqrt{x+12}-\sqrt{2x+1}=1$ , start by squaring both sides to work towards isolating $x$ :

$\displaystyle\\\left(\sqrt{x+12}-\sqrt{2x+1}\right)^2=\left(1\right)^2$

Recall $(a-b)^2=a^2-2ab+b^2$ and $\sqrt{a}\cdot \sqrt{b}=\sqrt{a\cdot b}$ :

$\displaystyle\\\left(\sqrt{x+12}-\sqrt{2x+1}\right)^2=\left(1\right)^2\\\implies x+12-2\sqrt{(x+12)(2x+1)}+2x+1=1$

Isolate the radical:

$\displaystyle\\x+12-2\sqrt{(x+12)(2x+1)}+2x+1=1\\\implies -2\sqrt{(x+12)(2x+1)}=-3x-12\\\implies \sqrt{(x+12)(2x+1)}=\frac{-3x-12}{-2}$

Square both sides:

$\displaystyle\\(x+12)(2x+1)=\left(\frac{-3x-12}{-2}\right)^2$

Expand using FOIL and $(a+b)^2=a^2+2ab+b^2$ :

$\displaystyle\\2x^2+25x+12=\frac{9}{4}x^2+18x+36$

Move everything to one side to get a quadratic:

$\displaystyle-\frac{1}{4}x^2+7x-24=0$

Solving using the quadratic formula:

A quadratic in $ax^2+bx+c$ has real solutions $\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ . In $\displaystyle-\frac{1}{4}x^2+7x-24$ , assign values:

$\displaystyle \\a=-\frac{1}{4}\\b=7\\c=-24$

Solving yields:

$\displaystyle\\x=\frac{-7\pm \sqrt{7^2-4\left(-\frac{1}{4}\right)\left(-24\right)}}{2\left(-\frac{1}{4}\right)}\\\\x=\frac{-7\pm \sqrt{25}}{-\frac{1}{2}}\\\\\begin{cases}x=\frac{-7+5}{-0.5}=\frac{-2}{-0.5}=\boxed{4}\\x=\frac{-7-5}{-0.5}=\frac{-12}{-0.5}=24 \:(\text{Extraneous})\end{cases}$