Answer:
2 real solutions
Step-by-step explanation:
We can use the determinant, which says that for a quadratic of the form ax² + bx + c, we can determine what kind of solutions it has by looking at the determinant of the form:
b² - 4ac
If b² - 4ac > 0, then there are 2 real solutions. If b² - 4ac = 0, then there is 1 real solution. If b² - 4ac < 0, then there are 2 imaginary solutions.
Here, a = 6, b = -20, and c = 1. So, plug these into the determinant formula:
b² - 4ac
(-20)² - 4 * 6 * 1 = 400 - 24 = 376
Since 376 is clearly greater than 0, we know this quadratic has 2 real solutions.
<em>~ an aesthetics lover</em>
Answer:
x-y=4
and
x+y=32 is <u>your</u><u> </u><u>equation</u>
<u>d</u><u>i</u><u>f</u><u>f</u><u>e</u><u>r</u><u>e</u><u>n</u><u>c</u><u>e</u><u> </u><u>o</u><u>f</u><u> </u><u>t</u><u>w</u><u>o</u><u> </u><u>n</u><u>u</u><u>m</u><u>b</u><u>e</u><u>r</u><u> </u><u>=</u><u>x-y</u><u> </u><u>is</u><u> </u><u>equal</u><u> </u><u>to</u><u> </u><u>4</u>
<u>s</u><u>u</u><u>m</u><u> </u><u>o</u><u>f</u><u> </u><u>t</u><u>w</u><u>o</u><u> </u><u>n</u><u>u</u><u>m</u><u>b</u><u>e</u><u>r</u><u>=</u><u>x</u><u>+</u><u>y</u><u> </u><u>i</u><u>s</u><u> </u><u>e</u><u>q</u><u>u</u><u>a</u><u>l</u><u> </u><u>t</u><u>o</u><u> </u><u>3</u><u>2</u>
<u>x</u><u>-</u><u>y</u><u>=</u><u>4</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>(</u><u>1</u><u>)</u>
<u>x</u><u>+</u><u>y</u><u>=</u><u>3</u><u>2</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>(</u><u>2</u><u>)</u>
<u>a</u><u>d</u><u>d</u><u>i</u><u>n</u><u>g</u><u> </u><u>equation</u><u> </u><u>1</u><u> </u><u>a</u><u>n</u><u>d</u><u> </u><u>2</u>
<u>x</u><u>-</u><u>y</u><u> </u><u>+</u><u>x</u><u>+</u><u>y</u><u>=</u><u>4</u><u>+</u><u>3</u><u>2</u>
<u>2</u><u>x</u><u>=</u><u>3</u><u>6</u>
<u>x</u><u>=</u><u>3</u><u>6</u><u>/</u><u>2</u><u>=</u><u>1</u><u>6</u><u>a</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u>
<u>s</u><u>u</u><u>b</u><u>s</u><u>t</u><u>i</u><u>t</u><u>u</u><u>t</u><u>i</u><u>n</u><u>g</u><u> </u><u>value</u><u> </u><u>of</u><u> </u><u>y</u><u> </u><u>in</u><u> </u><u>equation</u><u> </u><u>1</u>
<u>1</u><u>6</u><u>-</u><u>y</u><u>=</u><u>4</u>
<u>1</u><u>6</u><u>-</u><u>4</u><u>=</u><u>y</u>
<u>y</u><u>=</u><u>1</u><u>2</u><u> </u><u>a</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u>
The point slope form, , is useful in situations involving slope and the location of one or more points. The standard form, , is usually easier to use when we need to make algebraic calculations. When needs or knowledge change, we can convert an equation from one form into another.
You make 2 2/7 an improper fraction (16/7) and then you subtract 8/7 from 16/7 and you get 1 1/7 (or 8/7) as your final answer
Do pemdas and once you got your answer from that keep checking