The Quadrilateral ABCD with coordinates (3, 1), (4, 4), (7, 5), (6, 2) is a rhombus because its length and width are both square root of 10 units and adjacent sides are not perpendicular. This can be seen by plotting the points you can either plot it or by using programs.
I hope this helps, God bless, and have a great day!
Brainliest is always appreciated :)
THE ANSWER IS -10P.
Step-by-step explanation:
USING BY MULTIPLICATION RULE.
minus × plus = minus
then minus × minus = plus.
so we add the 4p+6p= -10p.
why -10p is come it means THE BIGGER NUMBER SIGN IS MINUS. So it's comes -10p.
<em>HOPE</em><em> </em><em>IT</em><em> </em><em>HELP</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>
Answer:
There are 15 halves in 7 1/2.
Answer:
Step-by-step explanation:
<u>Given</u>
- Q(s) = -4s^3 + 7s^2 - 24;
- s = -4 and 1
<u>Verifying zeroes</u>
Q(-4) =
- -4(-4)^3 + 7(-4)^2 - 24 =
- 256 + 112 - 24 =
- 344
- Incorrect as 344 ≠ 0
Q(1) =
- -4(1)^3 + 7(1)^2 - 24 =
- -12 + 7 - 24 =
- -29
- Incorrect as -29 ≠ 0
For a fraction, if the denominator turns to 0, the fraction becomes undefined, and therefore, that's a restriction on a rational.
now, what values of "x" makes the denominator 0? let's check,
x+2 = 0
x = -2
so, if "x" ever becomes -2, then you'd get

so, the domain, or values "x" can take on safely, are any real numbers EXCEPT -2.