Answer:
-3
Step-by-step explanation:
The length of a segment is
sqrt( ( y2-y1)^2 + (x2-x1) ^2) = 2 sqrt(10)
sqrt( ( a-4 - -1)^2 + (2a -4) ^2) = 2 sqrt(10)
sqrt( ( a-4 +1)^2 + (2a -4) ^2) = 2 sqrt(10)
Combine like terms
sqrt( ( a-3)^2 + (2a -4) ^2) = 2 sqrt(10)
Square each side
( a-3)^2 + (2a -4) ^2) = 4 *(10)
FOIL the left side
a^2 -6a +9 + 4a^2 -16a +16 = 40
Combine like terms
5a^2 -22a +25 = 40
Subtract 40 from each side
5a^2 -22a -15 =0
Factor
(a - 5) (5 a + 3) = 0
Using the zero product property
a-5 =0 5a +3 = 0
a = 5 5a = -3
a=5 a = -3/5
The product of the terms is
5 * -3/5 = -3
Answer:
The answer is -38
Step-by-step explanation:
Answer:
0.16666
Step-by-step explanation:
100/60
Answer:
See explanation below.
Step-by-step explanation:
Having students in the classroom who are at different levels of knowledge, interest, and ability can be managed by differentiated instruction. This method is a way of thinking that provides a framework where the instructor can set students with learning tasks that are at levels appropriate with the abilities and interests of each student. Each student can have a different type of class and different type of instruction with the differentiated instruction way of thinking.
A gifted and talented student might be assigned a higher math course, perhaps based on a math assessment for advanced placement. Then students that need to stay on the typical high school path of Algebra I, Geometry, Algebra II, and Trigonometry can do that.
Gifted students might take an alternate path with honors classes or trajectories involving Pre-Calculus or advanced placement Calculus, for example. In some instances, universities have allowed High School students to obtain college credit for some courses taken during High School.
Hope this helps! Have an Awesome Day!! :-)
(-1.2,-2.0) and (1.9,2.2) are the best approximations of the solutions to this system.
Option B
<u>Step-by-step explanation:</u>
Here, we have a graph of two functions from which we need to find the approximate value of common solutions. Let's find this:
First look at where we have intersection points, In first quadrant & in third quadrant.
<u>At first quadrant:</u>
Draw perpendicular lines from x-axis & y-axis from this point . After doing this we can clearly see that the perpendicular lines cut x-axis at x=1.9 and y-axis at y=2.2. So, one point is (1.9,2.2)
<u>At Third quadrant:</u>
Draw perpendicular lines from x-axis & y-axis from this point. After doing this we can clearly see that the perpendicular lines cut x-axis at x=-1.2 and y-axis at y= -2.0. So, other point is (-1.2,-2.0).
