Which is the graph of y = 3x + 1?
2 answers:
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Answer:
Hello! answer: x = 15
Step-by-step explanation:
Here is a visual i created of what the complementary angle would look like!
Actually Welcome to the Concept of the inequalities.
9p -9 <4p -2
===> 9p - 4p < -2 +9
===> 5p < 7
===> p < 7/5
hence the answer is P < 7/5
<h3>Answer: Choice C. 4*sqrt(6)</h3>
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Explanation:
Each cube has a side length of 4. Placed together like this, the total horizontal side combines to 4+8 = 8. This is the segment HP as shown in the diagram below. I've also added point Q to form triangle HPQ. This is a right triangle so we can find the hypotenuse QH
Use the pythagorean theorem to find QH
a^2 + b^2 = c^2
(HP)^2 + (PQ)^2 = (QH)^2
8^2 + 4^2 = (QH)^2
(QH)^2 = 64 + 16
(QH)^2 = 80
QH = sqrt(80)
Now we use segment QH to find the length of segment EH. Focus on triangle HQE, which is also a right triangle (right angle at point Q). Use the pythagorean theorem again
a^2 + b^2 = c^2
(QH)^2 + (QE)^2 = (EH)^2
(EH)^2 = (QH)^2 + (QE)^2
(EH)^2 = (sqrt(80))^2 + (4)^2
(EH)^2 = 80 + 16
(EH)^2 = 96
EH = sqrt(96)
EH = sqrt(16*6)
EH = sqrt(16)*sqrt(6)
EH = 4*sqrt(6), showing the answer is choice C
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A shortcut is to use the space diagonal formula. As the name suggests, a space diagonal is one that goes through the solid space (rather than stay entirely on a single face; which you could possibly refer to as a planar diagonal or face diagonal).
The space diagonal formula is
d = sqrt(a^2+b^2+c^2)
which is effectively the 3D version of the pythagorean theorem, or a variant of such.
We have a = HP = 8, b = PQ = 4, and c = QE = 4 which leads to...
d = sqrt(a^2+b^2+c^2)
d = sqrt(8^2+4^2+4^2)
d = sqrt(96)
d = sqrt(16*6)
d = sqrt(16)*sqrt(6)
d = 4*sqrt(6), we get the same answer as before
The space diagonal formula being "pythagorean" in nature isn't a coincidence. Repeated uses of the pythagorean theorem is exactly why this is.
Answer:
As per the statement:
The path that the object takes as it falls to the ground can be modeled by:
h =-16t^2 + 80t + 300
where
h is the height of the objects and
t is the time (in seconds)
At t = 0 , h = 300 ft
When the objects hit the ground, h = 0
then;
-16t^2+80t+300=0
For a quadratic equation: ax^2+bx+c=0 ......[1]
the solution for the equation is given by:
On comparing the given equation with [1] we have;
a = -16 ,b = 80 and c = 300
then;
Simplify:
= -2.5 sec and
= 7.5 sec
Time can't be in negative;
therefore, the time it took the object to hit the ground is 7.5 sec
