Will give brainliest! number 4 it’s the pythagorean theorem
2 answers:
You might be interested in
Y = mx + b. Use this equation for most linear functions. Enter the information that is given to you.
Multiply the -3 and 4 to get 12.
Now divide the -12 from both sides.
-12 cancels out on the right side and your left with
Now re-enter the info back into the formula.
Then you've got your formula
Multiply the -3 and 4 to get 12.
Now divide the -12 from both sides.
-12 cancels out on the right side and your left with
Now re-enter the info back into the formula.
Then you've got your formula
This is an interesting question. Wish there were more questions of this kind.
This question helps us recognize the use of the vertex form of a quadratic expression/function.
The vertex form is in the form
The extreme value of the function occurs when x=h, i.e. when the first term vanishes, which leaves the value of the function equal to k. I.e. the vertex of the function is at (h,k).
For example, when
f(x)=5(x-4)^2+7
at x=4, f(4)=5(4-4)^2+7=5(0)+7=7,
in other words, f(x) is at its minimum when x=4, with a value of 7,
even simpler, the vertex of the function is at (4,7).
How do we know if it is a maximum or minimum?
If the first parameter "a" is positive, then at any other value than x=h, the function has a greater value than the vertex, hence a>0 => minimum.
Similarly, if the first parameter "a" is negative, then whenever x does not equal h, the value of the function is smaller, hence a<0 => maximum.
For example, with f(x)=5(x-4)^2+7, a=+5 >0, so (4,7) is a minimum.
Check: f(0)=5(0-4)^2+7=16+7=23 > 7, or (0,23) verifies that (4,7) is a minimum.
For given situations A, B, C, & D, we see that only one function is in the vertex form, i.e. y=a(x-h)+k, namely situation B
B. y=-3(x-2)^2+5, where a=-3, h=2 and k=5.
This means that the function has a maximum at the vertex (2,5). It has a maximum because a=-3 < 0, as discussed above.
Oh yes, enjoy the rest of April Fools Day! lol
This question helps us recognize the use of the vertex form of a quadratic expression/function.
The vertex form is in the form
The extreme value of the function occurs when x=h, i.e. when the first term vanishes, which leaves the value of the function equal to k. I.e. the vertex of the function is at (h,k).
For example, when
f(x)=5(x-4)^2+7
at x=4, f(4)=5(4-4)^2+7=5(0)+7=7,
in other words, f(x) is at its minimum when x=4, with a value of 7,
even simpler, the vertex of the function is at (4,7).
How do we know if it is a maximum or minimum?
If the first parameter "a" is positive, then at any other value than x=h, the function has a greater value than the vertex, hence a>0 => minimum.
Similarly, if the first parameter "a" is negative, then whenever x does not equal h, the value of the function is smaller, hence a<0 => maximum.
For example, with f(x)=5(x-4)^2+7, a=+5 >0, so (4,7) is a minimum.
Check: f(0)=5(0-4)^2+7=16+7=23 > 7, or (0,23) verifies that (4,7) is a minimum.
For given situations A, B, C, & D, we see that only one function is in the vertex form, i.e. y=a(x-h)+k, namely situation B
B. y=-3(x-2)^2+5, where a=-3, h=2 and k=5.
This means that the function has a maximum at the vertex (2,5). It has a maximum because a=-3 < 0, as discussed above.
Oh yes, enjoy the rest of April Fools Day! lol