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3 years ago

The following table gives the result of a random sample of upper level students at Rocky Vista

Mathematics

1 answer:

AysviL [449]3 years ago

6 0

Answer:

nodgxdrfcvg I iiu v vvhgfrt

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Find the intersection points of the linear and quadratic functions shown below f(x)=2x-5 g(x)= x2+2x-21

Mice21 [21]

Answer:

$(-4,-13)$ and $(4,3)$ the intersection points.

Step-by-step explanation:

Intersection point of two functions is a common point which satisfies both the functions.

Given functions are,

$f(x)=2x-5$

$g(x)=x^2+2x-21$

For a common point of these functions,

$f(x)=g(x)$

$2x-5=x^2+2x-21$

$-5=x^2-21$

$0=x^2-16$

$x^2=16$

$x=-4,4$

For $x=-4$ ,

$f(-4)=g(-4)=2(-4)-5$

$=-13$

For $x=4$ ,

$f(4)=g(4)=2(4)-5$

$=3$

Therefore, $(-4,-13)$ and $(4,3)$ the intersection points.

3 0

3 years ago

1. Which of the following equations would NOT contain the point (4,12)?

vodka [1.7K]

Answer:

the answer is b hope this helps :)

6 0

3 years ago

Question 3

Katyanochek1 [597]

Answer:

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Step-by-step explanation:

5 0

3 years ago

What is the best math answer ever

BigorU [14]

Lol you like maths? Honestly, I don't like it much b r u h

6 0

3 years ago

Suppose you have some wolves and some dogs. Dogs are named and wolves are not. If you call a dog it will run to you. A wolf woul

Ira Lisetskai [31]

Answer:

4dogs and no wolves

Step-by-step explanation:

when a dog is taken care of (Trained) it could be let to run free in a definite region.

On a plane with one line we can place two dogs (one dog in each of the regions) and they will not fight as they will be separated by a line, and they will run free.

If we place three lines in general, then we would have four regions where 4 dogs can run free as they are separated by a line so as not to fight.

But for Wolves from the question it is deduced that: A wolf has to be caged all the time and no matter how well you train a wolve, it wont yield to instructions.

With these two, wolves wont be allowed on the plane at all.

8 0

3 years ago