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il63 [147K]
3 years ago
15

Find the x-intercepts of the parabola with vertex (-5,80) and y-intercept (0,-45). Write your answer in this form (x of 1 , y of

1), (x of 2, y of 2). If necessary round to the nearest hundereth
Mathematics
1 answer:
Ivan3 years ago
7 0
A parabola is a quadratic function, and a quadratic can be expressed in vertex form, which is:

y=a(x-h)^2+k, where (h,k) is the vertex (absolute maximum or minimum point of the quadratic)

In this case we are given that (h,k) is (-5,80) so we have so far:

y=a(x--5)^2+80

y=a(x+5)^2+80, we are also told that it passes through the point (0,-45) so:

-45=a(0+5)^2+80

-45=25a+80  subtract 80 from both sides

-125=25a  divide both sides by 25

-5=a, so now we know the complete vertex form is:

y=-5(x+5)^2+80

The x-intercepts occur when y=0 so:

0=-5(x+5)^2+80  add 5(x+5)^2 to both sides

5(x+5)^2=80  divide both sides by 5

(x+5)^2=16  take the square root of both sides

x+5=±√16  which is

x+5=±4  subtract 5 from both sides

x=-5±4 so the x-intercepts are:

x=-1 and -9


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1. Let a; b; c; d; n belong to Z with n &gt; 0. Suppose a congruent b (mod n) and c congruent d (mod n). Use the definition
lukranit [14]

Answer:

Proofs are in the explantion.

Step-by-step explanation:

We are given the following:

1) a \equi b (mod n) \rightarrow a-b=kn for integer k.

1) c \equi  d (mod n) \rightarrow c-d=mn for integer m.

a)

Proof:

We want to show a+c \equiv b+d (mod n).

So we have the two equations:

a-b=kn and c-d=mn and we want to show for some integer r that we have

(a+c)-(b+d)=rn. If we do that we would have shown that a+c \equiv b+d (mod n).

kn+mn   =  (a-b)+(c-d)

(k+m)n   =   a-b+ c-d

(k+m)n   =   (a+c)+(-b-d)

(k+m)n  =    (a+c)-(b+d)

k+m is is just an integer

So we found integer r such that (a+c)-(b+d)=rn.

Therefore, a+c \equiv b+d (mod n).

//

b) Proof:

We want to show ac \equiv bd (mod n).

So we have the two equations:

a-b=kn and c-d=mn and we want to show for some integer r that we have

(ac)-(bd)=tn. If we do that we would have shown that ac \equiv bd (mod n).

If a-b=kn, then a=b+kn.

If c-d=mn, then c=d+mn.

ac-bd  =  (b+kn)(d+mn)-bd

          =    bd+bmn+dkn+kmn^2-bd

          =           bmn+dkn+kmn^2

          =            n(bm+dk+kmn)

So the integer t such that (ac)-(bd)=tn is bm+dk+kmn.  

Therefore, ac \equiv bd (mod n).

//

3 0
3 years ago
Began with a temperature of -10 and ended after 8 minutes at a temperature of 0 which graph represents same relationship between
Alexus [3.1K]
Answer: The temperature, y, increases 5/4 or 1.25 degrees per minute, x.


(0,-10) (8,0)

use y1-y2/x1-x2 to find the slope

-10-0/0-8

= -10/-8

= 5/4
6 0
3 years ago
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