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RSB [31]
3 years ago
10

BONUS QUESTION: Earn up to 2 bonus points for this question.

Mathematics
1 answer:
svlad2 [7]3 years ago
8 0

Answer: I would say try and use your 150 stars you’ll save more money for that day and earn more stars.

Step-by-step explanation:Hope this helps!

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Does anyone know what the answer is..?
Leno4ka [110]
It should be 5 because that's the smallest number and A is the first point
3 0
3 years ago
#19<br> The distance between the points (4, 1) and (9, 1) is
kirill115 [55]

Hello!

Given the two points, (x_{1}, y_{1}) and (x_{2}, y_{2}), and to find the distance between these two points is found by using the formula:

d = \sqrt{(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}}

(x_{1}, y_{1}) is assigned to one the points, in this case, is (4, 1).

(x_{2}, y_{2}) is assigned to other point, which is (9, 1).

Then, plug in these values into the formula and solve.

d = \sqrt{(9-4)^{2}-(1-1)^{2}}

d = \sqrt{5^{2}-0^{2}}

d = \sqrt{25}

d = 5

Therefore, the distance between the two points is 5.

4 0
3 years ago
Consider the two functions. Which statement is true?
madreJ [45]

Answer:

Step-by-step explanation:

The Answer is  

D) The rate of change for function B is greater than the rate of change for function A.

7 0
3 years ago
Read 2 more answers
The sum of two consecutive even integers is 122. what are the integers?​
Volgvan

Answer:

If the numbers were the same, they'd both be 61

So, try 60 and 62.

x + x+2 = 122

2x = 120

x = 60

Hope that helps!

4 0
2 years ago
Read 2 more answers
Find an equation of the plane that contains the points p(5,−1,1),q(9,1,5),and r(8,−6,0)p(5,−1,1),q(9,1,5),and r(8,−6,0).
topjm [15]
Given plane passes through:
p(5,-1,1), q(9,1,5), r(8,-6,0)

We need to find a plane that is parallel to the plane through all three points, we form the vectors of any two sides of the triangle pqr:
pq=p-q=<5-9,-1-1,1-5>=<-4,-2,-4>
pr=p-r=<5-8,-1-6,1-0>=<-3,5,1>

The vector product pq x pr gives a vector perpendicular to both pq and pr.  This vector is the normal vector of a plane passing through all three points
pq x pr
=
  i   j   k
-4 -2 -4
-3  5  1
=<-2+20,12+4,-20-6>
=<18,16,-26>

Since the length of the normal vector does not change the direction, we simplify the normal vector as
N = <9,8,-13>

The required plane must pass through all three points.
We know that the normal vector is perpendicular to the plane through the three points, so we just need to make sure the plane passes through one of the three points, say q(9,1,5).

The equation of the required plane is therefore
Π :  9(x-9)+8(y-1)-13(z-5)=0
expand and simplify, we get the equation
Π  :  9x+8y-13z=24

Check to see that the plane passes through all three points:
at p: 9(5)+8(-1)-13(1)=45-8-13=24
at q: 9(9)+8(1)-13(5)=81+9-65=24
at r: 9(8)+8(-6)-13(0)=72-48-0=24
So plane passes through all three points, as required.

3 0
3 years ago
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