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Sever21 [200]
3 years ago
7

Which triangle would be most helpful in finding the distance between the points (–4, 3) and (1, –2) on the coordinate plane? On

a coordinate plane, a triangle has points (negative 4, 3), (1, negative 2), (5, 3). On a coordinate plane, a triangle has points (negative 4, 2), (negative 1, 4), (1, negative 3). On a coordinate plane, a triangle has points (negative 4, 3), (negative 4, negative 2), (1, negative 2). On a coordinate plane, a line is drawn between points (negative 4, negative 3) and (negative 1, 4).
Mathematics
2 answers:
Rainbow [258]3 years ago
8 0

Answer:

The last one is the answer

Step-by-step explanation:

I got took the test. ( :

strojnjashka [21]3 years ago
7 0

Answer:

On a coordinate plane, a triangle has points (negative 4, 3), (1, negative 2), (5, 3).

Step-by-step explanation:

This coordinate above gives us the best range of values

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Connor invested $8,000 in an account paying an interest rate of 82% compounded
fredd [130]

Answer:

Connor would have $83.37 more than Henry.

Step-by-step explanation:

Connor:

Initial investment= $8,000

Interest rate= 8.2% = 0.082/12= 0.00683

Number of periods= 12*5= 60 months

Henry:

Initial investment= $8,000

Interest rate= 8% = 0.08/365= 0.00022

Number of periods= 365*5= 1,825 days

<u>To calculate the future value, we need to use the following formula on each investment:</u>

FV= PV*(1+i)^n

Connor:

FV= 8,000*(1.00683^60)

FV= $12,035.35

Henry:

FV= 8,000*(1.00022^1,825)

FV= $11,951.98

Difference= 12,035.35 - 11,951.98= $83.37

Connor would have $83.37 more than Henry.

4 0
3 years ago
What are the possible numbers of positive, negative, and complex zeros of f(x)=x^6-x^5-x^4+4x^3-12x^2+12?
Vadim26 [7]
f(x)=x^6-x^5-x^4+4x^3-12x^2+12\\\\12:\{\pm1;\ \pm2;\ \pm3;\ \pm4;\ \pm6;\ \pm12\}\\1:\{\pm1\}\\\\Answer:\boxed{\{\pm1;\ \pm2;\ \pm3;\ \pm4;\ \pm6;\ \pm12\}}
8 0
3 years ago
JL is a diameter of circle K. If tangents to circle K are constructed through points L and J, what relationship would exist betw
Sergeu [11.5K]
I. Let t be the line tangent at point J. We know that a tangent line at a  point on a circle, is perpendicular to the diameter comprising that certain point.
So t is perpendicular to JL

let l be the tangent line through L. Then l is perpendicular to JL
 
ii. So t and l are 2 different lines, both perpendicular to line JL.

2 lines perpendicular to a third line, are parallel to each other, so the tangents t and l are parallel to each other.

Remark. Draw a picture to check the steps
6 0
3 years ago
Question
MrRa [10]
<h2>>> Answer </h2>

_______

\:

a10 = 10 + (10 - 1) × 1

a10 = 10 + 9 × 1

a10 = 10 + 9

a10 = 19

8 0
3 years ago
Pls help me I’m failing lolz :)
kaheart [24]
Answer: For no solution it would mean that there is no answer to the equation. For infinite solutions it would mean that any value for the variable would make the equation true.

Explanation: For example of no solution 6=5 as you can tell 6 is not the same to 5 so there is no solution. More examples are 2=3, 8=10, and 1=0. For infinite solutions examples are 2=2, 7=7, and 10=10 as we can see 2 does equal 2 and 7 equals 7 which makes it true and infinite.
6 0
3 years ago
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