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wel
3 years ago
6

The two equations will intersect at which of the following coordinates : y= -x + 3 y= 1/2x

Mathematics
1 answer:
kvv77 [185]3 years ago
4 0
Y=1/2 the points would intersect
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List all angles that are congruent to angle 1.
Tasya [4]
1, 3, 5, 7. 1 and 3 they are vertical meaning they are congruent. 3 and 5 are alternate interior angles meaning they are congruent. 7 and 5 are vertical so they are congruent.
8 0
3 years ago
WILL GIVE BRAINLIEST need help asap
Aleonysh [2.5K]

The location of the y value of R' after using the translation rule is -10

<h3>What will be the location of the y value of R' after using the translation rule? </h3>

The translation rule is given as:

(x + 4, y - 7)

The pre-image of R is located at (-17, -3)

Rewrite as

R = (-17, -3)

When the translation rule is applied, we have:

R' = (-17 + 4, -3 - 7)

Evaluate

R' = (-13, -10)

Remove the x coordinate

R'y = -10

Hence, the location of the y value of R' after using the translation rule is -10

Read more about translation at:

brainly.com/question/26238840

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5 0
2 years ago
Which is the sum of 14t/10t^2 + 5t^2+40t/10t^2
Softa [21]

Answer:

7t/5^3 + 5t^2 + 4t^3

Step-by-step explanation:

4 0
3 years ago
I’m having trouble on 10-11, if you can find the answer. Thank you!
PolarNik [594]

Answer:

10-11 is -1

Step-by-step explanation:

i don'T care much for attachments ;)

3 0
3 years ago
The first term of a geometric sequence is equal to a and the common ratio of the sequence is r.
ololo11 [35]

Answer: (a)  {a, ar, ar², ar³, ar⁴, ar⁵...}, (b)  arⁿ⁻¹

For part (a), the question gives us the first term a, and then asks us to apply the common ratio r six times.

In order for ar = a, the nth term of r will have to equal 0 (this implies that n is an exponent; thus giving us the first term a, as r = 1).

Since we use this method on the first term, we must use it for the next five, in which r gains an additional exponent for every consecutive value (nth term) thereafter.  

Ultimately getting: {a, ar, ar², ar³, ar⁴, ar⁵...}

For part (b), we first have to understand that the sequence does not start at 0, but at 1 for n. In order for ar = a, with n = 1, there needs to be subtraction of -1 within the exponent. So that arⁿ⁻¹

If we check and apply this, we can see that:

{ar¹⁻¹, ar²⁻¹, ar³⁻¹, ar⁴⁻¹, ar⁵⁻¹, ar⁶⁻¹...} = {a, ar, ar², ar³, ar⁴, ar⁵...} = arⁿ⁻¹ = Tn

4 0
3 years ago
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