All categories

2 years ago

Find the distance between the points (7,-5) and (2, -5).

Mathematics

2 answers:

vazorg [7]2 years ago

3 0

Answer:

15.62 units.

Step-by-step explanation:

To commence with, find for the vector.

Points = (7,-5) and (2,-5)

7 + 2

-5 + -5

-10

Vector = 12

-10

Magnitude = y² + x²

Points = (12,-10)

-10² + 12²

100 + 144

244

Square root of 244 is 15.62

Therefore, the distance between the points is 15.62 units.

Hope it helps.

iren2701 [21]2 years ago

3 0

Answer:

Step-by-step explanation:

√(x2 - x1)² + (y2 - y1)²

√(2 - 7)² + [-5 - (-5)]

√(-5)² + (0)²

√25

= 5

You might be interested in

Please answer as many as possible (DON'T ADD LINKS) If you have any questions put them in the comments. There will be another qu

DIA [1.3K]

1.G 2.A 3.I think A 4. F. 5. I hope F could be wrong.

3 0

2 years ago

A cell phone factory has a cost of production C(x) = 66x + 30,000 where x is the total number of attendees at the concert . The

dimulka [17.4K]

Answer:

8000-8500

Step-by-step explanation:

I can't manage to figure out the exact number without multiple choice answers. That is the closest I could get to finding out the answer. Good Luck!

8 0

2 years ago

Solve eqaution 4t+9=-8t-13

LiRa [457]

Simple.....

you have:

4t+9=-8t-13

isolate the variable...

4t+9=-8t-13
+8t +8t

12t+9=-13

12t+9=-13
-9 -9

12t=-22

t= $- \frac{11}{6}$

Thus, your answer.

3 0

2 years ago

A taxicab charges riders a flat fee of 3.75 and an additional 0.85 per mile of their trips.id Diego was charged 10.55 for his ta

alexandr402 [8]

Answer:

8 miles.

Step-by-step explanation:

As we know that the flat fee is 3.75, we can deduct it from the total, 10.55, leaving us with 6.80.

As we also know the rate per mile, we can divide the amount charged for the miles by the mile rate,

680/85 = 8

7 0

2 years ago

The expression (y^20) (y^-5)^2 is equivalent to y^n. What is the value of n? n = 27 n = 10 n = -200 n = 17

belka [17]

Answer:

n = 10

Step-by-step explanation:

Using the rules of exponents

$a^{m}$ × $a^{n}$ ⇔ $a^{(m+n)}$

$(a^{m}) ^{n}$ = $a^{mn}$

Given

( $y^{20}$ ) ( $(y^{-5}) ^{2}$

= $y^{20}$ × $y^{-10}$

= $y^{(20-10)}$

= $y^{10}$ → in the form $y^{n}$ with n = 10

8 0

2 years ago