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3 years ago

Coordinates A(2,5) B(6,4) and C(6,0) are connected to form ABC

Mathematics

1 answer:

Oduvanchick [21]3 years ago

6 0

.....Yep.

Or is there a second part to this problem?

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50 POINTS!!! jamie will choose between two catering companies for an upcoming party. company A chargers a set-up fee of $500 plu

Shtirlitz [24]

Answer:

60 guests

Step-by-step explanation:

We'll say that the number of guests is n.

We're looking for the n where both of these expressions are equal:

Company A (500 set-up fee, 25 for each guest)

500 + 25 * n

Company B (200 set-up fee, 30 for each guest)

200 + 30 * n

500 + 25n = 200 + 30n

Subtract 200 from both sides.

300 + 25n = 30n

Subtract 25n from both sides.

300 = 5n

We'll reverse the sides.

5n = 300

Divide both sides by 5.

n = 300/5 = 60

Thus, both companies are equal when n is 60, or when there are 60 guests.

5 0

3 years ago

Read 2 more answers

(easy subtraction) Cassie wrote a story with 945 words. While she was revising her work, she erased 138 words. How many words do

lara [203]

Answer:

the answer to your question is 807

Step-by-step explanation:

have a nice day

8 0

3 years ago

How can I simplify these?

Schach [20]

Well the square root of 72 is 8.485 and the square root of 45 is 6.708. hope this helps :)

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3 years ago

an architect is designing a rectangular greenhouse. Along one walks is a 7-ft storage area and 5 sections for different kinds of

galben [10]

17+2n+=21 +2n is no solution

7 0

3 years ago

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What is wrong with the following equation?x2+xâˆ’20/xâˆ’4=x+5a. (xâˆ’4)(xâˆ’5)â‰ x2+xâˆ’20b. The left-hand side is not defined f

joja [24]

Answer:

The answer to this question can be defined as follows:

In part (i), the answer is "option d".

In part (ii), the answer is "option 2".

Step-by-step explanation:

Given:

Part (a)

$\Rightarrow \bold{\frac{x^2+x-20}{x-4}=x+5}\\\\$

Solve the above equation:

$\Rightarrow x^2+x-20=(x+5)(x-4)\\\\\Rightarrow x^2+x-20=(x^2-4x+5x-20)\\\\\Rightarrow x^2+x-20=x^2-4x+5x-20\\\\\Rightarrow \boxed{x^2+x-20=x^2+x-20}\\$

Given:

Part (b)

$\Rightarrow \bold{ \lim_{x \to \3} \frac{x^2+x-12}{x-3}= \lim_{x \to 3} (x+4)}\\\\$

Solve the above equation:

factor of $\Rightarrow x^2+x-20 =(x-3)(x+4)$

$\Rightarrow \lim_{x \to \3} \frac{(x-3)(x+4)}{x-3}= \lim_{x \to 3} (x+4)\\\\\Rightarrow \lim_{x \to \3} (x+4)= \lim_{x \to 3} (x+4)\\\\$

apply limit value:

$\Rightarrow (3+4)= (3+4)\\\\\Rightarrow 7= 7$

6 0

3 years ago