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3 years ago

You have 40 minutes to exercise at the gym, and you want to burn 300 calories total using both

1 answer:

6 0

A and b are the minutes spent on machine A and machine B, respectively.

a+b = 40
b = 40-a

8a + 6b = 300
8a + 6(40-a) = 300
8a + 240 - 6a = 300
2a = 60
a = 30
b = 40-a = 10

30 minutes on machine A and 10 minutes on machine B

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What coordinates are 7 units away from (2,−7)

AlekseyPX

One coordinate that is 7 units away from (2,-7) is (2,0). (2,0) is 7 units to the right of (2,-7)

Another coordinate is (9,-7). This one also works.

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3 years ago

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Consider the functions f and g defined by \[f(x) = \sqrt{\dfrac{x+1}{x-1}}\qquad\qquad\text{and}\qquad\qquad g(x) = \dfrac{\sqrt

tino4ka555 [31]

Answer:

The given functions are not same because the domain of both functions are different.

Step-by-step explanation:

The given functions are

$f(x)= \sqrt{\dfrac{x+1}{x-1}}$

$g(x) = \dfrac{\sqrt{x+1}}{\sqrt{x-1}}$

First find the domain of both functions. Radicand can not be negative.

Domain of f(x):

$\dfrac{x+1}{x-1}>0$

This is possible if both numerator or denominator are either positive or negative.

Case 1: Both numerator or denominator are positive.

$x+1\geq 0\Rightarrow x\geq -1$

$x-1\geq 0\Rightarrow x\geq 1$

So, the function is defined for x≥1.

Case 2: Both numerator or denominator are negative.

$x+1\leq 0\Rightarrow x\leq -1$

$x-1\leq 0\Rightarrow x\leq 1$

So, the function is defined for x≤-1.

From case 1 and 2 the domain of the function f(x) is (-∞,-1]∪[1,∞).

Domain of g(x):

$x+1\geq 0\Rightarrow x\geq -1$

$x-1\geq 0\Rightarrow x\geq 1$

So, the function is defined for x≥1.

So, domain of g(x) is [1,∞).

Therefore, the given functions are not same because the domain of both functions are different.

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3 years ago

I’m having trouble answering e,f, and g

kodGreya [7K]

What’s is it about can you tell me.?

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3 years ago

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How many vertices have an odd degree?

sashaice [31]

did you find the answers :( ??

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Solve the following problem and show the work. (x - 6)^2 + (y + 1)^2 = 40

Taya2010 [7]

Answer:

see explanation

Step-by-step explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r is the radius

(x - 6)² + (y + 1)² = 40 ← is the equation of a circle in standard form

centre = (6, - 1) and r = $\sqrt{40}$ = 2 $\sqrt{10}$

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3 years ago