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vichka [17]
3 years ago
5

Jermaine is going off to college and needs to figure out how much he needs to pay. Jermaine will be getting $1,000 in savings, $

500 from work study, and $13,000 from grants. If the total cost of tuition is $63,000, how much remaining tuition does Jermaine have?
Mathematics
2 answers:
Rom4ik [11]3 years ago
8 0

Answer:

48950$

Step-by-step explanation:

Temka [501]3 years ago
6 0

Answer:

$48,500

Step-by-step explanation:

63000-1000-500-13000=x

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(-1)×(-2)×10=? (this is integers)​
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Answer:the first one

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2 years ago
Please please use explanation with answer
KIM [24]

Answer:

\large\boxed{\theta=-\dfrac{\pi}{4}+k\pi\ or\ \theta=k\pi}\ for\ k\in\mathbb{Z}

Step-by-step explanation:

\tan^2\theta+\tan\theta=0\\\\\tan\theta(\tan\theta+1)=0\iff\tan\theta=0\ \vee\ \tan\theta+1=0\\\\\tan\theta=0\Rightarrow \theta=k\pi\ \text{for}\ k\in\mathbb{Z}\\\\\tan\theta+1=0\qquad\text{subtract 1 from both sides}\\\\\tan\theta=-1\Rightarrow\theta=-\dfrac{\pi}{4}+k\pi\ \text{for}\ k\in\mathbb{Z}

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3 years ago
1) Let f(x)=6x+6/x. Find the open intervals on which f is increasing (decreasing). Then determine the x-coordinates of all relat
brilliants [131]

Answer:

1) increasing on (-∞,-1] ∪ [1,∞), decreasing on [-1,0) ∪ (0,1]

x = -1 is local maximum, x = 1 is local minimum

2) increasing on [1,∞), decreasing on (-∞,0) ∪ (0,1]

x = 1 is absolute minimum

3) increasing on (-∞,0] ∪ [8,∞), decreasing on [0,4) ∪ (4,8]

x = 0 is local maximum, x = 8 is local minimum

4) increasing on [2,∞), decreasing on (-∞,2]

x = 2 is absolute minimum

5) increasing on the interval (0,4/9], decreasing on the interval [4/9,∞)

x = 0 is local minimum, x = 4/9 is absolute maximum

Step-by-step explanation:

To find minima and maxima the of the function, we must take the derivative and equalize it to zero to find the roots.

1) f(x) = 6x + 6/x

f\prime(x) = 6 - 6/x^2 = 0 and x \neq 0

So, the roots are x = -1 and x = 1

The function is increasing on the interval (-∞,-1] ∪ [1,∞)

The function is decreasing on the interval [-1,0) ∪ (0,1]

x = -1 is local maximum, x = 1 is local minimum.

2) f(x)=6-4/x+2/x^2

f\prime(x)=4/x^2-4/x^3=0 and x \neq 0

So the root is x = 1

The function is increasing on the interval [1,∞)

The function is decreasing on the interval (-∞,0) ∪ (0,1]

x = 1 is absolute minimum.

3) f(x) = 8x^2/(x-4)

f\prime(x) = (8x^2-64x)/(x-4)^2=0 and x \neq 4

So the roots are x = 0 and x = 8

The function is increasing on the interval (-∞,0] ∪ [8,∞)

The function is decreasing on the interval [0,4) ∪ (4,8]

x = 0 is local maximum, x = 8 is local minimum.

4) f(x)=6(x-2)^{2/3} +4=0

f\prime(x) = 4/(x-2)^{1/3} has no solution and x = 2 is crtitical point.

The function is increasing on the interval [2,∞)

The function is decreasing on the interval (-∞,2]

x = 2 is absolute minimum.

5) f(x)=8\sqrt x - 6x for x>0

f\prime(x) = (4/\sqrt x)-6 = 0

So the root is x = 4/9

The function is increasing on the interval (0,4/9]

The function is decreasing on the interval [4/9,∞)

x = 0 is local minimum, x = 4/9 is absolute maximum.

5 0
3 years ago
Which phrase describes the variable expression m + 5
Vikentia [17]

b) five more than m

"m in addition to 5"= m+5

5 0
3 years ago
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