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iVinArrow [24]
2 years ago
12

Question 5 of 5

Mathematics
1 answer:
Ksju [112]2 years ago
8 0
Rotation about the origin
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79=11x+13 <br> Solve each equation by using the properties of equality and justify each step
neonofarm [45]

Solution: X=6

Detailed explanation:

In this problem we're asked to find "x" or the unknown.

The first step is to subtract 11x from both sides.

---\mapsto\boldsymbol{-11x+79=13}

Next, subtract 79 from both sides.

---\mapsto\boldsymbol{-11x=-66}

To finish this off, divide both sides by -11.

---\mapsto\boldsymbol{x=6}

Voila! There's our answer!

<h2><em>Cheers!! ^-^</em></h2>

___________________

Hope I helped.  Best wishes.

Reach far. Aim high. Dream big.

___________________

8 0
2 years ago
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One bag of pretzels costs $3 , $5 bags of pretzels costs $10
dalvyx [7]
If 5 bags of pretzels cost $10, then the pretzels are $2 per bag.
8 0
3 years ago
Students are selling raffle tickets for a school fundraiser. They collect $25 for every 10 raffle tickets sold. Rain equation th
Free_Kalibri [48]

Answer: m= $25/10 r

Step-by-step explanation:

Let m= money

r= raffle ticket

Then according to the statement

m= $25 for 10 tickets

so 10 tickets= $25

Or the equation goes,

m= $25/10 r

7 0
3 years ago
Consider the equation below. f(x) = 2x3 + 3x2 − 336x (a) Find the interval on which f is increasing. (Enter your answer in inter
Vaselesa [24]

We have been given a function f(x)=2x^3+3x^2-336x. We are asked to find the interval on which function is increasing and decreasing.

(a). First of all, we will find the critical points of function by equating derivative with 0.

f'(x)=2(3)x^{2}+3(2)x^1-336

f'(x)=6x^{2}+6x-336

6x^{2}+6x-336=0

x^{2}+x-56=0

x^{2}+8x-7x-56=0

(x+8)-7(x+8)=0

(x+8)(x-7)=0

x=-8,x=7

So x=-8,7 are critical points and these will divide our function in 3 intervals (-\infty,-8)U(-8,7)U(7,\infty).

Now we will find derivative over each interval as:

f'(x)=(x+8)(x-7)

f'(-9)=(-9+8)(-9-7)=(-1)(-16)=16

Since f'(9)>0, therefore, function is increasing on interval (-\infty,-8).

f'(x)=(x+8)(x-7)

f'(1)=(1+8)(1-7)=(9)(-6)=-54

Since f'(1), therefore, function is decreasing on interval (-8,7).

Let us check for the derivative at x=8.

f'(x)=(x+8)(x-7)

f'(8)=(8+8)(8-7)=(16)(1)=16

Since f'(8)>0, therefore, function is increasing on interval (7,\infty).

(b) Since x=-8,7 are critical points, so these will be either a maximum or minimum.

Let us find values of f(x) on these two points.

f(-8)=2(-8)^3+3(-8)^2-336(-8)

f(-8)=1856

f(7)=2(7)^3+3(7)^2-336(7)

f(7)=-1519

Therefore, (-8,1856) is a local maximum and (7,-1519) is a local minimum.

(c) To find inflection points, we need to check where 2nd derivative is equal to 0.

Let us find 2nd derivative.

f''(x)=6(2)x^{1}+6

f''(x)=12x+6

12x+6=0

12x=-6

\frac{12x}{12}=-\frac{6}{12}

x=-\frac{1}{2}

Therefore, x=-\frac{1}{2} is an inflection point of given function.

3 0
3 years ago
Triangles practice, need help.
Sliva [168]
Y is 15 and x is 14 mark me as brainless
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