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2 years ago

Will give, Branliest.Thanks, :)

Mathematics

2 answers:

diamong [38]2 years ago

6 0

Answer:

X= -9

Y=7

Z=119

X=-70

Y=0.8

z=4.667

Step-by-step explanation:

X=-18-6= -24/3= -9

Y=24+11/5= 35/5= 7

Z=10+7=17*7=119

X=-2-5=-7*10=-70

Y=19-7=12/15=0.8

Z=-2+4=2*7=14/3=4.667

olasank [31]2 years ago

4 0

Answer:

-8; 7; 119; 30; 4/5 (0.8); 14/3 (4.6666...)(4 2/3)

Step-by-step explanation:

> 3x + 6 = -18 > 5y - 11 = 24

3x = -18 - 6 5y = 24 + 11

3x = -24 5y = 35

x = -8 y = 7

> z/7 - 7 = 10 > x/10 + 5 = -2

z/7 = 10 + 7 x/10 = -2 + 5

z/7 = 17 x/10 = 3

z/7 = 17 x = 30

z = 119

> 15y + 7 = 19 > 3z/7 - 4 = -2

15y = 19 - 7 3z/7 = -2 +4

15y = 12 3z/7 = 2

y = 4/5 (0.8) z = 14/3 (4.6666...) ( 4 2/3)

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Heyyyyy! Need Help with this! I will give brainliest to correct answer!! <3

Alex787 [66]

Answer:

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Step-by-step explanation:

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2 years ago

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3 years ago

Carey is trying to decide between two guitars. The first had an original price of $160, but is on sale for $100. The second had

Zolol [24]

Subtract the sale price from the original price then divide the difference by the original price.

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2 years ago

Read 2 more answers

Thanks for the help!

Bogdan [553]

Hope you could understand.

If you have any query, feel free to ask.

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2 years ago

I am having trouble with this relative minimum of this equation.<br>

Norma-Jean [14]

Answer:

So the approximate relative minimum is (0.4,-58.5).

Step-by-step explanation:

Ok this is a calculus approach. You have to let me know if you want this done another way.

Here are some rules I'm going to use:

$(f+g)'=f'+g'$ (Sum rule)

$(cf)'=c(f)'$ (Constant multiple rule)

$(x^n)'=nx^{n-1}$ (Power rule)

$(c)'=0$ (Constant rule)

$(x)'=1$ (Slope of y=x is 1)

$y=4x^3+13x^2-12x-56$

$y'=(4x^3+13x^2-12x-56)'$

$y'=(4x^3)'+(13x^2)'-(12x)'-(56)'$

$y'=4(x^3)'+13(x^2)'-12(x)'-0$

$y'=4(3x^2)+13(2x^1)-12(1)$

$y'=12x^2+26x-12$

Now we set y' equal to 0 and solve for the critical numbers.

$12x^2+26x-12=0$

Divide both sides by 2:

$6x^2+13x-6=0$

Compaer $6x^2+13x-6=0$ to $ax^2+bx+c=0$ to determine the values for $a=6,b=13,c=-6$ .

$a=6$

$b=13$

$c=-6$

We are going to use the quadratic formula to solve for our critical numbers, x.

$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-13 \pm \sqrt{13^2-4(6)(-6)}}{2(6)}$

$x=\frac{-13 \pm \sqrt{169+144}}{12}$

$x=\frac{-13 \pm \sqrt{313}}{12}$

Let's separate the choices:

$x=\frac{-13+\sqrt{313}}{12} \text{ or } \frac{-13-\sqrt{313}}{12}$

Let's approximate both of these:

$x=0.3909838 \text{ or } -2.5576505$ .

This is a cubic function with leading coefficient 4 and 4 is positive so we know the left and right behavior of the function. The left hand side goes to negative infinity while the right hand side goes to positive infinity. So the maximum is going to occur at the earlier x while the minimum will occur at the later x.

The relative maximum is at approximately -2.5576505.

So the relative minimum is at approximate 0.3909838.

We could also verify this with more calculus of course.

Let's find the second derivative.

$f(x)=4x^3+13x^2-12x-56$