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3 years ago

A sno-cone machine priced at $139 is on sale for 20% off. What is the price of the sno-cone machine after the discount? *

1 answer:

7 0

So percents are just written as 20% = 0.2 so keep that in mind

139 divided by 0.2 is 95 , which would be the price after the discount :)

hope this helps!
give me brainliest

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Tim purchases 3,000 shares in company X at $2.49 per share. The company subsequently announces a profit warning, and the share p

anygoal [31]

Answer:

$750

Step-by-step explanation:

Tim's share price changes by $2.24 -2.49 = -0.25, so the change in the value of his investment is ...

(3000 shares)(-0.25/share) = -$750

Tim takes a loss of $750 when he sells.

3 0

3 years ago

The ratio of incorrect to correct answers on Marco's math test was 4 to 6. If Marco missed 39 problems, how many did he get corr

lesantik [10]

39/6= 6.5 then u multiply
6.5 times 4 = 26
which 26 is your answer

3 0

4 years ago

The vertex of this parabola is at (-3, 6). Which of the following could be its equation?

pishuonlain [190]

Since the vertex is at (-3, 6), the equation is given by
y = a(x + 3) + 6

Therefore, option A is the correct answer.

6 0

3 years ago

Factor completely 36x2− 121.

lesantik [10]

The expression 36x² - 121 is in the 'difference of two squares form' which is (a² - b²)

Factorising (a² - b²) gives (a+b)(a-b)

Factorising 36x² - 121 gives (6x+11)(6x-11)

6 0

3 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.

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