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kogti [31]
3 years ago
6

What is the solution to this inequality?

Mathematics
2 answers:
miss Akunina [59]3 years ago
8 0
X>12*4
X>48 is the correct answer :)
trasher [3.6K]3 years ago
6 0
X/4 < 12
multiply both sides by 4,
4x/4 < 12*4
x < 48
so, option A is your answer

hope this helps!
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Oranges are sold in a supermarket at £1.42 per KG.
liraira [26]

Answer:

1.5kg = £2.13

Step-by-step explanation:

1kg = $1.42

1000gm = $1.42

0.5kg = $0.71

500gm = $0.71

1. 42+ 0.71 = £2.13  = 1.5kg = 1500gm

5 0
3 years ago
What value does "h" have in this equation for this to be an infinite solutions equation. 6 (hx + 5) + 4 = -4(8 - 4(8 – 18x) 18x)
Sever21 [200]

We have the following:

6(hx+5)+4=-4(8-18x)+66

A system of linear equations has infinite solutions when the graphs are exactly the same line.

for this to happen, on both sides the value of x must be equal:

\begin{gathered} 6\cdot h=4\cdot18 \\ h=4\cdot\frac{18}{6} \\ h=12 \end{gathered}

replacing:

\begin{gathered} 6\cdot(12x+5)+4=-4\cdot(8-18x)+66 \\ 72x+30+4=-32+72x+66 \\ 7x+34=7x+34 \\ 0=0 \\ \text{true for all x} \end{gathered}

4 0
1 year ago
What is the net-price rate when a designer swimsuit, originally priced at $85 is discounted 25/60?
Gre4nikov [31]
Designer's swimsuit is originally priced at 85 dollars.
NOw, it applied a discount which is 25/60. First, let's find out how much the discount rate.
=> 25 / 60 = 0.42 * 100 = 42% is the discount.
=> 85 dollars * 0.42 = 35.7 is the discount.
=> 85 dollars - <span>35.7 dollars = 49.3 dollars is now the price.</span>
4 0
3 years ago
Can you define f(0, 0) = c for some c that extends f(x, y) to be continuous at (0, 0)? If so, for what value of c? If not, expla
Ahat [919]

(i) Yes. Simplify f(x,y).

\displaystyle \frac{x^2 - x^2y^2 + y^2}{x^2 + y^2} = 1 - \frac{x^2y^2}{x^2 + y^2}

Now compute the limit by converting to polar coordinates.

\displaystyle \lim_{(x,y)\to(0,0)} \frac{x^2y^2}{x^2+y^2} = \lim_{r\to0} \frac{r^4 \cos^2(\theta) \sin^2(\theta)}{r^2} = 0

This tells us

\displaystyle \lim_{(x,y)\to(0,0)} f(x,y) = 1

so we can define f(0,0)=1 to make the function continuous at the origin.

Alternatively, we have

\dfrac{x^2y^2}{x^2+y^2} \le \dfrac{x^4 + 2x^2y^2 + y^4}{x^2 + y^2} = \dfrac{(x^2+y^2)^2}{x^2+y^2} = x^2 + y^2

and

\dfrac{x^2y^2}{x^2+y^2} \ge 0 \ge -x^2 - y^2

Now,

\displaystyle \lim_{(x,y)\to(0,0)} -(x^2+y^2) = 0

\displaystyle \lim_{(x,y)\to(0,0)} (x^2+y^2) = 0

so by the squeeze theorem,

\displaystyle 0 \le \lim_{(x,y)\to(0,0)} \frac{x^2y^2}{x^2+y^2} \le 0 \implies \lim_{(x,y)\to(0,0)} \frac{x^2y^2}{x^2+y^2} = 0

and f(x,y) approaches 1 as we approach the origin.

(ii) No. Expand the fraction.

\displaystyle \frac{x^2 + y^3}{xy} = \frac xy + \frac{y^2}x

f(0,y) and f(x,0) are undefined, so there is no way to make f(x,y) continuous at (0, 0).

(iii) No. Similarly,

\dfrac{x^2 + y}y = \dfrac{x^2}y + 1

is undefined when y=0.

5 0
1 year ago
Who can help with probability
Aleks04 [339]

What's the question? I'll help with it as much as I can :)


7 0
3 years ago
Read 2 more answers
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