All categories

3 years ago

Find the maximum and minimum values by evaluating the equation

Mathematics

1 answer:

dezoksy [38]3 years ago

7 0

Answer:

min = -9

max =3

Step-by-step explanation:

C = x-3y

x ≥0

x≤3

y≥0

y≤3

The minimum will be be when x is smallest and y is at its max

x =0 and y = 3

C = 0 - 3(3)

C = 0-9 = -9

The minimum is -9

The maximum occurs when x is largest and y is smallest

x =3 and y = 0

C = 3 - 3(0)

C = 3-0 = 3

The max is 3

You might be interested in

Rationalise the denominator by 6 root 7 minus 1

nikklg [1K]

Answer:

Assuming the numerator was 1.

$(6\sqrt{7}+1)/251$

Step-by-step explanation:

Assuming the numerator was 1.

$1/(6\sqrt{7} -1)\\ (6\sqrt{7}+1)/((6\sqrt{7} - 1)(6\sqrt{7} + 1))\\ (6\sqrt{7}+1)/((6\sqrt{7})^2 + 6\sqrt{7} - 7\sqrt{7} - 1)\\(6\sqrt{7}+1)/((6\sqrt{7})^2 - 1)\\(6\sqrt{7}+1)/((36*7 - 1)\\(6\sqrt{7}+1)/251$

7 0

3 years ago

Estimate the limit A B C D

Lera25 [3.4K]

Answer:

C. $\lim_{x\rightarrow f(x)=2$

Step-by-step explanation:

Given problem is $\lim_{x\rightarrow \frac{\pi}{4}}\frac{tan(x)-1}{x-\frac{\pi}{4}}$ .

Now we need to evaluate the given limit.

If we plug $\x=\frac{\pi}{4}$ , into given problem then we will get 0/0 form which is an indeterminate form so we can apply L Hospitals rule

take derivative of numerator and denominator

$\lim_{x\rightarrow \frac{\pi}{4}}\frac{tan(x)-1}{x-\frac{\pi}{4}}$

$=\lim_{x\rightarrow \frac{\pi}{4}}\frac{sec^2(x)-0}{1-0}$

$=\frac{sec^2(\frac{\pi}{4})-0}{1-0}$

$=\frac{(\sqrt{2})^2}{1}$

$=\frac{2}{1}$

Hence choice C is correct.

5 0

3 years ago

Amy's math notebook weighs 1/2 pound, her science notebook weighs 7/8 pound, and her history notebook weighs 3/4 pound. What are

QveST [7]

1/2 is equal to 50%, 3/4 is equal to 75%, and 7/8 is equal to 87.5 percent. Therefore, the order is 1/2, 3/4, and then 7/8.

4 0

3 years ago

Read 2 more answers

the probability of a number cube landing on 4 is 1/6 if a number cube is tossed 12 times how many times can it be expected to la

oksano4ka [1.4K]

The answer is twice, because 1/6 of 12 is 2

7 0

3 years ago

How many arrangements can be made using 2 letters of the word HYPERBOLAS if no letter is to be used more than once?

alisha [4.7K]

Answer:

90 arrangements

Step-by-step explanation:

Since there are no repititions of letters, there are unique 10 letters in total.

THe number of arrangements would be 2 permutation 10. We need the formula for permutation. That is:

$nPr=\frac{n!}{(n-r)!}\\$

Now, n = 10 [total] and r is 2, so we have:

$nPr=\frac{n!}{(n-r)!}\\\\_{10}P_{2}=\frac{10!}{(10-2)!}\\=\frac{10!}{8!}\\=\frac{10*9*8!}{8!}\\=10*9\\=90$

So, there can be 90 arrangements

3 0

3 years ago