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daser333 [38]
3 years ago
6

What is the value of -5 x 6

Mathematics
1 answer:
erik [133]3 years ago
3 0

Answer:

-30

Step-by-step explanation:

I used a calculator

You might be interested in
What is the sum of the multiples of 6 between 6 and 999?
riadik2000 [5.3K]
Your method is completely correct. The first term will be 6 and each subsequent term can be obtained by adding 6 to the previous one, meaning the common difference is 6. The number of terms is given by the highest number that is divisible by 6 and dividing it by 6; that is 996/6 = 166
Then we simply apply the formula for arithmetic sequence sum:
S = n/2 [2a₁ + (n - 1)d]
S = 166/2 [ 2(6) + (166 - 1)6]
S = 83,166
8 0
4 years ago
Daniel borrowed $5000 from the bank for 2 years at 9% simple interest per annum. Calculate how much interest Daniel is charged.
RideAnS [48]

Answer:

Daniel is charged $900

Step-by-step explanation:

This is because 9 percent of 5000 is 450. Multiply 450x2 and get 900.

6 0
3 years ago
Zero and negative exponentswrite in simplest for without zero or negative exponents- 17 ⁰
Valentin [98]
-17^0=(-17)^0=1\begin{gathered} (-2)^2\times(-2)^{-5}=(-2)^{2-5} \\ =(-2)^{-3} \\ =\frac{1}{(-2)^3} \\ =\frac{1}{-8} \end{gathered}

Thus, the final answers are 1 and -1/8.

4 0
1 year ago
Answers for part one and part two please!
il63 [147K]

When a customer has a 6 pound Chihuahua, the cost that will be charged is $5.00.

<h3>How to calculate the cost?</h3>

a. If a customer has a 6 pound Chihuahua, how much would you charge?

It should be noted that from the information given, for dogs that weigh 0 to 15 pounds, the amount charged is $5.00.

b. If a customer has a 65 pound Labrador, how much would you charge?

It should be noted that for dogs over 45 pounds, the amount that's charged is $9.00

There, the amount charged will be $9.00.

Learn more about cost on:

brainly.com/question/25109150

#SPJ1

3 0
2 years ago
Does there exist a di↵erentiable function g : [0, 1] R such that g'(x) = f(x) for all x 2 [0, 1]? Justify your answer
agasfer [191]

Answer:

No; Because g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Step-by-step explanation:

Assuming:  the function is f(x)=x^{2} in [0,1]

And rewriting it for the sake of clarity:

Does there exist a differentiable function g : [0, 1] →R such that g'(x) = f(x) for all g(x)=x² ∈ [0, 1]? Justify your answer

1) A function is considered to be differentiable if, and only if  both derivatives (right and left ones) do exist and have the same value. In this case, for the Domain [0,1]:

g'(0)=g'(1)

2) Examining it, the Domain for this set is smaller than the Real Set, since it is [0,1]

The limit to the left

g(x)=x^{2}\\g'(x)=2x\\ g'(0)=2(0) \Rightarrow g'(0)=0

g(x)=x^{2}\\g'(x)=2x\\ g'(1)=2(1) \Rightarrow g'(1)=2

g'(x)=f(x) then g'(0)=f(0) and g'(1)=f(1)

3) Since g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Because this is the same as to calculate the limit from the left and right side, of g(x).

f'(c)=\lim_{x\rightarrow c}\left [\frac{f(b)-f(a)}{b-a} \right ]\\\\g'(0)=\lim_{x\rightarrow 0}\left [\frac{g(b)-g(a)}{b-a} \right ]\\\\g'(1)=\lim_{x\rightarrow 1}\left [\frac{g(b)-g(a)}{b-a} \right ]

This is what the Bilateral Theorem says:

\lim_{x\rightarrow c^{-}}f(x)=L\Leftrightarrow \lim_{x\rightarrow c^{+}}f(x)=L\:and\:\lim_{x\rightarrow c^{-}}f(x)=L

4 0
4 years ago
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