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3 years ago

What is the least number which is exactly divisible by 76 and 57

Mathematics

1 answer:

lesantik [10]3 years ago

4 0

Answer:

228

Step-by-step explanation:

Your phrasing is a little bit confusing but I'll assume that you're looking for the Least Common Multiple.

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Please explain to me what the Law Of Syllogism is. 100pts... Explain it well please...

3241004551 [841]

The Law Of Syllogism, which is also called Reasoning by Transitivity, is basically if a = b and b = c, then a = c, which is also the transitive property. Hence the alternate name. It can be used for mathematical deductions and is mainly used in the unit of logic statements, such as conditional and inverse statements. Here is an example:
If it is raining today, then I will wear my raincoat.
If I wear my raincoat today, then I will wear my boots.
When we use the Law of Syllogism, we get:
If it is raining today, then I will wear my boots.

Hope this helps and have a nice day!

7 0

3 years ago

Read 2 more answers

The water was pumped out of a backyard pond. What is the domain of this<br> graph?

nignag [31]

The correct answer to your question is C.

6 0

3 years ago

Read 2 more answers

What is bigger 1/5 or -4

kaheart [24]

Answer:

1/5

Step-by-step explanation:

The best way to think about this is on a number line. -4 would be on the left of 0 meaning it is less than 0 and $\frac{1}{5}$ is on the right side of 0 but it is less than 1.

8 0

3 years ago

I need help understanding square roots

Vlad [161]

I wonder if you mean to write $\sqrt[3]{256}$ in place of $3\sqrt{256}$ ...

If you meant what you wrote, then we have

$\sqrt{256}=\sqrt{16^2}=16$

$3\sqrt{256}=3\cdot16=48$

If you meant to write $\sqrt[3]{256}$ (the cube root of 256), then we could go on to have

$\sqrt[3]{256}=\sqrt[3]{16^2}=\sqrt[3]{(4^2)^2}=\sqrt[3]{4^4}=\sqrt[3]{4^3\cdot4}=4\sqrt[3]4$

6 0

3 years ago

Solve the system of equations using matrices. Use the Gauss- Jordan elimination method And find a solution set

KATRIN_1 [288]

$\begin{gathered} x+y+z=4 \\ x-y-z=0 \\ x-y+z=8 \\ \text{The system using matrix is} \\ \begin{bmatrix}{1} & {1} & {1}, & {4} \\ {1} & {-1} & {-1,} & {0} \\ {1} & {-1} & {1,} & {8}{}{}\end{bmatrix}\rightarrow F2=F2-F1=\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {-2} & {-2,} & {-4} \\ {1} & {-1} & {1,} & {8}{}{}\end{bmatrix} \\ \rightarrow F3=F3-F1=\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {-2} & {-2,} & {-4} \\ {0} & {-2} & {0,} & {4}{}{}\end{bmatrix}\rightarrow F2=-\frac{1}{2}F2 \\ =\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {1} & {1,} & {2} \\ {0} & {-2} & {0,} & {4}{}{}\end{bmatrix}\rightarrow F3=F3+2F2=\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {1} & {1,} & {2} \\ {0} & {0} & {2,} & {8}{}{}\end{bmatrix}\rightarrow F3=\frac{1}{2}F3 \\ =\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {1} & {1,} & {2} \\ {0} & {0} & {1,} & {4}{}{}\end{bmatrix}\rightarrow F2=F2-F3=\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {1} & {0,} & {-2} \\ {0} & {0} & {1,} & {4}{}{}\end{bmatrix} \\ \rightarrow F1=F1-F3=\begin{bmatrix}{1} & {1} & {0}, & {0} \\ {0} & {1} & {0,} & {-2} \\ {0} & {0} & {1,} & {4}{}{}\end{bmatrix}\rightarrow F1=F1-F2 \\ =\begin{bmatrix}{1} & {0} & {0}, & {2} \\ {0} & {1} & {0,} & {-2} \\ {0} & {0} & {1,} & {4}{}{}\end{bmatrix} \\ \text{Therefore, the solution is }x=2,\text{ y=-2 and z=4} \end{gathered}$

6 0

1 year ago