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3 years ago

If 65%=x/100 then x =?

1 answer:

5 0

Answer:
65
Why?
This can be rewritten as;
0.65 = x/100
It can be rewritten again to get x on the left side alone;
-x/100 = -0.65
Now, multiply each term by -100 to further distinguish x;
x = 0.65*100
Now, solve;
x = 65
There’s your answer.
Check:
0.65 = 65/100
0.65 = 0.65

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For the equation: D = -0.306c + 0.9979, what is the value of D when c= 1.67? Express answer with the correct number of sig figs?

Gekata [30.6K]

Answer:

Step-by-step explanation:

Let's solve your equation step-by-step.

d=(−0.306)(1.67)+0.9979

Step 1: Simplify both sides of the equation.

d=(−0.306)(1.67)+0.9979

d=−0.51102+0.9979

d=(−0.51102+0.9979) (Combine Like Terms)

d=0.48688

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What is the domain of the following function?

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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2 years ago

Read 2 more answers

Select all the expressions that are equivalent to -36x+54y−90.

CaHeK987 [17]

Answer:

b) -18(2x -3y + 5)

d) 18(-2x + 3y - 5)

e) -2(18x - 27y + 45)

Step-by-step explanation:

a) -9( 4x - 6y - 10) = -36x + 54y + 90 not equal

b) -18(2x - 3y + 5) = -36x + 54y - 90 equal

c) -6(6x + 9y - 15) = -36x - 54y + 90 not equal

d) 18(-2x + 3y - 5) = -36x + 54y - 90 equal

e) -2(18x - 27y + 45) = -36x + 54y - 90 equal

f) 2(-18x + 54y - 90) = -36x + 108y - 180 not equal

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2 years ago

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Determine whether the set of vectors is a basis for ℛ3. Given the set of vectors , decide which of the following statements is t

schepotkina [342]

Answer:

(A) Set A is linearly independent and spans $R^3$ . Set is a basis for $R^3$ .

Step-by-Step Explanation

<u>Definition (Linear Independence)</u>

A set of vectors is said to be linearly independent if at least one of the vectors can be written as a linear combination of the others. The identity matrix is linearly independent.

<u>Definition (Span of a Set of Vectors)</u>

The Span of a set of vectors is the set of all linear combinations of the vectors.

<u>Definition (A Basis of a Subspace).</u>

A subset B of a vector space V is called a basis if: (1)B is linearly independent, and; (2) B is a spanning set of V.

Given the set of vectors $A= \left(\begin{array}{[c][c][c][c]}1 & 0 & 0 & 0\\ 0 & 1 & 0 & 1\\ 0 & 0 & 1 & 1\end{array} \right)$ , we are to decide which of the given statements is true:

In Matrix $A= \left(\begin{array}{[c][c][c][c]}(1) & 0 & 0 & 0\\ 0 & (1) & 0 & 1\\ 0 & 0 & (1) & 1\end{array} \right)$ , the circled numbers are the pivots. There are 3 pivots in this case. By the theorem that The Row Rank=Column Rank of a Matrix, the column rank of A is 3. Thus there are 3 linearly independent columns of A and one linearly dependent column. $R^3$ has a dimension of 3, thus any 3 linearly independent vectors will span it. We conclude thus that the columns of A spans $R^3$ .

Therefore Set A is linearly independent and spans $R^3$ . Thus it is basis for $R^3$ .

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