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Anestetic [448]
3 years ago
7

Whats the answer to 1/5 ×25=

Mathematics
2 answers:
kiruha [24]3 years ago
7 0

Answer:

5

Step-by-step explanation:

\frac{1}{5}*25

=>\frac{1}{1}*5 ; [∵ 25÷5=5]

=>1*5

=>5

crimeas [40]3 years ago
5 0
5 is the correct answer
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. Solve each given equation and show your work. Tell whether each equation has one solution, an infinite number of solutions, or
DerKrebs [107]

No Solution


No Solution


All real numbers

Any number in place of x makes the equation true


6 0
3 years ago
After 5.75 hours, Kyler had finished painting 62.5% of his house. He needs to paint a total of 4,000 square feet on his house. A
MissTica
First, we are going to find the painted area of his house. We know that the total area he needs to paint is 4000 square meters, so 4000 is the area of this house. Now, to find the 62.5% of that area, we are going to multiply it by \frac{62.5}{100}:
Painted area= (4000)(\frac{62.5}{100})=2500

Now, to find <span>rate, in square feet per hour, we are going to divide the painted area by the number of hours: 
Rate=</span>\frac{2500}{5.75} =434.78

We can conclude that Kyler paints at a rate of 434.78 square feet per hour.
8 0
3 years ago
Is 2/3 greater than 10/18
SVEN [57.7K]
2/3 is greater than 10/18
4 0
3 years ago
Read 2 more answers
Let V be a vector space of dimension 4. Determine if each statement is true or false. (a) Any set of 5 vectors in V must be line
son4ous [18]

Answer:

a) True

b) False

c) False

d) False

e) True

Step-by-step explanation:

a) Each basis of V has four vectors. Then any set of 5 vectors must be linear dependent (LD).

b) Suppose that \{v_1,v_2,v_3,v_4\} is a basis of V. Considere the set A=\{v_1,\lambda_1v_1,\lambda_2v_1,v_2,v_3\} where \lambda_1, \lambda_2 are scalars. The set has 5 vectors but V\neq span(A) because v_4 is not belong to A and v_4  is linear independent of v_1

c)  Suppose that \{v_1,v_2,v_3,v_4\} is a basis of V. Considere the set A=\{v_1,\lambda_1v_1,\lambda_2v_1,\lambda_3v_1\} where \lambda_1, \lambda_2,\lambda_3 are scalars. A has four nonzero vectors but isn't a basis because is a LD set.

d)  Suppose that \{v_1,v_2,v_3,v_4\} is a basis of V. Considere the set A=\{v_1,\lambda_1v_1,\lambda_2v_1\} where \lambda_1, \lambda_2 are scalars. A has 3 nonzero vectors but isn't a basis because is a LD set.

e)  Since any basis of V must have 4 elements, then a set of three vectors cannot generate V.

4 0
3 years ago
What is the quadratic portion in this quadratic equation 7x2-12x+16=0
alexdok [17]

Answer:

x = 6/7 + (2 i sqrt(19))/7 or x = 6/7 - (2 i sqrt(19))/7

Step-by-step explanation:

Solve for x:

7 x^2 - 12 x + 16 = 0

Hint: | Write the quadratic equation in standard form.

Divide both sides by 7:

x^2 - (12 x)/7 + 16/7 = 0

Hint: | Solve the quadratic equation by completing the square.

Subtract 16/7 from both sides:

x^2 - (12 x)/7 = -16/7

Hint: | Take one half of the coefficient of x and square it, then add it to both sides.

Add 36/49 to both sides:

x^2 - (12 x)/7 + 36/49 = -76/49

Hint: | Factor the left hand side.

Write the left hand side as a square:

(x - 6/7)^2 = -76/49

Hint: | Eliminate the exponent on the left hand side.

Take the square root of both sides:

x - 6/7 = (2 i sqrt(19))/7 or x - 6/7 = -(2 i sqrt(19))/7

Hint: | Look at the first equation: Solve for x.

Add 6/7 to both sides:

x = 6/7 + (2 i sqrt(19))/7 or x - 6/7 = -(2 i sqrt(19))/7

Hint: | Look at the second equation: Solve for x.

Add 6/7 to both sides:

Answer:  x = 6/7 + (2 i sqrt(19))/7 or x = 6/7 - (2 i sqrt(19))/7

6 0
4 years ago
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