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umka21 [38]
3 years ago
14

How many fourths are there in 1 3/4?

Mathematics
2 answers:
Firlakuza [10]3 years ago
8 0

Answer:

12 one fourths

Step-by-step explanation:

There 3 × 1 1 / 4 = 3 × 4 = 12 one fourths in 3 .

kramer3 years ago
8 0

Answer:

7

Step-by-step explanation:

1 3/4 = (4+3)/4 = 7/4 = 7 fourths

You might be interested in
What are these questions ​
alina1380 [7]

Answer:

Theres quadrants so -3,9 would be lower about III

6,-6 would be IV

0,-1 1/2 That would be near the 0 obviously so II

3 0
2 years ago
Which of the following is not a possible number of solutions when solving a system of equations containing a quadratic and a lin
Kobotan [32]

Answer:

3

Step-by-step explanation:

We have a system with two equations, one equation is a quadratic function and the other equation is a linear function.

To solve this system we have to clear "y" in both equations, and then equal both equations, then we will have a quadratic function and equal it to zero:

ax^2+bx+c=0, a\neq 0

Then to resolve a quadratic equation we apply Bhaskara's formula:

x_{1}=\frac{-b+\sqrt{b^2-4ac} }{2a}

x_{2}=\frac{-b-\sqrt{b^2-4ac} }{2a}

It usually has two solutions.

But it could happen that \sqrt{b^2-4ac} then the equation doesn't have real solutions.

Or it could happen that there's only one solution, this happen when the linear equation touches the quadratic equation in one point.

And it's not possible to have more than 2 solutions. Then the answer ir 3.

For example:

In the three graphs the pink one is a quadratic function and the green one is a linear function.

In the first graph we can see that the linear function intersects the quadratic function in two points, then there are two solutions.

In the second graph we can see that the linear function intersects the quadratic function in only one point, then there is one solutions.

In the third graph we can see that the linear function doesn't intersect the quadratic function, then there aren't real solutions.

7 0
3 years ago
Consider the line represented by the equation 5x + 2y = 10. How is the slope of the line related to values of A, B, and C in sta
Flauer [41]

Step-by-step explanation:

5x + 2y = 10

Ax + By = C

A = 5

B = 2

C = 10

5 0
3 years ago
Drag the expressions into the boxes to correctly complete the table.
lora16 [44]

Answer:

SUMMARY:

x^4+\frac{5}{x^3}-\sqrt{x}+8                               →    Not a Polynomial

-x^5+7x-\frac{1}{2}x^2+9                           →    A Polynomial

x^4+x^3\sqrt{7}+2x^2-\frac{\sqrt{3}}{2}x+\pi              →    A Polynomial

\left|x\right|^2+4\sqrt{x}-2                                   →    Not a Polynomial

x^3-4x-3                                        →    A Polynomial

\frac{4}{x^2-4x+3}                                              →    Not a Polynomial

Step-by-step explanation:

The algebraic expressions are said to be the polynomials in one variable which consist of terms in the form ax^n.

Here:

n = non-negative integer

a = is a real number (also the the coefficient of the term).

Lets check whether the Algebraic Expression are polynomials or not.

Given the expression

x^4+\frac{5}{x^3}-\sqrt{x}+8

If an algebraic expression contains a radical in it then it isn’t a polynomial. In the given algebraic expression contains \sqrt{x}, so it is not a polynomial.

Also it contains the term \frac{5}{x^3} which can be written as 5x^{-3}, meaning this algebraic expression really has a negative exponent in it which is not allowed. Therefore, the expression x^4+\frac{5}{x^3}-\sqrt{x}+8 is not a polynomial.

Given the expression

-x^5+7x-\frac{1}{2}x^2+9

This algebraic expression is a polynomial. The degree of a polynomial in one variable is considered to be the largest power in the polynomial. Therefore, the algebraic expression is a polynomial is a polynomial with degree 5.

Given the expression

x^4+x^3\sqrt{7}+2x^2-\frac{\sqrt{3}}{2}x+\pi

in a polynomial with a degree 4. Notice, the coefficient of the term can be in radical. No issue!

Given the expression

\left|x\right|^2+4\sqrt{x}-2

is not a polynomial because algebraic expression contains a radical in it.

Given the expression

x^3-4x-3

a polynomial with a degree 3. As it does not violate any condition as mentioned above.

Given the expression

\frac{4}{x^2-4x+3}

\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}

Therefore, is not a polynomial because algebraic expression really has a negative exponent in it which is not allowed.

SUMMARY:

x^4+\frac{5}{x^3}-\sqrt{x}+8                               →    Not a Polynomial

-x^5+7x-\frac{1}{2}x^2+9                           →    A Polynomial

x^4+x^3\sqrt{7}+2x^2-\frac{\sqrt{3}}{2}x+\pi              →    A Polynomial

\left|x\right|^2+4\sqrt{x}-2                                   →    Not a Polynomial

x^3-4x-3                                        →    A Polynomial

\frac{4}{x^2-4x+3}                                              →    Not a Polynomial

3 0
3 years ago
The perimeter of a triangle is 140 cm the length of the second side is twice the length of the first side the length of the thir
kykrilka [37]

Perimeter = 140

Side 1 = x

Side 2 = 2x

Side 3 = 3x - 10

x + 2x + 3x - 10 = 140

6x - 10 = 140

6x = 150

x = 25

Side 1 = 25 cm

Side 2 = 50 cm

Side 3 = 65 cm

Hope this helps!! :)

6 0
3 years ago
Read 2 more answers
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