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3 years ago

What is 27/4 as a mixed number?

Mathematics

2 answers:

Helga [31]3 years ago

5 0

$\frac{27}{4}=\frac{24+3}{4}=\frac{24}{4}+\frac{3}{4}=6+\frac{3}{4}=6\frac{3}{4}$

rodikova [14]3 years ago

3 0

27/4
Divide 4 into 27
I know 4 x 6= 24 +3 left over =27
6 3/4

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You are graphing Square ABCDABCDA, B, C, D in the coordinate plane. The following are three of the vertices of the square: A(4,

Vsevolod [243]

Answer:

D(4,-3)

Step-by-step explanation:

Given three of the vertices of the square: A(4, -7), B(8, -7),C(8, -3)

Let the coordinate of the fourth vertex be D(x,y).

We know that diagonals of a square are perpendicular bisector. So, the midpoint of both diagonals is the same.

The diagonals are BD and AC

Midpoint of BD = Midpoint of AC

$\left(\dfrac{8+x}{2},\dfrac{-7+y}{2}\right) =\left(\dfrac{4+8}{2},\dfrac{-7+(-3)}{2}\right)\\ \left(\dfrac{8+x}{2},\dfrac{y-7}{2}\right) =\left(\dfrac{12}{2},\dfrac{-10}{2}\right)\\ \left(\dfrac{8+x}{2},\dfrac{y-7}{2}\right) =\left(6,-5\right)\\$Therefore$:\\\dfrac{8+x}{2}=6\\8+x=12\\x=12-8\\x=4\\$Similarly$\\\dfrac{y-7}{2}=-5\\y-7=-5*2\\y-7=-10\\y=-10+7=-3$

The coordinates of the fourth vertex is D(4,-3)

8 0

3 years ago

Rectangle ABCD has vertices A(–6, –2), B(–3, –2),

Tatiana [17]

Answer:

A. T–4, 3(x, y)

Step-by-step explanation:

We are given that the transformation changes the rectangle ABCD to the rectangle A'B'C'D'.

The changes in the co-ordinates are given by,

A = (-6,-2) = (-6,-2) + (-4,3) = (-10,1)

B = (-3,-2) = (-3,-2) + (-4,3) = (-7,-1)

C = (-3,-6) = (-3,-6) + (-4,3) = (-7,-3)

D = (-6,-6) = (-6,-6) + (-4,3) = (-10,-3)

So, we see that the rectangle ABCD is transformed by the co-ordinates (-4,3).

That is, it is translated 4 units to the left and 3 units upwards.

Thus, option A is correct.

3 0

3 years ago

Solve: x^+4/x-1 = 5/x-1

AlexFokin [52]

The first step for solving this equation is to determine the defined range.
$\frac{ x^{4} }{x-1} = \frac{5}{x-1}$ , x ≠ 1
Remember that when the denominators of both fractions are the same,, you need to set the numerators equal. This will look like the following:
$x^{4}$ = 5
Take the root of both sides of the equation and remember to use both positive and negative roots.
x +/- $\sqrt[4]{5}$
Separate the solutions.
x = $\sqrt[4]{5}$ , x ≠ 1
x = - $\sqrt[4]{5}$
Check if the solution is in the defined range.
x = $\sqrt[4]{5}$
x = - $\sqrt[4]{5}$
This means that the final solution to your question are the following:
x = $\sqrt[4]{5}$
x = - $\sqrt[4]{5}$
Let me know if you have any further questions.
:)

7 0

3 years ago

Legs 3 and 4 are present what is the hypotenuse?

andrew11 [14]

Answer:

4 is hypotenuse

Step-by-step explanation:

6 0

3 years ago

Need answer please !!

Misha Larkins [42]

Answer:

Step-by-step explanation:

Least common denominator is 10

there are 10/10 portions in the whole circle

1/2 = 5/10

2/5 = 4/10

10/10 - 5/10 - 4/10 = 1/10

3 0

2 years ago