All categories

3 years ago

What is 1 2/3 as an improper fraction

Mathematics

2 answers:

Troyanec [42]3 years ago

7 0

It is 5/3 ......................

makvit [3.9K]3 years ago

7 0

///He answer is 5/2///

You might be interested in

The intersection of roads M and N creates a 41.4 degree angle. What are the complimentary and supplementary angles of this inter

Keith_Richards [23]

Supplementary angles are two angles which sum up to 180 degrees while complementary angled are those the sum is 90 degrees. Therefore, the complementary of 41.4 is 90-41.4 = 48.6 and the supplement would be 180-41.4= 138.6.

3 0

3 years ago

841 rounded to the nearest 10 and 100

LUCKY_DIMON [66]

Answer:

rounded to the nearest 10 is 840

rounded to the nearest 100 is 800

Step-by-step explanation:

6 0

2 years ago

Find the standard form of the equation for the conic section represented by x^2 + 10x + 6y = 47.

Levart [38]

Answer:

The standard form of the equation for the conic section represented by $x^2\:+\:10x\:+\:6y\:=\:47$ is:

$4\left(-\frac{3}{2}\right)\left(y-12\right)=\left(x-\left(-5\right)\right)^2$

Step-by-step explanation:

We know that:

$4p\left(y-k\right)=\left(x-h\right)^2$ is the standard equation for an up-down facing Parabola with vertex at (h, k), and focal length |p|.

Given the equation

$x^2\:+\:10x\:+\:6y\:=\:47$

Rewriting the equation in the standard form

$4\left(-\frac{3}{2}\right)\left(y-12\right)=\left(x-\left(-5\right)\right)^2$

Thus,

The vertex (h, k) = (-5, 12)

Please also check the attached graph.

Therefore, the standard form of the equation for the conic section represented by $x^2\:+\:10x\:+\:6y\:=\:47$ is:

$4\left(-\frac{3}{2}\right)\left(y-12\right)=\left(x-\left(-5\right)\right)^2$

where

vertex (h, k) = (-5, 12)

7 0

2 years ago

Q3: Identify the graph of the equation and write and equation of the translated or rotated graph in general form. (Picture Provi

natta225 [31]

Answer:

b. circle; $2(x')^2+2(y')^2-5x'-5\sqrt{3}y'-6 =0$

Step-by-step explanation:

The given conic has equation;

$x^2-5x+y^2=3$

We complete the square to obtain;

$(x-\frac{5}{2})^2+(y-0)^2=\frac{37}{4}$

This is a circle with center;

$(\frac{5}{2},0)$

This implies that;

$x=\frac{5}{2},y=0$

When the circle is rotated through an angle of $\theta=\frac{\pi}{3}$ ,

The new center is obtained using;

$x'=x\cos(\theta)+y\sin(\theta)$ and $y'=-x\sin(\theta)+y\cos(\theta)$

We plug in the given angle with x and y values to get;

$x'=(\frac{5}{2})\cos(\frac{\pi}{3})+(0)\sin(\frac{\pi}{3})$ and $y'=--(\frac{5}{2})\sin(\frac{\pi}{3})+(0)\cos(\frac{\pi}{3})$

This gives us;

$x'=\frac{5}{4} ,y'=\frac{5\sqrt{3} }{4}$

The equation of the rotated circle is;

$(x'-\frac{5}{4})^2+(y'-\frac{5\sqrt{3} }{4})^2=\frac{37}{4}$

Expand;

$(x')^2+(y')^2-\frac{5\sqrt{3} }{2}y'-\frac{5}{2}x'+\frac{25}{4} =\frac{37}{4}$

Multiply through by 4; to get

$4(x')^2+4(y')^2-10\sqrt{3}y'-10x'+25 =37$

Write in general form;

$4(x')^2+4(y')^2-10x'-10\sqrt{3}y'-12 =0$

Divide through by 2.

$2(x')^2+2(y')^2-5x'-5\sqrt{3}y'-6 =0$

8 0

3 years ago

What's the circumference of a<br> circle with a diameter of 9 inches?<br> Use 3.14 for n.

Rainbow [258]

Answer:

28.26 in

Step-by-step explanation:

Radius = (1/2) diameter

Radius = (1/2)(9)

Radius = 4.5 in

Circumference = 2πr

Circumference = 2(3.14)(4.5)

Circumference = 28.26 in

8 0

3 years ago