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3 years ago

What is 3 1/3 x 2 1/4 ?

2 answers:

8 0

$3 \frac{1}{3} \times 2 \frac{1}{4} \\ = \frac{(3 \times 3) + 1}{3} \times \frac{(2 \times 4) + 1}{4} \\ = \frac{9 + 1}{3} \times \frac{8 + 1}{4} \\ = \frac{10}{3} \times \frac{9}{4} \\ = \frac{90}{12} \\ = \frac{15}{2}$
The answer is 15/2 = 7.5

swat323 years ago

6 0

You need to change the mixed number into a improper fraction:

$3\dfrac{1}{3}=\dfrac{3\cdot3+1}{3}=\dfrac{10}{3}\\\\2\dfrac{1}{4}=\dfrac{2\cdot4+1}{4}=\dfrac{9}{4}\\\\3\dfrac{1}{3}\cdot2\dfrac{1}{4}=\dfrac{10}{3}\cdot\dfrac{9}{4}\\\\=\dfrac{5}{3}\cdot\dfrac{9}{2}=\dfrac{5}{1}\cdot\dfrac{3}{2}=\dfrac{15}{3}=7\dfrac{1}{2}=7.5$

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A company charges a shipping fee that is 4% of the purchase price for all items

Readme [11.4K]

4 percent of 66
0.04 = 4 percent
66 * 0.04 = 2.64
The shipping fee is $2.64

7 0

3 years ago

Leo had $91, which is 7 times as much money as Alison had. How much

NARA [144]

Answer:

$7x=91$

$13.

Step-by-step explanation:

Let x represent money that Alison has.

We have been given that Leo has 7 times as much money as Alison had. So the amount of money that Leo has, would be $7x$ .

We are also told that Leo had $91, so we will equate $7x$ with 91 and solve for x as:

$7x=91$

To solve for x, we will divide both sides by 7:

$\frac{7x}{7}=\frac{91}{7}$

$x=13$

Therefore, Alison has $13.

8 0

3 years ago

Read 2 more answers

(11-6)^2+6x2-(3+3)^2 A. 0 B. 1 C. 16 D. 26

Dahasolnce [82]

Answer: =1

Step-by-step explanation:

$\left(11-6\right)^2+6\cdot \:2-\left(3+3\right)^2$

$=5^2+6\cdot \:2-\left(3+3\right)^2$

$=5^2+6\cdot \:2-6^2$

$=25+6\cdot \:2-6^2$

$=25+6\cdot \:2-36$

$=25+12-36$

$=1$

3 0

2 years ago

Please help me do this question i cant seem to get it

Zielflug [23.3K]

Answer:

2 < x < 24

Step-by-step explanation:

Given 2 sides of a triangle then the third side x is in the range

difference of 2 sides < x < sum of 2 sides , that is

13 - 11 < x < 13 + 11

2 < x < 24

6 0

3 years ago

Use Stokes' Theorem to evaluate C F · dr F(x, y, z) = xyi + yzj + zxk, C is the boundary of the part of the paraboloid z = 1 − x

Serggg [28]

I assume $C$ has counterclockwise orientation when viewed from above.

By Stokes' theorem,

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

so we first compute the curl:

$\vec F(x,y,z)=xy\,\vec\imath+yz\,\vec\jmath+xz\,\vec k$

$\implies\nabla\times\vec F(x,y,z)=-y\,\vec\imath-z\,\vec\jmath-x\,\vec k$

Then parameterize $S$ by

$\vec r(u,v)=\cos u\sin v\,\vec\imath+\sin u\sin v\,\vec\jmath+\cos^2v\,\vec k$

where the $z$ -component is obtained from

$1-(\cos u\sin v)^2-(\sin u\sin v)^2=1-\sin^2v=\cos^2v$

with $0\le u\le\dfrac\pi2$ and $0\le v\le\dfrac\pi2$ .

Take the normal vector to $S$ to be

$\vec r_v\times\vec r_u=2\cos u\cos v\sin^2v\,\vec\imath+\sin u\sin v\sin(2v)\,\vec\jmath+\cos v\sin v\,\vec k$

Then the line integral is equal in value to the surface integral,

$\displaystyle\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}(-\sin u\sin v\,\vec\imath-\cos^2v\,\vec\jmath-\cos u\sin v\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{\pi/2}\int_0^{\pi/2}\cos v\sin^2v(\cos u+2\cos^2v\sin u+\sin(2u)\sin v)\,\mathrm du\,\mathrm dv=\boxed{-\frac{17}{20}}$

6 0

3 years ago