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3 years ago

Karen began her tennis game at 9:30. She played for two and a half hours.What time did she finish?

Mathematics

1 answer:

Ilya [14]3 years ago

7 0

12:00 because 2:30+9:30= 11:60 but because that's not how time works its 12:00

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Which of the triangles are right triangles?

rewona [7]

Answer:

Number 1 is a right triangle

Number 2 is a right triangle

Number 3 is NOT a right triangle

Number 4 is NOT a right triangle

Step-by-step explanation:

Hope this helps :)

7 0

3 years ago

Quadrilateral abcd is a rhombus and points c, d, and e are on the same line. is quadrilateral abde a rhombus?

Step2247 [10]

No, because BC and BD are not the same lengths. If AB = BC, then AB and BD are not the same lengths -> ABDE is not a rhombus.

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3 years ago

How do you write 5.4 as a mixed number

mestny [16]

5 is a whole number so keep it as 5 then 4 is in the tenths place so that means have ten for the denominator so 5 and 4/10 is the mixed number then simplified to 5 and 2/5

3 0

2 years ago

Read 2 more answers

Ira went to the movies with $35.91 in

svetlana [45]

Answer:

$13.36

Step-by-step explanation:

12.5+6.25+3.8=22.55

35.91-22.55= 13.36

6 0

3 years ago

Read 2 more answers

Find the work required to move an object in the force field F = e^x + y (1, 1, z) along the straight line from A(0, 0, 0) to B(-

DENIUS [597]

$\vec F(x,y,z)=(e^x+y)(1,1,z)$

is conservative if we can find a scalar function $f$ such that $\nabla f=\vec F$ . This would require

$f_x=e^x+y$

$f_y=e^x+y$

$f_z=(e^x+y)z$

Integrating both sides of the first equation wrt $x$ gives

$f(x,y,z)=e^x+xy+g(y,z)$

Differentiating both sides of this wrt $y$ gives

$f_y=x+g_y=1\implies g_y=1-x\implies g(y,z)=y-xy+h(z)$

but we assumed $g$ was a function of $y$ and $z$ , independent of $x$ . So there is no such $f$ and $\vec F$ is not conservative.

To find the work, first parameterize the path (call it $C$ ) by

$\vec r(t)=(1-t)(0,0,0)+t(-4,5,-5)=(-4t,5t,-5t)$

for $0\le t\le1$ . Then

$\vec r'(t)=(-4,5,-5)$

and the work is given by the line integral,

$W=\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\int_0^1(e^{-4t}+5t,e^{-4t}+5t,-(e^{-4t}+5t)5t)\cdot(-4,5,-5)\,\mathrm dt$

$W=\displaystyle\int_0^1(125t^2+5t+(25t+1)e^{-4t})\,\mathrm dt=\boxed{\frac{2207}{48}-\frac{129}{16e^4}}$

8 0

3 years ago