2 years ago

Please help, answer asap

Mathematics

1 answer:

sergejj [24]2 years ago

7 0

Since it is 1/5 of original length (13cm)
13cm * 1/5 = 2.6 cm

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nadezda [96]

Answer:

yes

Step-by-step explanation:

maybe

4 0

3 years ago

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What is the area of the figure?<br><br>a 28 yd<br><br>b 40 yd<br><br>c 52 yd<br><br>d 64 yd

Vilka [71]

Answer:

Step-by-step explanation:

We have the square and the right triangle.

The formula of an area of a square:

$A_{\boxed{}}=a^2$

a - length of side

The formula of an area of a right triangle:

$A_\triangle=\dfrac{ab}{2}$

a,b - legs

We have:

SQUARE:

a = 4 yd

$A_{\boxed{}}=4^2=16\ yd^2$

RIGHT TRIANGLE:

a = 6 yd, b = 4 yd + 4 yd = 8 yd

$A_\triangle=\dfrac{(6)(8)}{2}=(3)(8)=24\ yd^2$

The area of the figure:

$A=A_{\boxed{}}+A_\triangle\\\\A=16\ yd^2+24\ yd^2=40\ yd^2$

8 0

2 years ago

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Latasha ran a mile in 7.03 minutes and Erik ran a mile in 6.8 minutes. How many more minutes did it take Latasha to run the mile

Murljashka [212]

7.03-6.8= 0.23 therefore your answer is 0.23

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3 years ago

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Finish this 5-9 all of it

4vir4ik [10]

There is no picture. What do we need to solve ?

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2 years ago

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Evaluate the surface integral ∫∫S F · dS for the given vector field F and the oriented surface S. In other words, find the flux

Strike441 [17]

Parameterize $S$ by

$\vec r(u,v)=u\,\vec\imath+v\,\vec\jmath+(7-u^2-v^2)\,\vec k$

with $0\le u\le 1$ and $0\le v\le1$ . Take the normal vector to be

$\vec r_u\times\vec r_v=2u\,\vec\imath+2v\,\vec\jmath+\vec k$

Then the flux across $S$ is

$\displaystyle\iint_S\vec F\cdot\mathrm d\vec S$

$\displaystyle=\int_0^1\int_0^1(uv\,\vec\imath+(7-u^2-v^2)v\,\vec\jmath+(7-u^2-v^2)u\,\vec k)\cdot(\vec r_u\times\vec r_v)\,\mathrm du\,\mathrm dv$

$\displaystyle\int_0^1\int_0^1(2u^2v+(u+2v^2)(7-u^2-v^2))\,\mathrm du\,\mathrm dv=\frac{1343}{180}$

8 0

2 years ago