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3 years ago

17.

Mathematics

2 answers:

zhannawk [14.2K]3 years ago

7 0

D. Square and rectangle

hram777 [196]3 years ago

4 0

C. square and rhombus

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I need help it’s homework and idk how to do it

Volgvan

Answer:

x = 40 1/2

Step-by-step explanation:

The hash marks tell you the triangles are similar and the smaller one is 1/2 the size of the larger one. Then x is half the length of the corresponding segment marked 81.

x = 81/2

x = 40 1/2

___

You may have to write it as 81/2 or as 40.5. Sometimes the answer needs to be in a particular form.

3 0

3 years ago

Which choice below lists all of the square roots of the number 15?

grigory [225]

Answer:

A is your answer

Step-by-step explanation:

6 0

3 years ago

Consider the force field and circle defined below.

grin007 [14]

By Green's theorem,

$\displaystyle\int_{x^2+y^2=9}\vec F(x,y)\cdot\mathrm d\vec r=\iint_D\left(\frac{\partial(xy)}{\partial x}-\frac{\partial(x^2)}{\partial y}\right)\,\mathrm dx\,\mathrm dy=\iint_Dy\,\mathrm dx\,\mathrm dy$

where $C$ is the circle $x^2+y^2=9$ and $D$ is the interior of $C$ , or the disk $x^2+y^2\le1$ .

Convert to polar coordinates, taking

$\begin{cases}x=r\cos\theta\\y=r\sin\theta\end{cases}\implies\mathrm dx\,\mathrm dy=r\,\mathrm dr\,\mathrm d\theta$

Then the work done by $\vec F$ on the particle is

$\displaystyle\iint_Dy\,\mathrm dx\,\mathrm dy=\int_0^{2\pi}\int_0^3(r\sin\theta)r\,\mathrm dr\,\mathrm d\theta=\left(\int_0^{2\pi}\sin\theta\,\mathrm d\theta\right)\left(\int_0^3r^2\,\mathrm dr\right)=\boxed0$

4 0

3 years ago

Simplify: 3x2 - 27<br> Helppppp

oksian1 [2.3K]

3x^2- 27 … you divide the whole thing by 3 to get x^2- 9 and that’s the furthest you can simplify

6 0

3 years ago

Answer with workings please

icang [17]

AB = 9 cm
BC = 6cm

CD = 7 cm
AE = 6 cm

3BC = AB
3ED = AE

  AB = AE
  BC ED
  ⁹/₃ = ⁶/ₓ
3 · 6 = 9 · x
18 = 9x
9 9
2 = x

ED = 2 cm

5 0

3 years ago