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2 years ago

I need help pleaseee

Mathematics

1 answer:

katen-ka-za [31]2 years ago

8 0

Answer:

Step-by-step explanation:

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Rudy measured an Italian restaurant and made a scale drawing. The scale of the drawing was 1 centimeter : 6 meters. In the drawi

vekshin1

So I’m assuming you’re saying that 1 cm=6 m so if the kitchen is 4 cm in the drawing it has to be scaled up by 6 for each centimeter. So 4*6=24 so the kitchen is 24 meters long. Hope this helps!

7 0

2 years ago

Given that A, O & B lie on a straight line segment, evaluate <br>x<br>.

Afina-wow [57]

Answer:

Step-by-step explanation:

It is the same way:

we know that

if A , O and B lie on a straight line segment

then

∡AOY+∡YOZ+∡ZOB=180°

3x+x+x=180-----> 5x=180--------> x=180/5--------> x=36°

the answer is

x=36°

<em>please mark me as brainliest</em>

4 0

2 years ago

What percentage of 128 is 96?

katrin [286]

So, since 100% = 128, x should equal 96, and we are trying to find x.
Multiply both sides of the equation by x.
(100/x) * x = (128/96) * x
Cancel out the x's on the left side so
100 = 1.333(x)
75 = x
96 is 75% of 128

7 0

3 years ago

Read 2 more answers

Henry invested $59,000 in an account paying an interest rate of 2% compounded quarterly. Assuming no deposits or withdrawals are

AfilCa [17]

Answer:

Step-by-step explanation:

$A=P(1+r/4)^{(4t)}\\ \\ A=P(1+.02/4)^{(4t)}\\ \\ A=59000(1.005)^{4(20)}\\ \\ A=\$ 87929.98$

5 0

3 years ago

Consider the following region R and the vector field F. a. Compute the two-dimensional curl of the vector field. b. Evaluate bo

Shalnov [3]

Looks like we're given

$\vec F(x,y)=\langle-x,-y\rangle$

which in three dimensions could be expressed as

$\vec F(x,y)=\langle-x,-y,0\rangle$

and this has curl

$\mathrm{curl}\vec F=\langle0_y-(-y)_z,-(0_x-(-x)_z),(-y)_x-(-x)_y\rangle=\langle0,0,0\rangle$

which confirms the two-dimensional curl is 0.

It also looks like the region $R$ is the disk $x^2+y^2\le5$ . Green's theorem says the integral of $\vec F$ along the boundary of $R$ is equal to the integral of the two-dimensional curl of $\vec F$ over the interior of $R$ :

$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\iint_R\mathrm{curl}\vec F\,\mathrm dA$

which we know to be 0, since the curl itself is 0. To verify this, we can parameterize the boundary of $R$ by

$\vec r(t)=\langle\sqrt5\cos t,\sqrt5\sin t\rangle\implies\vec r'(t)=\langle-\sqrt5\sin t,\sqrt5\cos t\rangle$

$\implies\mathrm d\vec r=\vec r'(t)\,\mathrm dt=\sqrt5\langle-\sin t,\cos t\rangle\,\mathrm dt$

with $0\le t\le2\pi$ . Then

$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\int_0^{2\pi}\langle-\sqrt5\cos t,-\sqrt5\sin t\rangle\cdot\langle-\sqrt5\sin t,\sqrt5\cos t\rangle\,\mathrm dt$

$=\displaystyle5\int_0^{2\pi}(\sin t\cos t-\sin t\cos t)\,\mathrm dt=0$

7 0

2 years ago

I need help pleaseee​

I need help pleaseee