All categories

3 years ago

Please help! I will mark brainliest!

Mathematics

2 answers:

Eva8 [605]3 years ago

4 0

Answer:

C. 9.6ft

D. Angle M

Step-by-step explanation:

Height of smaller surfboard = $\frac{12}{4}$ × 3.2ft

= 9.6ft

If you flip ABCD over, the positions of the angles that are corresponding to each other would be the same, so the angle that would correspond with Angle D would be Angle M since both would be at the top right.

Mekhanik [1.2K]3 years ago

3 0

C: small surf board is 9.6 ft tall

D: angle m corrosponds with d

You might be interested in

During a promotional weekend a state fair gives a free admission to every 175th person who enters the fair. On Saturday there we

aivan3 [116]

69 people got the free admission and that after that 154 more people came, so 20 people more and after than someone goes in for free!

7 0

3 years ago

Police response time to an emergency call is the difference between the time the call is first received by the dispatcher and th

Marizza181 [45]

Answer:

$\mu = 8.8\\\sigma = 1.9$

Now we are supposed to find probabilities that the response time is between 5 and 10 minutes i.e P(5<x<10)

Formula : $z= \frac{x-\mu}{\sigma}$

at x = 5

$z= \frac{5-8.8}{1.9}$

$z=-2$

at x = 10

$z= \frac{10-8.8}{1.9}$

$z=0.6315$

P(-2<z<0.6315)=P(z<0.6315)-P(z<-2)

Refer the z table

P(-2<z<10)=0.7357-0.0228=0.7129

So, the probability that response time is between 5 and 10 minutes is 0.7129

b)the response time is less than 5 minutes

at x = 5

$z= \frac{5-8.8}{1.9}$

$z=-2$

P(x<5)=P(z<-2)=0.0228

So, the probability that the response time is less than 5 minutes is 0.0228

c)the response time is more than 10 minutes

at x = 10

$z= \frac{10-8.8}{1.9}$

$z=0.6315$

P(x>10) = 1-P(x<10) = 1-P(z<0.63) = 1-0.7357 = 0.2643

So, The probability that the response time is more than 10 minutes is 0.2643

4 0

3 years ago

Write the function f(x) = 3x2 - 6x + 5 in vertex

weeeeeb [17]

3(x-1)^2 +2
This is your answer in vertex form, your h and k values are the vertex. Solving the function by using b/2a, we get that h is 1. ( in the equation 3 is your a, 6 is b and 5 is c.). ( 6/2(3)) = 1. We can then plug in 1 as x into the original equation and get positive 2 ( 3(1)^2 -6(1) +5) = 3-6+5 = 2. This is your vertex. In the function, your a value will always stay the same as this is your shrink or stretch. In this case, a is 3 so it will go outside the parenthesis. Put that all together and you get the function above.
Hope this helps :)

7 0

3 years ago

What is the equation of a line perpendicular to y= 3x +6 that crosses the ordered pair (3,2).

jasenka [17]

Answer:

y=- $\frac{1}{3}$ x+3

Step-by-step explanation:

5 0

3 years ago

Evaluate c (y + 7 sin(x)) dx + (z2 + 9 cos(y)) dy + x3 dz where c is the curve r(t) = sin(t), cos(t), sin(2t) , 0 ≤ t ≤ 2π. (hin

saw5 [17]

Treat $\mathcal C$ as the boundary of the region $\mathcal S$ , where $\mathcal S$ is the part of the surface $z=2xy$ bounded by $\mathcal C$ . We write

$\displaystyle\int_{\mathcal C}(y+7\sin x)\,\mathrm dx+(z^2+9\cos y)\,\mathrm dy+x^3\,\mathrm dz=\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r$

with $\mathbf f=(y+7\sin x,z^2+9\cos y,x^3)$ .

By Stoke's theorem, the line integral is equivalent to the surface integral over $\mathcal S$ of the curl of $\mathbf f$ . We have

$\nabla\times\mathbf f=(-2z,-3x^2,-1)$

so the line integral is equivalent to

$\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\mathrm d\mathbf S$
$=\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv$

where $\mathbf s(u,v)$ is a vector-valued function that parameterizes $\mathcal S$ . In this case, we can take

$\mathbf s(u,v)=(u\cos v,u\sin v,2u^2\cos v\sin v)=(u\cos v,u\sin v,u^2\sin2v)$

with $0\le u\le1$ and $0\le v\le2\pi$ . Then

$\mathrm d\mathbf S=\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv=(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv$

and the integral becomes

$\displaystyle\iint_{\mathcal S}(-2u^2\sin2v,-3u^2\cos^2v,-1)\cdot(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv$
$=\displaystyle\int_{v=0}^{v=2\pi}\int_{u=0}^{u=1}u-6u^4\sin^3v-4u^4\cos v\sin2v\,\mathrm du\,\mathrm dv=\pi$ <span />

4 0

3 years ago