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3 years ago

Given the triangle below find the measure of ZF and ZD.

Mathematics

1 answer:

kramer3 years ago

6 0

Answer: angle f is 31.788 and angle d is 58.212

Step-by-step explanation:I need one more brainliest for next rank I’d appreciate it

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Make D the subject of S=D/T<br> You may use the flowchart to help you if you wish.

hammer [34]

Given the equation, S = D/T:

To make the D the subject of the equation, we can start by multiplying both sides by T to isolate D:

(T) S = D/T (T)

TS = D

Therefore, the final answer is D = TS

8 0

3 years ago

How do I do this?<br>*Look at the directions in the photo*

lora16 [44]

Answer:

$Area\ of\ material\ required\ for\ the\ first\ box=384\ inches^2\\Area\ of\ material\ required\ for\ the\ second\ box=486\ inches^2\\Area\ of\ material\ required\ for\ the\ first\ box=600\ inches^2\\Total\ Area\ of\ material\ required=1470\ inches^2$

Step-by-step explanation:

$We\ are\ given:\\Diameter\ of\ the\ first\ volleyball=8\ inches \\Diameter\ of\ the\ second\ volleyball=9\ inches\\Diameter\ of\ the\ third\ volleyball= 10\ inches.\\Hence,\\We\ know\ that,\\If\ the\ side\ of\ the\ cube\ box\ is\ s, it's\ Total\ Surface\ Area\ =No.\ of\\ faces\ in\ a\ regular\ polyhedron\ *Area\ of\ each\ face\ of\ the\ polyhedron=6*s^2=6s^2\\Hence,\\Lets\ apply\ this\ equation\ in\ finding\ the\ area\ of\ material\ required\ for\ the\\ three\ cases.\\$

$As\ the\ volleyball\ should\ wholly\ fit\ into\ the\ box,\ the\ diameter\ of\ the\\ volleyballs\ would\ be\ the\ side\ of\ the\ cube\ box.\\Hence,\\For\ the\ first\ volleyball,\\Diameter\ of\ the\ first\ volleyball=8\ inches\\Hence,\\Side\ of\ the\ cubical\ box\ for\ the\ first\ volleyball=8\ inches.\\Hence,\\The\ Total\ Surface\ Area\ of\ the\ first\ box=6s^2=6*8*8=384\ inches^2$

$For\ the\ second\ volleyball,\\Diameter\ of\ the\ second\ volleyball=9\ inches\\Hence,\\Side\ of\ the\ cubical\ box\ for\ the\ second\ volleyball=9\ inches.\\Hence,\\The\ Total\ Surface\ Area\ of\ the\ second\ box=6s^2=6*9*9=486\ inches^2$

$For\ the\ third\ volleyball,\\Diameter\ of\ the\ third\ volleyball=8\ inches\\Hence,\\Side\ of\ the\ cubical\ box\ for\ the\ third\ volleyball=10\ inches.\\Hence,\\The\ Total\ Surface\ Area\ of\ the\ third\ box=6s^2=6*10*10=600\ inches^2$

$Hence,\\If\ you\ are\ asked\ the\ Total\ Area\ to\ make\ all\ the\ boxes,\\ you\ just\ add\ them\ together.\\Hence,\\Total\ Area\ of\ Material\ required\ to\ make\ the\ three\ boxes=384+486+600=1470\ inches^2$

7 0

3 years ago

Plz help ima fail my class if I don’t get this one

Fofino [41]

Answer:

its the first one

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

When Samuel commutes to work, the amount of time it takes him to arrive is normally distributed with a mean of 32 minutes and a

k0ka [10]

Answer:

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

And we can find the nnumber of deviations from the mean for each limit given:

$z_1 = \frac{17-32}{5} = -3$

$z_2= \frac{47-32}{5} = 3$

So we are 3 deviation from the mean and using the empirical rule we know that within 3 deviations from the mean we have 99.7% of the values

Step-by-step explanation:

Let X the random variable that represent the amount of time it takes him to arrive, and for this case we know the distribution for X is given by:

$X \sim N(32,5)$

Where $\mu=32$ and $\sigma=5$

We want to find this probability:

$P(17$

And we can use the z score formula given by:

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

And we can find the nnumber of deviations from the mean for each limit given:

$z_1 = \frac{17-32}{5} = -3$

$z_2= \frac{47-32}{5} = 3$

So we are 3 deviation from the mean and using the empirical rule we know that within 3 deviations from the mean we have 99.7% of the values

8 0

3 years ago

Question 20

Alexxandr [17]

Answer:

The camera had to cover the greatest angle is CAMERA 3 because it had the largest angle of 71.47°

Step-by-step explanation:

From the above question,

We have:

Camera 1 = Angle A

Camera 2= Angle B

Camera 3 = Angle C

A = 210ft

B = 234ft

C = 260ft

We need to find Angle A( angle of camera 1) using the cosine rule

A=(B² + C² - 2BCCosA)

210² = 234² + 260² - 2 × 234 × 260 × CosA

210² = 122356 - 121680CosA

Square both sides

210² = 122356 - 121680CosA

44100 = 122356 - 121680CosA

121680CosA = 122356 - 44100

121680CosA = 78256

Cos A = 78256/121680

Cos A = 0.6431295201

A = arc cos (0.6431295201)

A = 49.974422249°

Angle A approximately = 49.97°

Using the Sine rule to find the Angle B

A/Sine A = B/Sine B

210ft/Sine 49.97° = 234ft/Sine B

210ft × Sine B = 49.97° × 234ft

Sine B =( Sine 49.97° × 234ft)/210ft

B = arc sin (0.8532172354)

Angle B = 58.56334

Approximately = 58.56°

Angle C = 180 - (49.97 + 58.56)°

Angle C = 71.47°

Therefore, the camera had to cover the greatest angle is Camera 3 because it had the largest angle of 71.47°

6 0

3 years ago