All categories

3 years ago

I need your help

Mathematics

2 answers:

wariber [46]3 years ago

5 0

Your going to make the expression 7*3*2*4 which equals 168

likoan [24]3 years ago

4 0

Answer:

the answer is a. 168

Step-by-step explanation:

You might be interested in

What is the range of the data?

e-lub [12.9K]

64 I think I’m not sure

3 0

3 years ago

ms.morgan bought 3.5 pounds of bananas at $0.51 a pound and 4.5 pounds of pineapple at $1.19 a pound.how much did she pay for th

Tanzania [10]

3.5*0.51 = 1.785
4.5*1.19 = 5.355
1.785 + 5.355 = 7.14
Ms Morgan paid $7.14c for the bananas and pineapples.

4 0

3 years ago

Read 2 more answers

What is the slope and y-intercept for x-axis, 1,2,3,4 and y-axis 11,7,3,-1

Alona [7]

36 is the slope hopefully it helps

4 0

4 years ago

Select the correct text in the passage.

VikaD [51]

Answer:

not be able to raise enough funds for his work

Step-by-step explanation:

7 0

4 years ago

Read 2 more answers

You stand a known distance from the base of the tree, measure the angle of elevation the top of the tree to be 15â—¦ , and then

gogolik [260]

Answer:

The maximum possible error of in measurement of the angle is $d\theta_1 =(14.36p)^o$

Step-by-step explanation:

From the question we are told that

The angle of elevation is $\theta_1 = 15 ^o = \frac{\pi}{12}$

The height of the tree is h

The distance from the base is D

h is mathematically represented as

$h = D tan \theta$ Note : this evaluated using SOHCAHTOA i,e

$tan\theta = \frac{h}{D}$

Generally for small angles the series approximation of $tan \theta \ is$

$tan \theta = \theta + \frac{\theta ^3 }{3}$

So given that $\theta = 15 \ which \ is \ small$

$h = D (\theta + \frac{\theta^3}{3} )$

$dh = D (1 + \theta^2) d\theta$

=> $\frac{dh}{h} = \frac{1 + \theta ^2}{\theta + \frac{\theta^3}{3} } d \theta$

Now from the question the relative error of height should be at most

$\pm p$ %

=> $\frac{dh}{h} = \pm p$

=> $\frac{1 + \theta ^2}{\theta + \frac{\theta^3}{3} } d \theta = \pm p$

=> $d\theta = \pm \frac{\theta + \frac{\theta^3}{3} }{1+ \theta ^2} * \ p$

So for $\theta_1$

$d\theta_1 = \pm \frac{\theta_1 + \frac{\theta^3_1 }{3} }{1+ \theta_1 ^2} * \ p$

substituting values

$d [\frac{\pi}{12} ] = \pm \frac{[\frac{\pi}{12} ] + \frac{[\frac{\pi}{12} ]^3 }{3} }{1+ [\frac{\pi}{12} ] ^2} * \ p$

=> $d\theta_1 = 0.25 p$

Converting to degree

$d\theta_1 = (0.25* 57.29) p$

$d\theta_1 =(14.36p)^o$

4 0

3 years ago