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3 years ago

Can a tutor help me please

Mathematics

1 answer:

DanielleElmas [232]3 years ago

6 0

I’m pretty sure it would be 16

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What is the area of this figure?

NARA [144]

Answer:

through lengthy mafs, your answer should be <u>307 mi (wrong somehow)</u>

Step-by-step explanation:

if you want an explanation, please, dont be afraid to comment ;)

(to other people who look this up the answer is actually 318 i dont know, but i got it wrong

7 0

2 years ago

Read 2 more answers

Find the limit. use l'hospital's rule if appropriate. if there is a more elementary method, consider using it. lim x→1 1 − x + l

AfilCa [17]

$\displaystyle\lim_{x\to1}\frac{1-x+\ln x}{1+\cos5\pi x}$

Evaluating the limand directly at $x=1$ gives an indeterminate form $\dfrac00$ . Apply L'Hospital's rule once and we get

$\displaystyle\lim_{x\to1}\frac{-1+\frac1x}{-5\pi\sin5\pi x}$

Again, plugging in $x=1$ returns $\dfrac00$ . Apply the rule once more:

$\displaystyle\lim_{x\to1}\frac{-\frac1{x^2}}{-25\pi^2\cos5\pi x}=\frac1{25\pi^2}\lim_{x\to1}\frac1{x^2\cos5\pi x}$

Now, in the denominator, when $x=1$ we get $x^2\cos5\pi x=-1$ , so the limit is $-\dfrac1{25\pi^2}$ .

3 0

3 years ago

Please help with this

snow_lady [41]

Answer:

B. $- \frac{9}{8}$

Step-by-step explanation:

$\huge\frac{ - 2 {a}^{ - 3} {b}^{2} }{a} \\ \\ \huge = \frac{ - 2 {b}^{2} }{a \times{a}^{ 3} } \\ \\ \huge = \frac{ - 2 {b}^{2} }{{a}^{ 4} } \\ \\ \huge = \frac{ - 2 \times {3}^{2} }{ {( - 2)}^{4} } \\ \\ \huge= \frac{ - 2 \times 9}{16} \\ \\ \huge \red{= - \frac{9}{8} }$

8 0

2 years ago

Any help would be great!

Virty [35]

Answer:

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

A 9,000-lb load is suspended from the roof in a shopping mall with a 16-ft-long solid aluminum rod. The modulus of elasticity of

Westkost [7]

Answer:

Step-by-step explanation:

Given:

elongation, x = 0.50 in

Force, f = 9000 lb

Young modulus, E = 10,000,000 psi

Maximum Stress, Sm = 30000 psi

Length, L = 16 ft

Converting ft to in,

12 in = 1 ft

=16 × 12 = 192 in

Young modulus, E = stress/strain

Stress = force/area, A

Strain = elongation, x/Length, L

E = f × L/A × E

1 × 10^7 = stress/(0.5/16)

= 26041.7 psi

Minimum stress = 26041.7 psi

Maximum stress = 30,000 psi

Stress = force/area

Area = 9000/26041.7

= 0.3456 in^2

Stress = force/area

Area = 9000/30000

= 0.3 in^2

Using minimum area of 0.3 in^2,

A = (pi/4)(d^2)

0.3 in^2 = (pi/4)(d^2)

d = 0.618 inches

diameter, d = 0.618 inches

7 0

3 years ago