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3 years ago

Can anyone help me with this one please

Mathematics

1 answer:

bulgar [2K]3 years ago

7 0

A. 314 sq cm

Explanation: pi•r^2
3.14 x 10^2
3.14 x 100 =314

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The first term of an arithmetic sequence is -5, and the tenth term is 13. Find the common difference.

zzz [600]

Hello,

$u_{1} =-5\\
 u_{2} =-5+d\\
 u_{3} =-5+2d\\
 u_{4} =-5+3d\\
...\\
 u_{10} =-5+9d=13\\
==\ \textgreater \ 9d=18 \\
==\ \textgreater \ d=2

$

6 0

4 years ago

PLEASE HELP ILL GIVE THANKS, 5.0, AND BRAINLIEST TO WHOEVER ANSWERS CORRECTLY PLEASE HELP AND ILL GIVE A HIGH POINT COUNT

Allushta [10]

Answer:

<h3>FOR THIS PROBLEM I GOT..... (20/r)-(20/s)</h3>

Step-by-step explanation:

pls brainlest mee

6 0

2 years ago

Suppose an opaque jar contains 3 red marbles and 10 green marbles. The following exercise refers to the experiment of picking tw

kow [346]

Answer:

$\frac{5}{13}$

Step-by-step explanation:

Given,

Red marbles = 3,

Green marbles = 10,

So, the total marbles = 3 + 10 = 13,

$\because \text{Probability}=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}$

Since, here replacement is not allowed,

Thus, the probability of getting a green marble and a red marble

= first red and second green + first green second red

$=\frac{3}{13}\times \frac{10}{12}+\frac{10}{13}\times \frac{3}{12}$

$=2\times \frac{3}{13}\times \frac{10}{12}$

$=\frac{30}{78}$

$=\frac{5}{13}$

Note : The probability of getting a green marble first and a red marble second

= $\frac{10}{13}\times \frac{3}{12}$

$=\frac{30}{156}$

$=\frac{5}{26}$

4 0

3 years ago

Someone help me please:)

Kryger [21]

Answer:

a. The time to strike the ground is 5 seconds

b. The time for the ball to be over than 96 ft is 1 second

Step-by-step explanation:

a. When the ball thrown vertically up ward to reach its maximum height at velocity = 0 then complete its motion vertically down ward to strike the ground its displacement = zero

∴ s(t) = 0 ⇒ $80t-16t^{2}=0$

∴ $16t(5 - t) = 0$

5 - t = 0⇒ ∴ t = 5 seconds

b. To tind the time that the ball will be over 96 ft over the ground substitute in $s(t)=80t-16t^{2}$

$96=80t-16t^{2}$

$16t^{2}-80t+96=0$

$16(t^{2}-5t+6)=0$

$16(t-3)(t-2)=0$

t = 3 sec. and t = 2 sec.

The ball will be at height 96 ft at time 2 sec. and 3 sec.

So the ball will be at height over than 96 = 3 - 2 = 1 sec.

8 0

3 years ago

Question 20: Please help what are sin x and cos x?

sweet [91]

Answer:

A is the correct answer.

Step-by-step explanation:

$\tan x = \frac{6}{8} = \frac{opposite \: side \: to \: \angle \: x}{adjacent \: side \: to \: \angle \: x} \\ \\ hypotenuse \\ = \sqrt{ {(opp. \: side)}^{2} + {(adj. \: side)}^{2} } \\ = \sqrt{ {6}^{2} + {8}^{2} } \\ = \sqrt{36 + 64} \\ = \sqrt{100} \\ = 10 \\ \\ \because \sin (x) = \frac{opposite \: side \: to \: \angle \: x}{hypotenuse} \\ \\ \huge{ \red{ \boxed{\therefore \: \sin (x) = \frac{6}{10}}}} \\ \\ \cos (x) = \frac{adjacent \: side \: to \: \angle \: x}{hypotenuse} \\ \\ \huge{ \purple{ \boxed{\therefore \: \cos (x) = \frac{8}{10} }}} \\$

Thus, option A is the correct answer.

8 0

3 years ago