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2 years ago

WILL GIVE BRAINLIEST

Mathematics

1 answer:

Svet_ta [14]2 years ago

3 0

Answer: The answer is A

Step-by-step explanation:

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Find an equation of the circle that satisfies the given conditions.

hodyreva [135]

The equation of a circle:
$(x-h)^2+(y-k)^2=r^2$
(h,k) - the coordinates of the centre
r - the radius

The midpoint of the diameter is the centre of a circle.
The coordinates of the midpoint:
$(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$
(x₁,y₁), (x₂,y₂) - the coordinates of endpoints

$P(-1,1) \\
x_1=-1 \\ y_1=1 \\ \\ Q(5,9) \\ x_2=5 \\ y_2=9 \\ \\
\frac{x_1+x_2}{2}=\frac{-1+5}{2}=\frac{4}{2}=2 \\ \frac{y_1+y_2}{2}=\frac{1+9}{2}=\frac{10}{2}=5$

The centre of the circle is (2,5).

The radius is the distance between an endpoint of the diameter and the centre.
The formula for distance:
$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$(-1,1) \\ x_1=-1 \\ y_1=1 \\ \\ (2,5) \\ x_2=2 \\ y_2=5 \\ \\ d=\sqrt{(2-(-1))^2+(5-1)^2}=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$

The radius is 5.

$(x-2)^2+(y-5)^2=5^2 \\
\boxed{(x-2)^2+(y-5)^2=25}$

4 0

2 years ago

Help me this question

lubasha [3.4K]

Answer:

(a) 218.6 N

(b) 97.14 N

Step-by-step explanation:

When the system is in equilibrium, the net torque on the system is zero.

AC = 1.5 m, CD = 2.3 m, DB = 5 - 1.5 - 2.3 = 1.2 m

Let the centre of gravity of plank is at G.

AG = 2.5 m, CG = 2.5 - 1.5 = 1 m, GB = 2.5 m

(a) Let the reaction at C is R and at D is R'.

R + R' = 29 x 9.8 = 284.2 N ... (1)

Take the torque about C.

29 x 9.8 x CG = R' x GD

29 x 9.8 x 1 = R' x 1.3

R' = 218.6 N

(b) Take the torque about D.

6 x 9.8 x AD = R x CD

6 x 9.8 x (1.5 + 2.3) = R x 2.3

R = 97.14 N

8 0

3 years ago

Plz help me ASAP<br> I need this answer fast please help me

liubo4ka [24]

$1200
explanation- 360/30%=$1200 !:)

8 0

2 years ago

Read 2 more answers

A construction company delivered 40 loads of concrete in 5 days.

AnnZ [28]

Well 40 divided by 5 is 8 and you don't have options so I would say 8 loads a day :)

3 0

3 years ago

Integrate this two questions

tankabanditka [31]

Simplify the integrands by polynomial division.

$\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)$

$\dfrac t{1 + 4t} = \dfrac14 \left(1 - \dfrac1{1 + 4t}\right)$

Now computing the integrals is trivial.

$\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}$

where we use the power rule,

$\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)$

and a substitution to integrate the last term,

$\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)$

$\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}$

using the same approach as above.

5 0

2 years ago