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3 years ago

Please help me I don’t understand

Mathematics

2 answers:

kvv77 [185]3 years ago

6 0

Answer:

Step-by-step explanation:

arc AB = 49°

arc BAC = 107°+49° = 156°

major arc ABC = 204°

major arc ACB = 360°-49° = 311°

Brilliant_brown [7]3 years ago

5 0

Ajaksajakakakkaakajakkaak

You might be interested in

PLEASE HELP!!!!!

Katena32 [7]

Answer:6018

Step-by-step explanation:

Given Sequence

$-282-266-250-234.....$

It represent an A.P. with

first term $a=-282$

common difference $d=16$

So sum of 51 term

$S_n=\frac{n}{2}[2a+(n-1)d]$

$S_{51}=\frac{51}{2}\times [2\times (-282)+(51-1)16]$

$S_{51}=\frac{51}{2}\times [-564+800]$

$S_{51}=\frac{51}{2}\times [236]$

$S_{51}=51\times 118$

$S_{51}=6018$

7 0

3 years ago

Complete the statements and equations to find the product of 6 and 13

Alina [70]

Answer:

Step-by-step explanation:

the product of 6 and 13 =78

8 0

3 years ago

Which ordered pair is a solution of the equation?<br> 7z -5 = 4y -6<br> Choose 1 answer:

ELEN [110]

Step-by-step explanation:

7z-5=4y-6

-4y=-6-7z+5

-4y=-1-7z

y=1/4+7/4z

y=7/4z+1/4

7z=4y-6+5

7z=4y-1

z=4/7y-1/7

3 0

2 years ago

In the previous part, we obtained dy dx = 3t2 − 27 −2t . Next, find the points where the tangent to the curve is horizontal. (En

mina [271]

Answer:

(27.55, 7.22), (-11.3, 3.21).

Step-by-step explanation:

When is the tangent to the curve horizontal?

The tangent curve is horizontal when the derivative is zero.

The derivative is:

$\frac{dy}{dx} = 3t^{2} - 2t - 27$

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

In this question:

$3t^{2} - 2t - 27 = 0$

$a = 3, b = -2, c = -27$

Then

$\bigtriangleup = b^{2} - 4ac = (-2)^{2} - 4*3*(-27) = 328$

$t_{1} = \frac{-(-2) + \sqrt{328}}{2*3} = 3.35$

$t_{2} = \frac{-(-2) - \sqrt{328}}{2*3} = -2.685$

Enter your answers as a comma-separated list of ordered pairs.

We found values of t, now we have to replace in the equations for x and y.

t = 3.35

$x = t^{3} - 3t = (3.35)^{3} - 3*3.35 = 27.55$

$y = t^{2} - 4 = (3.35)^2 - 4 = 7.22$

The first point is (27.55, 7.22)

t = -2.685

$x = t^{3} - 3t = (-2.685)^3 - 3*(-2.685) = -11.3$

$y = t^{2} - 4 = (-2.685)^2 - 4 = 3.21$

The second point is (-11.3, 3.21).

8 0

3 years ago

Simplify

ivolga24 [154]

The correct answer is: [A]: " $\frac{8}{729}$ " .
_________________________________________________________
Explanation:
_________________________________________________________
We are asked to simplify:

  → $(\frac{2}{9})^3$

_________________________________________________________
Note the following property of exponents:
_________________________________________________________

$( \frac{a}{b})^n$   = $\frac{(a^n)}{(b^n)}$ ; { b $\neq$ 0 } .

_________________________________________________________
As such:

→ $( \frac{2}{9})^3$ ;

= $\frac{2^3}{9^3}$ = $\frac{(2*2*2)}{(9*9*9)}$ = $\frac{8}{729}$ .

_____________________________________________________________

The answer is: " $\frac{8}{729}$ " ;

→ which is: "Answer choice: [A]:  " $\frac{8}{729}$ " .
_____________________________________________________________
Hope this answer—and explanation—is of some help to you.
Please comment below or otherwise respond to me if you have any questions or if I can be of further assistance.
Best wishes!

5 0

3 years ago

Read 2 more answers