All categories

4 years ago

Plz help 40 pointsplus brainliest

Mathematics

1 answer:

crimeas [40]4 years ago

5 0

I think it’s: Concurrency of Angle Bisectors of a Triangle.

You might be interested in

Solve the equation for a.

nataly862011 [7]

Answer:

a=-7.68

Step-by-step explanation:

3a+13.61=-9.43

-13.61

_____________

-23.04

3a=-23.04

-23.04/3= -7.68

7 0

3 years ago

8. The area of the bigger square at the picture is 32. What is the area of

Karolina [17]

Answer:

16 is the awnser or at least I belive

8 0

3 years ago

Use implicit differentiation to find an equation of the tangent line to the curve at the given point. x2/3 + y2/3 = 4 (−3 3 , 1)

vovikov84 [41]

Answer with Step-by-step explanation:

We are given that an equation of curve

$x^{\frac{2}{3}}+y^{\frac{2}{3}}=4$

We have to find the equation of tangent line to the given curve at point $(-3\sqrt3,1)$

By using implicit differentiation, differentiate w.r.t x

$\frac{2}{3}x^{-\frac{1}{3}}+\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=0$

Using formula : $\frac{dx^n}{dx}=nx^{n-1}$

$\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=-\frac{2}{3}x^{-\frac{1}{3}}$

$\frac{dy}{dx}=\frac{-\frac{2}{3}x^{-\frac{1}{3}}}{\frac{2}{3}y^{-\frac{1}{3}}}$

$\frac{dy}{dx}=-\frac{x^{-\frac{1}{3}}}{y^{-\frac{1}{3}}}$

Substitute the value x= $-3\sqrt3,y=1$

Then, we get

$\frac{dy}{dx}=-\frac{(-3\sqrt3)^{-\frac{1}{3}}}{1}$

$\frac{dy}{dx}=-(-3^{\frac{3}{2}})^{-\frac{1}{3}}=-\frac{1}{-(3)^{\frac{3}{2}\times \frac{1}{3}}}=\frac{1}{\sqrt3}$

Slope of tangent=m= $\frac{1}{\sqrt3}$

Equation of tangent line with slope m and passing through the point $(x_1,y_1)$ is given by

$y-y_1=m(x-x_1)$

Substitute the values then we get

The equation of tangent line is given by

$y-1=\frac{1}{\sqrt3}(x+3\sqrt3)$

$y-1=\frac{x}{\sqrt3}+3$

$y=\frac{x}{\sqrt3}+3+1$

$y=\frac{x}{\sqrt3}+4$

This is required equation of tangent line to the given curve at given point.

8 0

4 years ago

In ΔOPQ, o = 58 cm, q = 13 cm and ∠Q=52°. Find all possible values of ∠O, to the nearest degree.

PSYCHO15rus [73]

Answer:

There is no possible triangles

Step-by-step explanation:

\frac{\sin A}{a}=\frac{\sin B}{b}

sinA

sinB

From the reference sheet (reciprocal version).

\frac{\sin O}{58}=\frac{\sin 52}{13}

sinO

sin52

Plug in values.

\sin O=\frac{58\sin 52}{13}\approx 3.5157403

sinO=

58sin52

≈3.5157403O=\sin^{-1}(3.5157403)= \text{ERROR}

O=sin

−1

(3.5157403)=ERROR

Sine cannot be greater than 1.

{No Possible Triangles}

No Possible Triangles

{}

7 0

3 years ago

Kendall graphed 3 points from the equation x = 4 on a coordinate grid. She graphed (4, 0), (4, 1), and (4, 3). She drew a straig

devlian [24]

A) (4,6) would also be on this line. Let me know if you have any other questions!

6 0

3 years ago

Plz help 40 pointsplus brainliest​

Plz help 40 pointsplus brainliest