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3 years ago

Given: bisects ∠BAC; AB = AC Which congruence theorem can be used to prove △ABR ≅ △ACR?

Mathematics

2 answers:

astraxan [27]3 years ago

8 0

Answer: SAS

Step-by-step explanation:

Here ABC is a triangle, In which angle BAC is bisected by line AR where, $R\in BC$ .

Thus in triangles ABR and ACR,

AB= AC ( given)

Therefore $AB\cong AC$

$\angle BAR\cong \angle CAR$ ( by the property of angle bisector)

And, $AR\cong AR$ (reflexive)

Thus by SAS postulate,

△ABR ≅ △ACR.

Therefore, Option D is correct.

mote1985 [20]3 years ago

5 0

Answer:

The answer is SAS

Step-by-step explanation:

I got it right on edge

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A pitcher throws a 9 kg baseball in a straight line. Its momentum is 2.1 kg-m/s. What is the velocity of the

Zina [86]

Answer:

$0.23\:\mathrm{m/s}$

Step-by-step explanation:

The momentum of an object is given by $p=mv$ , where $m$ is the mass of the object and $v$ is the velocity of the object.

What we're given:

object's mass $m$
object's momentum $p$

Substituting given values, we can solve for $v$ :

$2.1=9v,\\v=\frac{2.1}{9}\approx \boxed{0.23\:\mathrm{m/s}}$

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3 years ago

Find the equation of the tangent line to the curve (a lemniscate)

olya-2409 [2.1K]

Answer:

$m=\frac{9}{13}$ and $b=\frac{40}{13}$

Step-by-step explanation:

The equation of curve is

$2(x^2+y^2)^2=25(x^2-y^2)$

We need to find the equation of the tangent line to the curve at the point (-3, 1).

Differentiate with respect to x.

$2[2(x^2+y^2)\frac{d}{dx}(x^2+y^2)]=25(2x-2y\frac{dy}{dx})$

$4(x^2+y^2)(2x+2y\frac{dy}{dx})=25(2x-2y\frac{dy}{dx})$

The point of tangency is (-3,1). It means the slope of tangent is $\frac{dy}{dx}_{(-3,1)}$ .

Substitute x=-3 and y=1 in the above equation.

$4((-3)^2+(1)^2)(2(-3)+2(1)\frac{dy}{dx})=25(2(-3)-2(1)\frac{dy}{dx})$

$40(-6+2\frac{dy}{dx})=25(-6-2\frac{dy}{dx})$

$-240+80\frac{dy}{dx})=-150-50\frac{dy}{dx}$

$80\frac{dy}{dx}+50\frac{dy}{dx}=-150+240$

$130\frac{dy}{dx}=90$

Divide both sides by 130.

$\frac{dy}{dx}=\frac{9}{13}$

If a line passes through a points $(x_1,y_1)$ with slope m, then the point slope form of the line is

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

The slope of tangent line is $\frac{9}{13}$ and it passes through the point (-3,1). So, the equation of tangent is

$y-1=\frac{9}{13}(x-(-3))$

$y-1=\frac{9}{13}(x)+\frac{27}{13}$

Add 1 on both sides.

$y=\frac{9}{13}(x)+\frac{27}{13}+1$

$y=\frac{9}{13}(x)+\frac{40}{13}$

Therefore, $m=\frac{9}{13}$ and $b=\frac{40}{13}$ .

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3 years ago

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HACTEHA [7]

£17.50 is 50% of what number
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2 years ago

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Need help on all these questions!!

Yuki888 [10]

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Now, you have the midpoint M at the correct spot.

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