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3 years ago

6

How does labeling the diagram help us show that we can use right triangle congruency (SAS in this case)?

1 answer:

dem82 [27]3 years ago

4 0

Answer:

Step-by-step explanation:

S A S congruence rule: Two triangles are congruent if two sides and the included angle of one triangle are equal to the two sides and the included angle of other triangle.

If two triangles are congruent by SAS, and the angle is 90, then by CPCT{corresponding part of congruent triangle) the hypotenuse of the two triangles will be congruent.

Now, as hypotenuse and one leg are congruent in both triangles, we can apply RHS congruent

You might be interested in

Plot a point on the coordinate plane to represent each of the ratio values in the table. Time (h) Distance (km) able displaying

Effectus [21]

Answer:

Refer the attached graph.

Step-by-step explanation:

Given : Time (h) Distance (km) able displaying Time in hours in one column and Distance in kilometers in the other. The time column contains 1, 3, 6, and 9. The distance column contains 2, 6, 12, and 18.

To plot : The points on the coordinate plane.

Solution :

Time is independent variable so it is taken on x-axis.

Distance is taken on y-axis.

x y

1 2

3 6

6 12

9 18

When we plot these points in the coordinate plane, we get a linear equation.

The linear equation form is $y=2x$

Refer the attached graph.

4 0

3 years ago

Y equals the quotient of the quantity x squared plus 4 times x and the quantity x cubed minus 5.

koban [17]

Given:

Consider the completer question is "Find the derivative $\dfrac{dy}{dx}$ for $y=\dfrac{x^2-4x}{x^3-5}$ ."

To find:

The derivative $\dfrac{dy}{dx}$ .

Solution:

Chain rule: $\dfrac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)$

Quotient rule: $\dfrac{d}{dx}\dfrac{f(x)}{g(x)}=\dfrac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}$

We have,

$y=\dfrac{x^2-4x}{x^3-5}$

Differentiate with respect to x.

$\dfrac{dy}{dx}=\dfrac{d}{dx}\left(\dfrac{x^2-4x}{x^3-5}\right)$

Using chain rule and quotient rule, we get

$\dfrac{dy}{dx}=\dfrac{(x^3-5)\dfrac{d}{dx}(x^2-4x)-(x^2-4x)\dfrac{d}{dx}(x^3-5)}{(x^3-5)^2}$

$\dfrac{dy}{dx}=\dfrac{(x^3-5)(2x-4)-(x^2-4x)(3x^2)}{(x^3-5)^2}$

$\dfrac{dy}{dx}=\dfrac{2x^4-4x^3-10x+20-3x^4+12x^3}{(x^3-5)^2}$

$\dfrac{dy}{dx}=\dfrac{-x^4+8x^3-10x+20}{(x^3-5)^2}$

Therefore, the required answer is $\dfrac{dy}{dx}=\dfrac{-x^4+8x^3-10x+20}{(x^3-5)^2}$ .

4 0

2 years ago

Pls help me with this

DedPeter [7]

Answer:

the volume of the cone would be approximately 490.09 m^3.

Step-by-step explanation:

The volume of a cone is pir^2h/3

So it would be pi*6^2*13/3

=pi*36*13/3

=pi*156

=490.09

Therefore, the volume of the cone would be approximately 490.09 m^3.

I hope this helped and have a good rest of your day!

7 0

2 years ago

Suppose the population of a town is 8,200 and is growing 4% each year, what is the population expected to be in 10 years

miv72 [106K]

Answer:

12,138 people

Step-by-step explanation:

Create an equation using the basic exponential growth equation:

y = a(1 + r)^t, where a is the initial amount, r is the growth rate as a decimal, and t is the amount of time in years.

Plug in the values we know

y = a(1 + r)^t

y = 8200(1 + 0.04)^10

y = 8200(1.04)^10

= 12,138 people

8 0

3 years ago

Help solve algebra 15a=6a-90

AleksandrR [38]

15a =6a - 90

Move all the numbers with variables to one side

15a - (6a) = 6a -(6a) -90
9a = -90
9a/9 = -90/9
a = -9

Hope this helps

5 0

3 years ago