What is the first step in solving the equation 2/3x x 1/3x+2=5?

By which rule are these triangles congruent A. AAS B. ASA C. SAS D. SSS

Allushta [10]

Answer:

option D

Step-by-step explanation:

this trianle will be congurent by SSS method easily

3 0

2 years ago

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A superhero is trying to leap over a tall building. The function f(x)=-16x^2+200x gives the superhero's height in feet as a func

Gemiola [76]

Answer:

Since $\bigtriangleup \geq 0$ , the superhero makes it over the building.

Step-by-step explanation:

The height is given by the following function:

$f(x) = -16x^{2} + 200x$

Will the superhero make it over the building?

We have to find if there is values of x for which f(x) = 612.

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:

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

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

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

If $\bigtriangleup < 0$ , the polynomial has no solutions.

In this question:

$f(x) = -16x^{2} + 200x$

$-16x^{2} + 200x = 612$

$16x^{2} - 200x + 612 = 0$

We have to find $\bigtriangleup$

We have that $a = 16, b = -200, c = 612$ . So

$\bigtriangleup = (-200)^{2} - 4*16*612 = 832$

Since $\bigtriangleup \geq 0$ , the superhero makes it over the building.

7 0

3 years ago

Let T: set of real numbers R Superscript nℝnright arrow→set of real numbers R Superscript mℝm be a linear transformation, and

Klio2033 [76]

Answer:

$\{T(v_1), T(v_2), T(v_3)\}$ is linearly dependent set.

Step-by-step explanation:

Given: $\{v_1,v_2,v_3\}$ is a linearly dependent set in set of real numbers R

To show: the set $\{T(v_1), T(v_2), T(v_3)\}$ is linearly dependent.

Solution:

If $\{v_1,v_2,v_3,...,v_n\}$ is a set of linearly dependent vectors then there exists atleast one $k_i:i=1,2,3,...,n$ such that $k_1v_1+k_2v_2+k_3v_3+...+k_nv_n=0$

Consider $k_1T(v_1)+k_2T(v_2)+k_3T(v_3)=0$

A linear transformation T: U→V satisfies the following properties:

1. $T(u_1+u_2)=T(u_1)+T(u_2)$

2. $T(au)=aT(u)$

Here, $u,u_1,u_2$ ∈ U

As T is a linear transformation,

$k_1T(v_1)+k_2T(v_2)+k_3T(v_3)=0\\T(k_1v_1)+T(k_2v_2)+T(k_3v_3)=0\\T(k_1v_1+k_2v_2+k_3v_3)=0\\$

As $\{v_1,v_2,v_3\}$ is a linearly dependent set,

$k_1v_1+k_2v_2+k_3v_3=0$ for some $k_i\neq 0:i=1,2,3$

So, for some $k_i\neq 0:i=1,2,3$

$k_1T(v_1)+k_2T(v_2)+k_3T(v_3)=0$

Therefore, set $\{T(v_1), T(v_2), T(v_3)\}$ is linearly dependent.

6 0

3 years ago

HELP ME, PLEASE!

andrey2020 [161]

It is clear that (AB) is a median, and it is obvious that the angle is 90, thus it is an altitude and a perpendicular bisector

3 0

2 years ago

Read 2 more answers

PLEASEEE HURRYYY

Masteriza [31]

Answer:

(2,-5) and (2,1).

Step-by-step explanation:

We need to find the find all the points having an x coordinate of 2 whose distance from the point (-2,-4) is 5.

A circle with center (-2,-4) and radius 5 represents all the points whose distance from the point (-2,-4) is 5.

Standard form of a circle is

$(x-h)^2+(y-k)^2=r^2$

where, (h,k) is center and r is the radius.

$(x-(-2))^2+(y-(-4))^2=(5)^2$

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

Now, put x=2.

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

$(4)^2+(y+2)^2=25$

$(y+2)^2=25-16$

$(y+2)^2=9$

Taking square root on both sides.

$y+2=\pm\sqrt{9}$

$y+2=\pm3$

$y+2=-3\text{ or }y+2=3$

$y=-5\text{ or }y=1$

So, the y-coordinates of the points are -5 and 1.

Therefore, the required points are (2,-5) and (2,1).

8 0

2 years ago