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2 years ago

Question 4 of 10 What else would need to be congruent to show that ABC= AXYZ by SAS?

Mathematics

1 answer:

Alja [10]2 years ago

5 0

Answer:

Step-by-step explanation:

The correct answer is D. Answered by Gauthmath

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Give all solutions if the nonlinear system of equations, including those with no real complex components.

konstantin123 [22]

$\left\{\begin{array}{ccc}y=x^2+6x\\4x-y=-24\end{array}\right\\\\substitute:\\\\4x-(x^2+6x)=-24\\\\4x-x^2-6x+24=0\\\\-x^2-2x+24=0\\\\a=-1;\ b=-2;\ c=24\\\\\Delta=b^2-4ac$

$\Delta=(-2)^2-4\cdot(-1)\cdot24=4+96=100\\\\x_1=\frac{-b-\sqrt\Delta}{2a};\ x_2=\frac{-b+\sqrt\Delta}{2a}\\\\\sqrt\Delta=\sqrt{100}=10\\\\x_1=\frac{2-10}{2\cdot(-1)}=\frac{-8}{-2}=4;\ x_2=\frac{2+10}{2\cdot(-1)}=\frac{12}{-2}=-6\\\\y_1=4^2+6\cdot4=16+24=40;\ y_2=(-6)^2+6\cdot(-6)=36-36=0\\\\Answer:\\x=4\ and\ y=40\ or\ x=-6\ and\ y=0$

4 0

3 years ago

Read 2 more answers

Nicoli jokic plays for the Denver nuggets in one game, his field-goal total was 26 points made up of 2 point and 3 point baskets

mel-nik [20]

9514 1404 393

Answer:

4 3-point baskets
7 2-point baskets

Step-by-step explanation:

Let x represent the number of 2-point baskets Nicoli made. Then (x -3) is the number of 3-point baskets. The total number of points is ...

2x +3(x -3) = 5x -9 = 26

5x = 35 . . . . . . add 9

x = 7 . . . . . . . divide by 5

(x -3) = 4 . . . find the number of 3-point baskets

Jokic made 7 2-point baskets and 4 3-point baskets.

7 0

2 years ago

Andy wrote the equation of a line that has a slope of Step 2: y plus 2 equals StartFraction 3 Over 4 EndFraction x minus StartFr

nlexa [21]

answer is C

He made an error in Step 3. He should have subtracted 2 from both sides.

4 0

3 years ago

Read 2 more answers

Name/ Uid:1. In this problem, try to write the equations of the given surface in the specified coordinates.(a) Write an equation

Gemiola [76]

To find:

(a) Equation for the sphere of radius 5 centered at the origin in cylindrical coordinates

(b) Equation for a cylinder of radius 1 centered at the origin and running parallel to the z-axis in spherical coordinates

Solution:

(a) The equation of a sphere with center at (a, b, c) & having a radius 'p' is given in cartesian coordinates as:

$(x-a)^{2}+(y-b)^{2}+(z-c)^{2}=p^{2}$

Here, it is given that the center of the sphere is at origin, i.e., at (0,0,0) & radius of the sphere is 5. That is, here we have,

$a=b=c=0,p=5$

That is, the equation of the sphere in cartesian coordinates is,

$(x-0)^{2}+(y-0)^{2}+(z-0)^{2}=5^{2}$

$\Rightarrow x^{2}+y^{2}+z^{2}=25$

Now, the cylindrical coordinate system is represented by $(r, \theta,z)$

The relation between cartesian and cylindrical coordinates is given by,

$x=rcos\theta,y=rsin\theta,z=z$

$r^{2}=x^{2}+y^{2},tan\theta=\frac{y}{x},z=z$

Thus, the obtained equation of the sphere in cartesian coordinates can be rewritten in cylindrical coordinates as,

$r^{2}+z^{2}=25$

This is the required equation of the given sphere in cylindrical coordinates.

(b) A cylinder is defined by the circle that gives the top and bottom faces or alternatively, the cross section, & it's axis. A cylinder running parallel to the z-axis has an axis that is parallel to the z-axis. The equation of such a cylinder is given by the equation of the circle of cross-section with the assumption that a point in 3 dimension lying on the cylinder has 'x' & 'y' values satisfying the equation of the circle & that 'z' can be any value.

That is, in cartesian coordinates, the equation of a cylinder running parallel to the z-axis having radius 'p' with center at (a, b) is given by,

$(x-a)^{2}+(y-b)^{2}=p^{2}$

Here, it is given that the center is at origin & radius is 1. That is, here, we have, $a=b=0,p=1$ . Then the equation of the cylinder in cartesian coordinates is,

$x^{2}+y^{2}=1$

Now, the spherical coordinate system is represented by $(\rho,\theta,\phi)$

The relation between cartesian and spherical coordinates is given by,

$x=\rho sin\phi cos\theta,y=\rho sin\phi sin\theta, z= \rho cos\phi$

Thus, the equation of the cylinder can be rewritten in spherical coordinates as,

$(\rho sin\phi cos\theta)^{2}+(\rho sin\phi sin\theta)^{2}=1$

$\Rightarrow \rho^{2} sin^{2}\phi cos^{2}\theta+\rho^{2} sin^{2}\phi sin^{2}\theta=1$

$\Rightarrow \rho^{2} sin^{2}\phi (cos^{2}\theta+sin^{2}\theta)=1$

$\Rightarrow \rho^{2} sin^{2}\phi=1$ (As $sin^{2}\theta+cos^{2}\theta=1$ )

Note that $\rho$ represents the distance of a point from the origin, which is always positive. $\phi$ represents the angle made by the line segment joining the point with z-axis. The range of $\phi$ is given as $0\leq \phi\leq \pi$ . We know that in this range the sine function is positive. Thus, we can say that $sin\phi$ is always positive.