What is the apparent solution to the system of equations?

A table and 8 chairs weigh 234.68 lb together. If the table weighs 157.84 lb, what is the weight of one chair in pounds?

Trava [24]

You subtract 234.68 and 157.84 and you get 76.84lb

5 0

3 years ago

Find equations of the spheres with center(3, −4, 5) that touch the following planes.a. xy-plane b. yz- plane c. xz-plane

postnew [5]

Answer:

(a) (x - 3)² + (y + 4)² + (z - 5)² = 25

(b) (x - 3)² + (y + 4)² + (z - 5)² = 9

(c) (x - 3)² + (y + 4)² + (z - 5)² = 16

Step-by-step explanation:

The equation of a sphere is given by:

(x - x₀)² + (y - y₀)² + (z - z₀)² = r² ---------------(i)

Where;

(x₀, y₀, z₀) is the center of the sphere

r is the radius of the sphere

Given:

Sphere centered at (3, -4, 5)

=> (x₀, y₀, z₀) = (3, -4, 5)

(a) To get the equation of the sphere when it touches the xy-plane, we do the following:

i. Since the sphere touches the xy-plane, it means the z-component of its centre is 0.

Therefore, we have the sphere now centered at (3, -4, 0).

Using the distance formula, we can get the distance d, between the initial points (3, -4, 5) and the new points (3, -4, 0) as follows;

$d = \sqrt{(3-3)^2+ (-4 - (-4))^2 + (0-5)^2}$

$d = \sqrt{(3-3)^2+ (-4 + 4)^2 + (0-5)^2}$

$d = \sqrt{(0)^2+ (0)^2 + (-5)^2}$

$d = \sqrt{(25)}$

d = 5

This distance is the radius of the sphere at that point. i.e r = 5

Now substitute this value r = 5 into the general equation of a sphere given in equation (i) above as follows;

(x - 3)² + (y - (-4))² + (z - 5)² = 5²

(x - 3)² + (y + 4)² + (z - 5)² = 25

Therefore, the equation of the sphere when it touches the xy plane is:

(x - 3)² + (y + 4)² + (z - 5)² = 25

(b) To get the equation of the sphere when it touches the yz-plane, we do the following:

i. Since the sphere touches the yz-plane, it means the x-component of its centre is 0.

Therefore, we have the sphere now centered at (0, -4, 5).

Using the distance formula, we can get the distance d, between the initial points (3, -4, 5) and the new points (0, -4, 5) as follows;

$d = \sqrt{(0-3)^2+ (-4 - (-4))^2 + (5-5)^2}$

$d = \sqrt{(-3)^2+ (-4 + 4)^2 + (5-5)^2}$

$d = \sqrt{(-3)^2 + (0)^2+ (0)^2}$

$d = \sqrt{(9)}$

d = 3

This distance is the radius of the sphere at that point. i.e r = 3

Now substitute this value r = 3 into the general equation of a sphere given in equation (i) above as follows;

(x - 3)² + (y - (-4))² + (z - 5)² = 3²

(x - 3)² + (y + 4)² + (z - 5)² = 9

Therefore, the equation of the sphere when it touches the yz plane is:

(x - 3)² + (y + 4)² + (z - 5)² = 9

(b) To get the equation of the sphere when it touches the xz-plane, we do the following:

i. Since the sphere touches the xz-plane, it means the y-component of its centre is 0.

Therefore, we have the sphere now centered at (3, 0, 5).

Using the distance formula, we can get the distance d, between the initial points (3, -4, 5) and the new points (3, 0, 5) as follows;

$d = \sqrt{(3-3)^2+ (0 - (-4))^2 + (5-5)^2}$

$d = \sqrt{(3-3)^2+ (0+4)^2 + (5-5)^2}$

$d = \sqrt{(0)^2 + (4)^2+ (0)^2}$

$d = \sqrt{(16)}$

d = 4

This distance is the radius of the sphere at that point. i.e r = 4

Now substitute this value r = 4 into the general equation of a sphere given in equation (i) above as follows;

(x - 3)² + (y - (-4))² + (z - 5)² = 4²

(x - 3)² + (y + 4)² + (z - 5)² = 16

Therefore, the equation of the sphere when it touches the xz plane is:

(x - 3)² + (y + 4)² + (z - 5)² = 16

3 0

3 years ago

Help Please! And THANK YOU!

Semenov [28]

Answer:

$100$

Step-by-step explanation:

$\frac{3.75 \times 100}{1.50 \times 100} = \frac{375}{150}$

<h3>so the dividend and divisor were both multiplied by 100.</h3>

6 0

3 years ago

Given congruent triangles name the corresponding sides and corresponding angles

Alborosie

Answer:

See explanation

Step-by-step explanation:

Given $\triangle ABC\cong \triangle ADC$

According to the order of the vertices,

side AB in triangle ABC (the first and the second vertices) is congruent to side AD in triangle ADC (the first and the second vertices);

side BC in triangle ABC (the second and the third vertices) is congruent to side DC in triangle ADC (the second and the third vertices);

side AC in triangle ABC (the first and the third vertices) is congruent to side AC in triangle ADC (the first and the third vertices);

angle BAC in triangle ABC is congruent to angle DAC in triangle ADC (the first vertex in each triangle is in the middle when naming the angles);

angle ABC in triangle ABC is congruent to angle ADC in triangle ADC (the second vertex in each triangle is in the middle when naming the angles);

angle BCA in triangle ABC is congruent to angle DCA in triangle ADC (the third vertex in each triangle is in the middle when naming the angles);

6 0

3 years ago

He vertices of square pqrs are p -4,0 q 4,3 r 7,-5 and s -1,-18.Show that the diagonals of square pqrs are congruent perpendicul

Anit [1.1K]

Answer:

Step-by-step explanation:

The vertices of the square given are P(-4, 0), Q(4, 3), R(7, -5) and, S(-1, -18)

For this diagonal to be right angle the slope of the diagonal must be m1=-1/m2

So let find the slope of diagonal 1

The two points are P and R

P(-4, 0), R(7, -5)

Slope is given as

m1=∆y/∆x

m1=(y2-y1)/(x2-x1)

m1=-5-0/7--4

m1=-5/7+4

m1=-5/11

Slope of the second diagonal

Which is Q and S

Q(4, 3), S(-1, -18)

m2=∆y/∆x

m2=(y2-y1)/(x2-x1)

m2=(-18-3)/(-1-4)

m2=-21/-5

m2=21/5

So, slope of diagonal 1 is not equal to slope two

This shows that the diagonal of the square are not diagonal.

But the diagonal of a square should be perpendicular, this shows that this is not a square, let prove that with distance between two points

Given two points

(x1,y1) and (x2,y2)

Distance between the two points is

D=√(y2-y1)²+(x2-x1)²

For line PQ

P(-4, 0), Q(4, 3)

PQ=√(3-0)²+(4--4)²

PQ=√(3)²+(4+4)²

PQ=√9+8²

PQ=√9+64

PQ=√73

Also let fine RS

R(7, -5) and, S(-1, -18)

RS=√(-18--5)+(-1-7)

RS=√(-18+5)²+(-1-7)²

RS=√(-13)²+(-8)²

RS=√169+64

RS=√233

Since RS is not equal to PQ then this is not a square, a square is suppose to have equal sides

But I suspect one of the vertices is wrong, vertices S it should have been (-1,-8) and not (-1,-18)

So using S(-1,-8)

Let apply this to the slope

Q(4, 3), S(-1, -8)

m2=∆y/∆x

m2=(y2-y1)/(x2-x1)

m2=(-8-3)/(-1-4)

m2=-11/-5

m2=11/5

Now,

Let find the negative reciprocal of m2

Reciprocal of m2 is 5/11

Then negative of it is -5/11

Which is equal to m1

Then, the square diagonal is perpendicular

6 0

3 years ago