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2 years ago

What is 6+3(-8x-2)=12

Mathematics

2 answers:

stepladder [879]2 years ago

6 0

Answer:

=12

Step-by-step explanation:

6+3(-8x-2)=12

ladessa [460]2 years ago

3 0

3(−8x−2)=63(-8x-2)=6

3(−8x−2)=6

3 Divide both sides by

−8x−2=63-8x-2=\frac{6}{3}

−8x−2=

4 Simplify

63\frac{6}{3}

−8x−2=2-8x-2=2

−8x−2=2

5 Add

2 to both sides.

−8x=2+2-8x=2+2

−8x=2+2

6 Simplify

2+22+2

2+2 to

−8x=4-8x=4

−8x=4

7 Divide both sides by

−8-8

−8.

x=−48x=-\frac{4}{8}

x=−

8 Simplify

x=−1/2

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The thickness of eight pads designed for use in aircraft engine mounts are measured. The results, in mm, are 41.8, 40.9, 42.1, 4

nika2105 [10]

Answer:

We accept H₀ we don´t have enough evidence to support that the mean thickness is greater than 41 mm

Step-by-step explanation:

Sample Information:

Results:

41.8

40.9

42.1

41.2

40.5

41.1

42.6

40.6

From the table we get:

sample mean : x = 41.35

sample standard deviation s = 0.698

Hypothesis Test:

Null Hypothesis H₀ x = 41

Alternative Hypothesis Hₐ x > 41

The test is a one-tail test

If significance level is 0.01 and n = 8 we need to use t-student distribution

From t-table α = 0.01 and degree of freedom df = n - 1 df = 8 - 1

df = 7 t(c) = 2.998

To calculate t(s) = ( x - 41 ) / s/√n

t(s) = ( 41.35 - 41 ) / 0.698/√8

t(s) = 0.35 * 2.83/ 0.698

t(s) = 1.419

Comparing t(s) and t(c)

t(s) < t(c)

t(s) is in the acceptance region we accept H₀

7 0

2 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

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

Consider the system of linear equations y=3x+5 y=Mx+b Which values an M & B will create a system of linear equations with no

OLEGan [10]

Answer:

value of 'm' equal to 5 will produce no solution as they'll be parallel lines

value of 'b' equal to 5 will produce no solution because it will be the same line

Step-by-step explanation:

5 0

2 years ago

R^2 -9tw +3wr -3tr =

Tju [1.3M]

The answer would be....(r + 3w) × ( r - 3t)

6 0

2 years ago

(Fog) (x) f(x)=6x-5 <br> g(x) =6x^2-3x

algol [13]

Answer:

see below

Step-by-step explanation:

put g(x) in for the x in f(x)

$f(x)= 6x-5\\g(x)= 6x^{2} -3x\\f(g(x)= 6(6x^{2} -3x)-5$

5 0

2 years ago