3(×-4)+6=5(×-1)+1 is it a one solution or infinitely many solution or no solution

Solve equation. k+3=14

lorasvet [3.4K]

To solve the equation, subtract 3 from both sides. The solution is k = 11.

A linear equation is an equation of which the highest degree of its variables is 1.

Steps to solve linear equations:

Simplify the expressions. If there are parentheses, remove them by multiplying the corresponding terms.
Combine the same terms.
Isolate the variable on one side by using subtraction of addition.
Use division or multiplication to find the value of the variable.

Recall that if we subtract or add the same number to both sides of an equation, it does not change the equality.

The given problem:

Solve k + 3 = 14.

The variable is k. Subtract 3 from both sides:

k + 3 - 3 = 14 - 3

k = 11 (solved)

Learn more about solving an equation here:

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5 0

1 year ago

Find b for the following right angled triange where a=5 and c=13

lilavasa [31]

You would use the Pythagorean there on which is a^2+b^2=c^2, from there you plug int what they gave you and solve for b, and b=12

4 0

3 years ago

Please help me it’s a formative grade

sergejj [24]

Answer:

Step-by-step explanation: 800/x=100/17

(800/x)*x=(100/17)*x - we multiply both sides of the equation by x

800=5.88235294118*x - we divide both sides of the equation by (5.88235294118) to get x

800/5.88235294118=x

136=x

x=136

83% of 800=664

5 0

3 years ago

33. Suppose you were on a planet where the

tatyana61 [14]

Answer:

a) 3.675 m

b) 3.67m

Explanation:

We are given acceleration due to gravity on earth = $9.8ms^-2$

And on planet given = $2.0ms^-2$

A) Since the maximum jump height is given by the formula

$\mathrm{H}=\frac{\left(\mathrm{v} 0^{2} \times \sin 2 \emptyset\right)}{2 \mathrm{g}}$

Where H = max jump height,

v0 = velocity of jump,

Ø = angle of jump and

g = acceleration due to gravity

Considering velocity and angle in both cases

$\frac{\mathrm{H} 1}{\mathrm{H} 2}=\frac{\mathrm{g} 2}{\mathrm{g} 1}$

Where H1 = jump height on given planet,

H2 = jump height on earth = 0.75m (given)

g1 = 2.0 $ms^-2$ and

g2 = 9.8 $ms^-2$

Substituting these values we get H1 = 3.675m which is the required answer

B) Formula to find height of ball thrown is given by

$\mathrm{h}=(\mathrm{v} 0 * \mathrm{t})+\frac{\mathrm{a} *\left(t^{2}\right)}{2}$

which is due to projectile motion of ball

Now h = max height,

v0 = initial velocity = 0,

t = time of motion,

a = acceleration = g = acceleration due to gravity

Considering t = same on both places we can write

$\frac{\mathrm{H} 1}{\mathrm{H} 2}=\frac{\mathrm{g} 1}{\mathrm{g} 2}$

where h1 and h2 are max heights ball reaches on planet and earth respectively and g1 and g2 are respective accelerations

substituting h2 = 18m, g1 = 2.0 $ms^-2$ and g2 = 9.8 $ms^-2$

We get h1 = 3.67m which is the required height

6 0

4 years ago

What is the solution of 5/6 x 4/9.

Sphinxa [80]

The answer is 10/27

please see the attached picture for full solution

hope it will helps

Good luck on your assignment

8 0

3 years ago

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