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3 years ago

How many solutions does this equation have 3x-2x+4=2+x+2

Mathematics

1 answer:

ch4aika [34]3 years ago

8 0

If you simply it down you get
$x + 4 = x + 4$
so X can technically be any number

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Help me help me help me help me help me help me help me help me help me

kirza4 [7]

Answer:

a) x=7

b) x=4

c) x= 1/9

I hope this helps! Do you need more help?

6 0

3 years ago

Read 2 more answers

Fiona solved the equation shown 1/2-1/3(6x-3)=-13/2 what is the missing step of her solution?

V125BC [204]

Answer:

Summation of the non variable Expression within the quality sign

1/2 + 1-2x= -13/2

3/2-2x= -13/2

Step-by-step explanation:

1/2-1/3(6x-3)=-13/2

First step

Using the distributive property to simply

1/2-(6x/3)+(3/3)=-13/2

1/2 -2x +1 = -13/2

Second step

Summation of the non variable Expression within the quality sign

1/2 + 1-2x= -13/2

3/2-2x= -13/2

Third step

Isolating the variable Expression by using the addition property of equality

-2x = -13/2 - 3/2

-2x = -16/2

Fourth step

Isolating the variable by using the division property of equality

-2x = -16/2

X = -16/2 * -1/2

X = -16/-4

X= 4

4 0

3 years ago

ashley is baking a cake shaped as a right triangle. Two of the angles are measurement of 5x and 10x. What is the value of x?

DENIUS [597]

Answer:

A. 6

Step-by-step explanation:

Since, cake is of right triangled shape, so measure of its one angle would be 90° and the sum of the measures of the remaining two angles will also be equal 90°.

Therefore,

5x +10x = 90

15x = 90

x = 90/15

x = 6

8 0

3 years ago

Read 2 more answers

If g(t) = 2(t)-6, find g(10)

ElenaW [278]

All good. Fill in t with 10 and solve for the expression g(t). 10 = t so it fill in

4 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

3 years ago