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vivado [14]
4 years ago
5

Solve each system by substitution.4x+2y=10x-y=13​

Mathematics
2 answers:
emmainna [20.7K]4 years ago
5 0

x-y=13

x=13+y--(1)

4x+2y=10

2x+y=5

26+2y+y=5

3y=-21

y=-7

Substituting value of y in equation (1)

x=13-7

x=6

Bond [772]4 years ago
3 0

Answer:

<h2>x = 6 and y = -7 → (6, -7)</h2>

Step-by-step explanation:

\left\{\begin{array}{ccc}4x+2y=10\\x-y=13&\text{add y to both sides}\end{array}\right\\\\\left\{\begin{array}{ccc}4x+2y=10&(1)\\x=y+13&(2)\end{array}\right\\\\\\\text{Substitute (2) to (1):}\\\\4(y+13)+2y=10\qquad\text{use the distributive property}\\\\(4)(y)+(4)(13)+2y=10\\\\4y+52+2y=10\qquad\text{subtract 52 from both sides}\\\\6y=-42\qquad\text{divide both sides by 6}\\\\\boxed{y=-7}\\\\\text{Put the value of y to (2):}\\\\x=-7+13\\\\\boxed{x=6}

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Stuck on this it's Algebra 2 12x^{3} -3x^{2} =0
olchik [2.2K]

Answer:

Exact Form:

x = 0, 1/4

Decimal Form:

x = 0, 0.25

Step-by-step explanation:

<u>Step 1: Factor 3x^2 out of 12x^3 - 3x^2 </u>

Factor 3x^2 out of 12x^3:

<em> 3x^2 (4x) - 3x^2 = 0</em>

Factor 3x^2 out of -3x^2:

<em> 3x^2 (4x) + 3x^2 (-1)  = 0</em>

Factor 3x^2 out of -3x^2 <em>(4x) + 3x^2 (-1) </em>:

<em> 3x^2 (4x-1) = 0</em>

<em />

<u>Step 2: Divide each term by 3 and simpify</u>

<u></u>

divide each term in 3x^2 (4x-1) = 0 by 3.

3x^2 (4x-1) / 3 = 0 / 3

<em>simplify 3x^2 (4x-1) / 3.</em>

<em />

<em>Cancel the common factors.</em>

<u><em>3</em></u><em> </em>x^2 (4x -1) / <em><u>3</u></em> = 0 / 3

divide x^2 (4x-1) by 1.

x^2 (4x-1) / 3 = 0 / 3

Apply the Distributive Property

Reorder.

Rewrite using the commutative property of multiplication.

4x^2 x + x^2 · -1 = 0 / 3

Move -1  to the left of  x^2

4x^2 x -1 · x^2  = 0 / 3

Simplify each term

multiply x^2 by x^2 by adding the exponents.

Move x

4 (x · x^2) -1 · x^2= 0 / 3

Multiply x by x^2

Rase x to the power of 1.

4 (x^1 · x^2) -1 · x^2= 0 / 3

Use the power rule a^m a^n = a^m+n to combine exponents

4x^1+2 -1 · x^2= 0 / 3

Add 1 and 2.

4x^3 -1 · x^2= 0 / 3

Rewrite -1x^2  as -x^2.

4x^3 -x^2= 0 / 3

Divide 0 by 3

4x^3 -x^2= 0

<u>Step 3: Factor x^2 out of 4x^3 -x^2.</u>

Factor x^2 out of 4x^3

x^2 (4x) -x^2 = 0

Factor x^2 out of -x^2

x^2 (4x) x^2 · -1 = 0

Factor x^2 out of x^2 (4x) x^2 · -1

x^2 (4x -1) = 0

If any individual factor on the left side of the equation is equal to 0,  the entire expression will be equal to  0

x^2 = 0

4x -1 = 0

Set x^2 equal to 0 and solve for x

x = 0

Set 4x -1 equal to 0 and solve for x

x = 1/4

The final solution is all the values that make x^2 (4x-1) = 0 true

x= (0, 1/4)

6 0
3 years ago
DO NOT PRANK!! I NEED HELP PLS (will give brainly)
Nitella [24]

Answer:

b

Step-by-step explanation:

hiiii

3 0
3 years ago
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V(t)=t2−3t what is the meaning of the quantity v(4) ?
GalinKa [24]
V(4)=4²-3×4=16-12=4 is the result of substituting t=4 in the function.
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A triangle has a side lengths of 34in., 28in., and 42in. is the triangle an acute, right, or obtuse?
jenyasd209 [6]
It would be an acute
3 0
3 years ago
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The temperature, H, in degrees Celsius, of a cup of coffee placed on the kitchen counter is given by H = f(t), where t is in min
Deffense [45]

Answer:

a

    f'(t) is  negative

b

 The  unit is degC/min

c

At <u>35</u> minutes after the coffee was put on the counter, its temperature is <u>68</u> and will <u>decrease</u> by about <u>0.75</u> in the next 30 seconds.          

Step-by-step explanation:

From the question we are told we are told that  

   The  equation for the temperature of a cup of coffee place on a counter is

         H= f(t)

Here  t is the time in minutes the  coffee was put on the counter

Generally  f(t)'  is the derivative of  f(t) at it represents the change of the temperature of the cup of coffee with respect to time

 Generally for the coffee placed on the counter after a couple of minutes the temperature will decrease hence  f(t)' is  negative

Generally the unit of the temperature is in degrees Celsius while that of time is  in minutes

  So the change in temperature with time f(35)' will be in degrees Celsius/minutes (i.e  degC/min)

  From the question we are told that

        |f(35)'| =  1.5

i.e  the rate of change of temperature after 35 minutes is  1.5

and

         |f(35)| =  68

i.e the temperature of the cup of coffee after 35 minutes is  67 degC

Now after another   t=  30  seconds =  \frac{30}{60}= 0.5 \ minutes the  rate  of change of the temperature of the cup of  coffee is  

          R =  |f(35)'| *  t

=>        R = 1.5  *  0.5

=>        R = 0.75    

So

At <u>35</u> minutes after the coffee was put on the counter, its temperature is <u>68</u> and will <u>decrease</u> by about <u>0.75</u> in the next 30 seconds.  

     

5 0
3 years ago
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