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Tresset [83]
1 year ago
13

Question 5 of 10

Mathematics
1 answer:
Harman [31]1 year ago
5 0

Answer:

D

Step-by-step explanation:

6x + c = k ( isolate the term in x by subtracting c from both sides )

6x = k - c ( isolate x by dividing both sides by 6 )

x = \frac{k-c}{6}

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Find the solution to the following system using substitution or elimination: y = 3x + 2 y = -2x - 8 O A. (-5,-) OB. (-2,-4) O C.
mr Goodwill [35]

Answer:

B. (-2,-4)

Explanation

Given equations:

   y = 3x + 2

   y = -2x - 8

Solving both equations will yield the values of x and y;

Solution:

   y = 3x + 2    ----- (i)

   y = -2x - 8   ------ (ii)

Using substitution method, input equation i, into ii

    3x + 2 = -2x - 8

Collect like terms and solve;

     3x + 2x = -8 -2

         5x  = -10

            x  = -2

Then put x = -2 into i, to find y

      y = (-2 x 3) + 2

       y = -6 + 2 = -4

So, the solution of the equation is B. (-2,-4)

6 0
2 years ago
Solve 3x-5=16<br> i need the answer this math
Juliette [100K]
The correct answer is 7. You add five to both sides then divide by three and you get 7 for x

4 0
3 years ago
Suppose a white dwarf star has a diameter of approximately 1.8083 to the power of 4 km. Use the formula 4n to the power of 2 to
aivan3 [116]

ANSWER:

The surface area of the star is 3.2700 x 10^{8} square kilometres.

EXPLANATIONS:

Diameter of the star = 1.8083 x 10^{4} Km.

Surface area of the star = 4n^{2}

Where n is the radius of the star.

So that;

n = \frac{ 1.8083 * 10^{4} }{2}

   = 0.90415 x 10^{4}

n =  0.90415 x 10^{4} Km

Thus,

Surface area = 4 x (0.90415*10^{4} )^{2}

                     = 326994889

Surface area = 3.2700 x 10^{8} km^{2}

Therefore, the surface area of the star is 3.2700 x 10^{8} square kilometres.

4 0
3 years ago
A student is reading a book at about 370 words per minute. convert this rate to words per hour
Vlad1618 [11]
This would convert to 22,200 words an hour.
3 0
3 years ago
Determine whether the following equation defines y as a function of x.
Aleks04 [339]

Answer:

The equation can define y as a function of x and it also can define x as a function of y.

Step-by-step explanation:

A relation is a function if and only if each value in the domain is mapped into only one value in the range.

So, if we have:

f(x₀) = A

and, for the same input x₀:

f(x₀) = B

Then this is not a function, because it is mapping the element x₀ into two different outputs.

Now we want to see it:

x + y = 27

defines y as a function of x.

if we isolate y, we get:

y = f(x) = 27 - x

Now, this is a linear equation, so for each value of x we will find an unique correspondent value of y, so yes, this is a function.

Now we also want to check if:

x +  y = 27

defines x as a function of y.

So now we need to isolate x to get:

x = f(y) = 27 - y

Again, this is a linear equation, there are no values of y such that f(y) has two different values. Then this is a function.

3 0
3 years ago
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