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Hunter-Best [27]
2 years ago
12

Leftover Expression

Mathematics
1 answer:
Mumz [18]2 years ago
4 0

The second leftover expression is not o(a+b). It is 6(a + b). I have attached the correct question to depict that.

Answer:

The equivalent expressions are;

8a + 2 and 6a + 6b

Step-by-step explanation:

The two leftover expressions are given as;

2(4x + 1) and 6(a + b)

In algebra, equivalent expressions are simply those expressions which when simplified, give the same resulting expression as the initial one.

Thus simply means expanding or collecting like times to make it clearer.

Now, in our question, like terms have already been collected. This means that to find an equivalent expression, we will just expand the bracket.

Thus;

2(4x + 1) will be expanded by using the 2 outside the bracket to multiply the terms inside the bracket. This gives;

8x + 2

Similarly,

6(a + b) will be expanded by using the 2 outside the bracket to multiply the terms inside the bracket. This gives;

6a + 6b

Thus;

The equivalent expressions are;

8a + 2 and 6a + 6b

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tankabanditka [31]

Step 1

<u>Find the slope of the line AB</u>

we know that

the formula to find the slope between two points is equal to

m=\frac{(y2-y1)}{(x2-x1)}  

we have

A(-2,4)\\B(0,-4)

substitute in the formula

m=\frac{(-4-4)}{(0+2)}  

m=\frac{(-8)}{(0+2)}  

m1=-4  

Step 2

<u>Find the slope of the line that passes through point Z and is perpendicular to line AB</u>

we know that

If two lines are perpendicular, then the product of their slopes is equal to minus one

so

m1*m2=-1

m2=-\frac{1}{m1}

we have

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substitute

m2=\frac{1}{4}

Step 3

<u>Find the equation of the line</u> <u>that passes through point Z and is perpendicular to line AB</u>

we have

Z(0,2)\\m2=\frac{1}{4}

we know that

the equation of the line into point -slope form is equal to

y-y1=m*(x-x1)

substitute

y-2=\frac{1}{4}*(x-0)

y=\frac{1}{4}x+2

<u>Verify which point is on the line that passes through point Z and is perpendicular to line AB</u>

we know that  

If a point is on the line, then the coordinates of the point must be satisfy the equation of the line

we are going to proceed to verify each point to determine the solution

Substitute the values of x and y of the point in the equation of the line. If the equation is true, then the point is on the line

Step 4

point A(-4,1)

1=\frac{1}{4}*(-4)+2

1=1 --------> the equation is true

therefore

the point A(-4,1) is on the line

Step 5

point B(1,-2)

-2=\frac{1}{4}*(1)+2

-2=\frac{9}{4} --------> the equation is not true

therefore

the point B(1,-2) is not on the line

Step 6

point C(2,0)

0=\frac{1}{4}*(2)+2

0=\frac{5}{2} --------> the equation is not true

therefore

the point C(2,0) is not on the line

Step 7

point D(4,4)

4=\frac{1}{4}*(4)+2

4=3 --------> the equation is not true

therefore

the point D(4,4) is not on the line

<u>the answer is the option</u>

A(-4,1)

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