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3 years ago

Is 20/100 equivalent to 1/5

2 answers:

8 0

Yes! if you check by doing cross multiplication, you will get the same answer! OR if you simplify 20/100 and reduce it, you will get 1/5

egoroff_w [7]3 years ago

6 0

The correct answer is: Yes, 20/100 equivalent to 1/5

Explanation:

The given fraction is:

$\frac{20}{100}$

Now, cancel one zero of 20 (in numerator) with one zero of 100 (in denominator), you would get:

$\frac{2}{10}$

Now 2 * 5 = 10, therefore after cancellation, the fraction will become:

$\frac{1}{5}$

Hence, we can say thaat 20/100 is equivalent to 1/5.

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3 years ago

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The product of 3p and q - 3.

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To find your answer. You have to distribute the 3p to by individually multiplying it by q and -3

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3 years ago

Find m(x) = f(x) * h(x) Values - f(x) = -3x + 6; g(x) = 7x -5; h(x) = -2

Lisa [10]

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Step-by-step explanation:

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2 years ago

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The sum of two numbers is 52. The greater number is 4 more than the smaller number. Which equation can be used to solve for the

Arisa [49]

Answer:

x+(x+4)=52

Step-by-step explanation:

Let's name the smaller number x.

The greater number would then be (x+4).

The sum of two numbers means we are adding them together.

The equation we could then set up would be:

x+(x+4)=52

Now we can solve the equation to find the two numbers if needed.

Here is how:

x+x+4=52

Combine like terms.

2x+4=52

Subtract 4 from both sides.

2x=48

Divide both sides by 2

x=24

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6 0

3 years ago

Part B

alisha [4.7K]

Question:

Consider ΔABC, whose vertices are A (2, 1), B (3, 3), and C (1, 6); let the line segment AC represent the base of the triangle.

(a) Find the equation of the line passing through B and perpendicular to the line AC

(b) Let the point of intersection of line AC with the line you found in part A be point D. Find the coordinates of point D.

Answer:

$y = \frac{1}{5}x + \frac{12}{5}$

$D = (\frac{43}{26},\frac{71}{26})$

Step-by-step explanation:

Given

$\triangle ABC$

$A = (2,1)$

$B = (3,3)$

$C = (1,6)$

Solving (a): Line that passes through B, perpendicular to AC.

First, calculate the slope of AC

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Where:

$A = (2,1)$ --- $(x_1,y_1)$

$C = (1,6)$ --- $(x_2,y_2)$

The slope is:

$m = \frac{6- 1}{1 - 2}$

$m = \frac{5}{-1}$

$m = -5$

The slope of the line that passes through B is calculated as:

$m_2 = -\frac{1}{m}$ --- because it is perpendicular to AC.

So, we have:

$m_2 = -\frac{1}{-5}$

$m_2 = \frac{1}{5}$

The equation of the line is the calculated using:

$m_2 = \frac{y_2 - y_1}{x_2 - x_1}$

Where:

$m_2 = \frac{1}{5}$

$B = (3,3)$ --- $(x_1,y_1)$

$(x_2,y_2) = (x,y)$

So, we have:

$\frac{1}{5} = \frac{y - 3}{x - 3}$

Cross multiply

$5(y-3) = 1(x - 3)$

$5y - 15 = x - 3$

$5y = x - 3 + 15$

$5y = x +12$

Make y the subject

$y = \frac{1}{5}x + \frac{12}{5}$

Solving (b): Point of intersection between AC and $y = \frac{1}{5}x + \frac{12}{5}$

First, calculate the equation of AC using:

$y = m(x - x_1) + y_1$

Where:

$A = (2,1)$ --- $(x_1,y_1)$

$m = -5$

So:

$y=-5(x - 2) + 1$

$y=-5x + 10 + 1$

$y=-5x + 11$

So, we have:

$y=-5x + 11$ and $y = \frac{1}{5}x + \frac{12}{5}$

Equate both to solve for x

i.e.

$y = y$

$-5x + 11 = \frac{1}{5}x + \frac{12}{5}$

Collect like terms

$-5x -\frac{1}{5}x = \frac{12}{5} - 11$

Multiply through by 5

$-25x-x = 12 - 55$

Collect like terms

$-26x = -43$

Solve for x

$x = \frac{-43}{-26}$

$x = \frac{43}{26}$

Substitute $x = \frac{43}{26}$ in $y=-5x + 11$

$y = -5 * \frac{43}{26} + 11$

$y = \frac{-5 *43}{26} + 11$

Take LCM

$y = \frac{-5 *43+11 * 26}{26}$

$y = \frac{71}{26}$

Hence, the coordinates of D is:

$D = (\frac{43}{26},\frac{71}{26})$

3 0

3 years ago