20/40 - 21/40 x 80/7 Reduce answer in lowest terms. HELP.

2 answers:

7 0

$\frac{20}{40}-\frac{21}{40}\cdot \frac{80}{7}=\frac{\not20^1 }{\not40^2} - \frac{\not21^3}{\not40^1} \cdot \frac{ \not80^2}{\not7^1} = \frac{1}{2}- 6=-5\frac{1}{2}$

LUCKY_DIMON [66]3 years ago

3 0

$\frac{20}{40} - \frac{21}{40} \cdot \frac{80}{7}= \frac{1}{2} - \frac{3\cdot7}{40}\cdot\frac{2\cdot40}{7}= \frac{1}{2} - \frac{3}{1}\cdot\frac{2}{1}= \frac{1}{2} -6=-5 \frac{1}{2}$

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PLEASE HELP 8th GRADE MATH QUESTION OVER HERE!!!

patriot [66]

The side AB measures option 2. $\sqrt{20}}$ units long.

Step-by-step explanation:

Step 1:

The coordinates of the given triangle ABC are A (4, 5), B (2, 1), and C (4, 1).

The sides of the triangle are AB, BC, and CA. We need to determine the length of AB.

To calculate the distance between two points, we use the formula $d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}.$

where ( $x_{1},y_{1}$ ) are the coordinates of the first point and ( $x_{2},y_{2}$ ) are the coordinates of the second point.

Step 2:

For A (4, 5) and B (2, 1), ( $x_{1},y_{1}$ ) = (4, 5) and ( $x_{2},y_{2}$ ) = (2, 1). Substituting these values in the distance formula, we get

$d=\sqrt{\left(2-4\right)^{2}+\left(1-5}\right)^{2}} = \sqrt{\left(2\right)^{2}+\left(4}\right)^{2}}=\sqrt{20}}.$