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3 years ago

1 3/5 x 2 7/8 please multiply then write in simplest form

Mathematics

1 answer:

prohojiy [21]3 years ago

3 0

Answer:

4 3/5

Step-by-step explanation:

Convert each term into an improper fraction, 8/5 x 15/8. Then find common denominator, 40. Then your equasion would be 64/40 x 75/40. Multiply your terms, then simplify the answer.

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Write an equation through the points (-3,7) and (1,-1)

Aliun [14]

The slope is -2.

The slope is the rise over the run. Because the line goes down from left to right, it makes the value negative. For every one the it moves to the right, the line goes down two units.

8 0

3 years ago

Triangle ABC with vertices A (-1, 4), B( -4, 1), and C(-1, 1) is reflected across the y-axis and then translated 4 units to the

Anna71 [15]

Given:

Triangle ABC with vertices A (-1, 4), B( -4, 1), and C(-1, 1) is reflected across the y-axis and then translated 4 units to the left to form triangle A’B’C’.

To find:

The coordinates of the vertices of triangle A’B’C’.

Solution:

If triangle ABC reflected across y-axis, then

$(x,y)\to (-x,y)$

$A(-1,4)\to A_1(1,4)$

$B(-4,1)\to B_1(4,1)$

$C(-1,1)\to C_1(1,1)$

Coordinates of image after reflection across y-axis are $A_1(1,4),B_1(4,1),C_1(1,1).$

Then triangle translated 4 units to the left to form triangle A’B’C’.

$(x,y)\to (x-4,y)$

$A_1(1,4)\to A'(-3,4)$

$B_1(4,1)\to B'(0,1)$

$C_1(1,1)\to C'(-3,1)$

Therefore, the vertices of triangle A’B’C’ are A’ (-3, 4), B’ (0, 1), C’ (-3, 1).

Hence, the correct option is B.

6 0

3 years ago

Hey does any body know how i can approximate 13 like this had this homework but it got late on me.

Illusion [34]

Answer:

(see attachment)

To approximate the square root of 13:

Working from the top down...

Enter the number you are trying to approximate in the top box: $\boxed{\sf \sqrt{13}}$

Find the perfect squares directly below and above 13.

Perfect squares: 1, 4, 9, 16, 25, 36, ...

Therefore, the perfect squares below and above 13 are: 9 and 16

Enter these with square root signs in the next two boxes: $\boxed{\sf \sqrt{9}}$ and $\boxed{\sf \sqrt{16}}$

Carry out the operation and enter $\boxed{\sf 3}$ and $\boxed{\sf 4}$ in the next two boxes.

Enter the number you are trying to square root (13) in the top left box, the perfect square above it (16) in the box below, then the perfect square below it (9) in the two boxes to the right of these. Carry out the subtractions and place the numbers in the boxes to the right.

$\dfrac{\boxed{\sf 13}-\boxed{\sf 9}}{\boxed{\sf 16}-\boxed{\sf 9}}=\dfrac{\boxed{\sf 4}}{\boxed{\sf 7}}$

Now enter the number you are trying to square root (13) under the square root sign. Place the square root of the perfect square below it (3) in the box to the right. Copy the fraction from above (4/7). Finally, enter this mixed number into a calculator and round to the nearest hundredth.

$\sf \sqrt{13}=\boxed{\sf3}\dfrac{\boxed{\sf 4}}{\boxed{\sf 7}}=\boxed{\sf3.57}$