All categories

3 years ago

Put the following products in order from greatest to least 6x 2/5 , 3 2/3 x 2/5, 2/7 X 2/5, 2/2 x2/5

Mathematics

1 answer:

DedPeter [7]3 years ago

6 0

6x2/5>3 2/3x 2/5>2/2x2/5>2/7x2/5

You might be interested in

Evaluate 7m+2n-8p/n m=4, n=2, and p=1.5

Firlakuza [10]

Steps to solve:

7m + 2n - 8p/n

~Substitute

7(4) + 2(2) - 8(1.5)/2

~Simplify

28 + 4 - 12/2

~Simplify

28 + 4 - 6

~Add

~Subtract

Best of Luck!

4 0

3 years ago

6.1.3 quiz pythagorean theorem question 9

Olegator [25]

Answer: 14.73

Step-by-step explanation:

The given triangle is a right angle triangle.

EF^2 + DF^2 = ED^2

The hypotenuse is |ED| while the two shorter legs are |EF| and |DF|.

We can then apply the Pythagoras Theorem to find the length of EF.

(EF)^2 + (DF)^2 = (ED)^2

(EF)^2 + (12)^2 = (19)^2

(EF)^2 + 144 = 361

(EF)^2 = 361 - 144

(EF)^2 = 217

EF = 14.73

6 0

2 years ago

Find the length of arc ac in terms of pi

aniked [119]

=theta/360°×2pi ×r
theta=180-60=120°
r=24÷2=12

120/360 × 2 × 22/7 × 12=
1/3 × 2 × 22/7 × 12=
176/7=
25.14

3 0

3 years ago

.. Which of the following are the coordinates of the vertices of the following square with sides of length a?

atroni [7]

Option A: $O(0,0), S(0,a), T(a,a), W(a,0)$

Option D: $O(0,0), S(a,0), T(a,a), W(0,a)$

Step-by-step explanation:

Option A: $O(0,0), S(0,a), T(a,a), W(a,0)$

To find the sides of a square, let us use the distance formula,

$d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

Now, we shall find the length of the square,

$\begin{array}{l}{\text { Length } O S=\sqrt{(0-0)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a} \\{\text { Length } S T=\sqrt{(a-0)^{2}+(a-a)^{2}}=\sqrt{a^{2}}=a} \\{\text { Length } T W=\sqrt{(a-a)^{2}+(0-a)^{2}}=\sqrt{a^{2}}=a} \\{\text { Length } O W=\sqrt{(a-0)^{2}+(0-0)^{2}}=\sqrt{a^{2}}=a}\end{array}$

Thus, the square with vertices $O(0,0), S(0,a), T(a,a), W(a,0)$ has sides of length a.

Option B: $O(0,0), S(0,a), T(2a,2a), W(a,0)$

Now, we shall find the length of the square,

$\begin{aligned}&\text { Length } O S=\sqrt{(0-0)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a\\&\text {Length } S T=\sqrt{(2 a-0)^{2}+(2 a-a)^{2}}=\sqrt{5 a^{2}}=a \sqrt{5}\\&\text {Length } T W=\sqrt{(a-2 a)^{2}+(0-2 a)^{2}}=\sqrt{2 a^{2}}=a \sqrt{2}\\&\text {Length } O W=\sqrt{(a-0)^{2}+(0-0)^{2}}=\sqrt{a^{2}}=a\end{aligned}$

This is not a square because the lengths are not equal.

Option C: $O(0,0), S(0,2a), T(2a,2a), W(2a,0)$

Now, we shall find the length of the square,

$\begin{array}{l}{\text { Length OS }=\sqrt{(0-0)^{2}+(2 a-0)^{2}}=\sqrt{4 a^{2}}=2 a} \\{\text { Length } S T=\sqrt{(2 a-0)^{2}+(2 a-2 a)^{2}}=\sqrt{4 a^{2}}=2 a} \\{\text { Length } T W=\sqrt{(2 a-2 a)^{2}+(0-2 a)^{2}}=\sqrt{4 a^{2}}=2 a} \\{\text { Length } O W=\sqrt{(2 a-0)^{2}+(0-0)^{2}}=\sqrt{4 a^{2}}=2 a}\end{array}$

Thus, the square with vertices $O(0,0), S(0,2a), T(2a,2a), W(2a,0)$ has sides of length 2a.

Option D: $O(0,0), S(a,0), T(a,a), W(0,a)$

Now, we shall find the length of the square,

$\begin{aligned}&\text { Length OS }=\sqrt{(a-0)^{2}+(0-0)^{2}}=\sqrt{a^{2}}=a\\&\text { Length } S T=\sqrt{(a-a)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a\\&\text { Length } T W=\sqrt{(0-a)^{2}+(a-a)^{2}}=\sqrt{a^{2}}=a\\&\text { Length } O W=\sqrt{(0-0)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a\end{aligned}$

Thus, the square with vertices $O(0,0), S(a,0), T(a,a), W(0,a)$ has sides of length a.

Thus, the correct answers are option a and option d.

8 0

2 years ago

If a liquid weighs 2 pounds and has a capacity of 3 gallons, what is its density?

Illusion [34]

5 gallons would be the right answer

8 0

3 years ago