Which is greater,20 gallons or 80 liters?

Find the perimeter of the following shapes 9cm and 4.1?

mario62 [17]

Answer: a) 26.2 cm

Step-by-step explanation:

Perimeter of rectangle = 2 (l + w)

2 (l + w) = 2 (9 + 4.1) = 2 (13.1) = 26.2 cm

Hope this helps!

7 0

2 years ago

Two and two fifths divided by one and a half

worty [1.4K]

Answer:

8/5 or 1 3/5

Step-by-step explanation:

2 2/5 = 12/5

1 1/2 = 3/2

Now that we have fractions that are easier to divide, one must understand that the opposite of dividing is multiplying. So if you multiply 12/5 by the reciprocal* of 3/2, which would be 2/3, you would get your answer so...

12 3 12 2 24 8 3

— ÷ — = — × — = — = — <em>or</em> 1 —

5 2 5 3 15 5 5

*the reciprocal is the opposite of the fraction you have. So the denominator would become the numerator, and vise versa.

6 0

3 years ago

What is 27.773 to its nearest tenth

Olenka [21]

Check the hundredth position after the decimal point.

This = 7 so we add 1 to the tenth position so 7 becomes 7+1 = 8

Answer is 27.8 to nearest tenth

7 0

3 years ago

!!!HELP PLEASE !!!!!

Fofino [41]

Tree 3 diagram is the answer

8 0

3 years ago

Solve the following equation by factoring:9x^2-3x-2=0

olya-2409 [2.1K]

Answer:

The two roots of the quadratic equation are

$x_1= - \frac{1}{3} \text{ and } x_2= \frac{2}{3}$

Step-by-step explanation:

Original quadratic equation is $9x^{2}-3x-2=0$

Divide both sides by 9:

$x^{2} - \frac{x}{3} - \frac{2}{9}=0$

Add $\frac{2}{9}$ to both sides to get rid of the constant on the LHS

$x^{2} - \frac{x}{3} - \frac{2}{9}+\frac{2}{9}=\frac{2}{9}$ ==> $x^{2} - \frac{x}{3}=\frac{2}{9}$

Add $\frac{1}{36}$ to both sides

$x^{2} - \frac{x}{3}+\frac{1}{36}=\frac{2}{9} +\frac{1}{36}$

This simplifies to

$x^{2} - \frac{x}{3}+\frac{1}{36}=\frac{1}{4}$

Noting that (a + b)² = a² + 2ab + b²

If we set a = x and b = $\frac{1}{6}\right)$ we can see that

$\left(x - \frac{1}{6}\right)^2$ = $x^2 - 2.x. (-\frac{1}{6}) + \frac{1}{36} = x^{2} - \frac{x}{3}+\frac{1}{36}$

So

$\left(x - \frac{1}{6}\right)^2=\frac{1}{4}$

Taking square roots on both sides

$\left(x - \frac{1}{6}\right)^2= \pm\frac{1}{4}$

So the two roots or solutions of the equation are

$x - \frac{1}{6}=-\sqrt{\frac{1}{4}}$ and $x - \frac{1}{6}=\sqrt{\frac{1}{4}}$

$\sqrt{\frac{1}{4}} = \frac{1}{2}$

So the two roots are

$x_1=\frac{1}{6} - \frac{1}{2} = -\frac{1}{3}$

and

$x_2=\frac{1}{6} + \frac{1}{2} = \frac{2}{3}$

7 0

1 year ago