Put the phone numbers in order from least to greatest. 1.26, 1.3, 1.2

2 answers:

5 0

1.2;1.26;1.3
You can add a zero to the end of each decimal, so it can change to make it easier.
1.26 doesn't change, 1.3 can be 1.30, and 1.2 can be 1.20.

So you can compare them like this
1.26
1.30
1.20
Just line them up and make sure they are all the same length, you can only do this to a decimal.

andriy [413]3 years ago

3 0

Least 1.2, 1.26, 1.3 greatest

You might be interested in

Can you help me please?

melisa1 [442]

Answer:

it is a it the first one. 7/13

5 0

3 years ago

Read 2 more answers

Write a definite integral that represents the area of the region. (Do not evaluate the integral.)

joja [24]

Answer:

$\int_{-1}^{1}7(x^{3}-x)\:dx$

Step-by-step explanation:

1) The other curve is $y=0$ then the common points of both curves are x-intercepts, the roots of $y=7(x^{3}-x)$

$y=7(x^{3}-x)\Rightarrow 7(x^3-x)=0 \Rightarrow 7(x^{3}-x)=7x(x-1)(x-1)\Rightarrow \\S=\left ( 0,0 \right ),\left ( -1,0 \right ),\left ( 1,0 \right )$

2). Then those intersection points are the upper and the lower limits. Plugging in to this formula for they belong to the interval [-1,1]:

$\int_{a}^{b}|f(x)-g(x)dx$

$\int_{a}^{b}|f(x)-g(x)dx \Rightarrow \int_{-1}^{1}|7(x^{3}-x)-0|dx \Rightarrow \int_{-1}^{1}7(x^{3}-x)\:dx$

8 0

3 years ago

Choose the correct word that completes each statement about inscribing a square in a circle.

Anna11 [10]

Answer:

1. Perpendicular

2. Isosceles

3. Never

Step-by-step explanation:

1. AC ⊥ BD because diameter of a square are perpendicular bisector of each other.

2. In Δ AOB , By using pythagoras : AB² = OA² + OB² .......( 1 )

In Δ COB , By using pythagoras : BC² = OC² + OB² ..........( 2 )

But, OA = OC because both are radius of same circle

So, by using equations ( 1 ) and ( 2 ), We get AB = BC ≠ AC

⇒ ABC is a triangle having two equal sides so ABC is an isosceles triangle.

3. The side can never be equal to radius of circle because the side of the square will be chord for the circle and in a circle chord can never be equal to its radius