All categories

3 years ago

Help Dana find the sum 346 421 152

1 answer:

8 0

To add and get the sum
346
421
+152

We start on the ones place 6 + 1 + 2 = 9, the sum of the digits in the ones place is not up to 10, we will not regroup this ones place.

So, basically the answer to 13a is NO.
Since we are not able to regroup the ones place there is no regrouped tens to add, therefore the answer to 13b is also NO.

Next is the tens place, 4 +2 +5 = 11, the sum of the digits in the ones place is up to 10 so we regroup the tens place.

Thus, the answer to 13c is YES.

Since, we regrouped the ones place, we add the regrouped hundred, thus the answer to 13d is YES.

Finally, for the hundred place, 1 + 3 + 4 + 1 = 9

Therefore, the sum of
346
421
+ 152
is 919

You might be interested in

The formula for the area of a circle is A = r 2. Classify the area of a circle with a radius of 0.75 centimeters as either ratio

OlgaM077 [116]

First:The area of a circle is πr squared. Not r2
Second. The area of the circle HAS to be irrational because π is an irrational number. The answer would be, if solved for a circle with the radius of 0.75, the area would be 0.5625π, which is irrational.

6 0

4 years ago

Two students factored 2x^2 + 6x - 20. Leuko said that the factorization was (2x - 4)(x + 5). Ray gave the factorization as (x -

shutvik [7]

Answer:

See explanation

Step-by-step explanation:

If you FOIL both of them, you see that they are both correct, but they both need to factor out a 2 and then they will both have the same thing.

2(x - 2)(x + 5)

6 0

3 years ago

Read 2 more answers

How old am I if 200 reduced by 2 times my age is 6?

padilas [110]

You would be 12. 6*2=12 then divide by 200

4 0

3 years ago

Read 2 more answers

vitfil [10]

Answer:

45-(-15)

430-(-180)

11,560-(-185)

Step-by-step explanation:

just add the two numbers together for diffrence.

6 0

3 years ago

Verify that each equation is an identity (1 - sin^(2)((x)/(2)))/(1+sin^(2)((x)/(2)))= (1+cosx)/(3-cosX)

Allisa [31]

Answer:

Given that we have;

$sin \left (\dfrac{x}{2} \right ) = \sqrt{\dfrac{1 - cos (x)}{2} }$

By the application of the law of indices and algebraic process of adding a and subtracting a fraction from a whole number, we have;

$\therefore \dfrac{\left ( 1 - sin^2 \left (\dfrac{x}{2} \right ) \right )}{\left ( 1 + sin^2 \left (\dfrac{x}{2} \right ) \right )} =\dfrac{\left ( \dfrac{1 + cos (x)}{2} \right)}{\left (\dfrac{3 - cos (x)}{2} \right ) } =\dfrac{\left ( 1 + cos (x))}{(3 - cos (x))}$

Step-by-step explanation:

An identity is a valid or true equation for all variable values

The given equation is presented as follows;

$\dfrac{\left ( 1 - sin^2 \left (\dfrac{x}{2} \right ) \right )}{\left ( 1 + sin^2 \left (\dfrac{x}{2} \right ) \right )} =\dfrac{\left ( 1 + cos (x))}{(3 - cos (x))}$

From trigonometric identities, we have;

$sin \left (\dfrac{x}{2} \right ) = \sqrt{\dfrac{1 - cos (x)}{2} }$

$\therefore sin^2 \left (\dfrac{x}{2} \right ) = \dfrac{1 - cos (x)}{2}$

$1 - sin^2 \left (\dfrac{x}{2} \right ) = 1 - \dfrac{1 - cos (x)}{2} = \dfrac{2 - (1 - cos (x))}{2} = \dfrac{1 + cos (x))}{2}$

$1 + sin^2 \left (\dfrac{x}{2} \right ) = 1 + \dfrac{1 - cos (x)}{2} = \dfrac{2 + 1 - cos (x))}{2} = \dfrac{3 - cos (x))}{2}$

$\therefore \dfrac{\left ( 1 - sin^2 \left (\dfrac{x}{2} \right ) \right )}{\left ( 1 + sin^2 \left (\dfrac{x}{2} \right ) \right )} =\dfrac{\left ( 1 + cos (x))}{(3 - cos (x))}$

3 0

3 years ago