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ValentinkaMS [17]
3 years ago
12

23. solve for x: -5=45

Mathematics
1 answer:
yuradex [85]3 years ago
3 0
That's the answer for no 23

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What is the result when like terms are combined in the expression
pentagon [3]

Answer:Use addition and subtraction within 20 to solve word problems involving situations of adding to, taking from, putting together, taking apart, and comparing, with unknowns in all positions, e.g., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem.1

CCSS.MATH.CONTENT.1.OA.A.2

Solve word problems that call for addition of three whole numbers whose sum is less than or equal to 20, e.g., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem.

Understand and apply properties of operations and the relationship between addition and subtraction.

CCSS.MATH.CONTENT.1.OA.B.3

Apply properties of operations as strategies to add and subtract.2 Examples: If 8 + 3 = 11 is known, then 3 + 8 = 11 is also known. (Commutative property of addition.) To add 2 + 6 + 4, the second two numbers can be added to make a ten, so 2 + 6 + 4 = 2 + 10 = 12. (Associative property of addition.)

CCSS.MATH.CONTENT.1.OA.B.4

Understand subtraction as an unknown-addend problem. For example, subtract 10 - 8 by finding the number that makes 10 when added to 8.

Add and subtract within 20.

CCSS.MATH.CONTENT.1.OA.C.5

Relate counting to addition and subtraction (e.g., by counting on 2 to add 2).

CCSS.MATH.CONTENT.1.OA.C.6

Add and subtract within 20, demonstrating fluency for addition and subtraction within 10. Use strategies such as counting on; making ten (e.g., 8 + 6 = 8 + 2 + 4 = 10 + 4 = 14); decomposing a number leading to a ten (e.g., 13 - 4 = 13 - 3 - 1 = 10 - 1 = 9); using the relationship between addition and subtraction (e.g., knowing that 8 + 4 = 12, one knows 12 - 8 = 4); and creating equivalent but easier or known sums (e.g., adding 6 + 7 by creating the known equivalent 6 + 6 + 1 = 12 + 1 = 13).

Work with addition and subtraction equations.

CCSS.MATH.CONTENT.1.OA.D.7

Understand the meaning of the equal sign, and determine if equations involving addition and subtraction are true or false. For example, which of the following equations are true and which are false? 6 = 6, 7 = 8 - 1, 5 + 2 = 2 + 5, 4 + 1 = 5 + 2.

Step-by-step explanation:

8 0
3 years ago
Which number should each side of the equation 3/4x = 9 be multiplied by to produce the equivalent equation of x =12
S_A_V [24]
3/4x = 9....multiply both sides by 4 <==
3x = 9 * 4
3x = 36
x = 36/3
x = 12

5 0
3 years ago
Read 2 more answers
What is the quotient in simplest form?
enyata [817]
The third one 02 2/5 is the answer
8 0
3 years ago
three sister are getting new outfits. shirts cost 11 each,skirts coat 25 each and shows cost 44 a pair what is the total coast o
MatroZZZ [7]

Answer:

240.

Step-by-step explanation:

Here is the correct question: Three sister are getting new outfits. shirts cost 11 each,skirts coat 25 each and shoes cost 44 a pair. what is the total cost of the new outfits for all of the sisters​?

Given: Cost of each shirt is 11

           Cost of each skirts coat is 25

           Cost of each pair of shoes cost is 44.

Now, finding total cost of shirt, skirt coat and shoes as cost of one new outfit.

= 11+25+44 = 80

∴ Total cost of one new outfit is 80.

Next, finding cost of new outfit for all, three sisters.

Cost of new outfit for three sister = \textrm { cost of one outfit}\times \textrm{ number of sisters}

∴ Cost of new outfit for three sister = 80\times 3= 240

240 is the total cost of new outfit for all the sisters.

         

3 0
3 years ago
Suppose m = 2 + 6i, and | m + n | = 3√10, where n is a complex number.
Ksju [112]

Answer: a) √50

b) n = 1 + 7i

Step-by-step explanation:

first, the modulus of a complex number z = a + bi is

IzI = √(a^2 + b^2)  

The fact that n is complex does not mean that n doesn't has a real part, so we must write our numbers as:

m = 2 + 6i

n = a  + bi

Im + nI = 3√10

Im + n I = √(a^2 + b^2 + 2^2 + 6^2)= 3√10

            = √(a^2 + b^2 + 40) = 3√10

             a^2 + b^2 + 40 = 3^2*10 = 9*10 = 90

             a^2 + b^2 = 90 - 40 = 50

            √(a^2 + b^2 ) = InI = √50

The modulus of n must be equal to the square root of 50.

now we can find any values a and b such a^2 + b^2 = 50.

for example, a = 1 and b = 7

1^2 + 7^2 = 1 + 49  = 50

Then a possible value for n is:

n = 1 + 7i

6 0
3 years ago
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