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Dvinal [7]
3 years ago
10

I need help on this question can anyone help?

Mathematics
1 answer:
inn [45]3 years ago
7 0

Answer:

The answer is A. the range is the set of all real numbers

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What is the least common multiple of 49 and 28?
USPshnik [31]
Least common mulitplue is the smallest number that has 49 and 28 as factors

so to do this
find the factor so 49 and 28 and gropu all of them so both are in it

49=7 times 7
28=2 times 2 times 7
therefor yo need at least two 7's and two 2's
that equals
7 times 7 times 2 itme s2=49 times 4=196

answer is 196
7 0
3 years ago
Write an algebraic expression for the word phrase.
vovangra [49]

Answer:

0.30 m

Step-by-step explanation:

<u>Explanation</u>:-

Given data is 30 % of m

Algebraic expression:-

The form of the algebraic expression is ax+ by +c

Given data is 30 % of m

here 'of' meaning is multiplied of given term

therefore \frac{30}{100} X m

The algebraic expression is 0.30 m

6 0
3 years ago
Read 2 more answers
Find the sum of 19x3+1(14x+4x3)
Step2247 [10]
Find the sum of 19x3+1(14x+4x) 

The answer is 19x3+4x3=4x.
5 0
3 years ago
Factor f(x) = 15x^3 - 15x^2 - 90x completely and determine the exact value(s) of the zero(s) and enter them as a comma separated
Illusion [34]

Answer:

x=-2,0,3

Step-by-step explanation:

We have been given a function f(x)=15x^3-15x^2-90x. We are asked to find the zeros of our given function.

To find the zeros of our given function, we will equate our given function by 0 as shown below:

15x^3-15x^2-90x=0

Now, we will factor our equation. We can see that all terms of our equation a common factor that is 15x.

Upon factoring out 15x, we will get:

15x(x^2-x-6)=0

Now, we will split the middle term of our equation into parts, whose sum is -1 and whose product is -6. We know such two numbers are -3\text{ and }2.

15x(x^2-3x+2x-6)=0

15x((x^2-3x)+(2x-6))=0

15x(x(x-3)+2(x-3))=0

15x(x-3)(x+2)=0

Now, we will use zero product property to find the zeros of our given function.

15x=0\text{ (or) }(x-3)=0\text{ (or) }(x+2)=0

15x=0\text{ (or) }x-3=0\text{ (or) }x+2=0

\frac{15x}{15}=\frac{0}{15}\text{ (or) }x-3=0\text{ (or) }x+2=0

x=0\text{ (or) }x=3\text{ (or) }x=-2

Therefore, the zeros of our given function are x=-2,0,3.

7 0
3 years ago
Can anyone answer this question ?​
kondaur [170]
X1=radical 6--->2.44
X2=-radical 6/3--->-0.816
5 0
3 years ago
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