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mylen [45]
3 years ago
12

-mb-m^2 for m=3.48 and b=96.52

Mathematics
1 answer:
AlekseyPX3 years ago
3 0
I- huh I don’t understand the question
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What is the answer to (( 5 x 12)/3)+30-50
SpyIntel [72]

I believe the correct answer is 0

Step-by-step explanation:

To solve this question, we use an abbreviation formula called BODMAS which is:

B = Brackets

O = Of

D = Division

M = Multiplication

A = Addition

S = Subtraction

We solve each element that is available in the order of the abbreviated letter.

1. We solve the brackets:

(5\times12) = 60

2. We solve the second bracket:

(60\div3) = 20

The equation now is 20+30-50

3. We now solve addition first:

20+30=50

4. Now we solve the subtraction:

50-50=0

The answer then = 0

8 0
3 years ago
Read 2 more answers
Solve the equation. Then check your solution. – two-fifths + a = One-fifth a. Three-fifths c. StartFraction 3 Over 10 EndFractio
VLD [36.1K]

Answer:

a = Negative one-fifth

Step-by-step explanation:

The given equation is :

\dfrac{2}{5}+a=\dfrac{1}{5}

We need to find the value of a.

Subtract 2/5 from both sides.

\dfrac{2}{5}+a-\dfrac{2}{5}=\dfrac{1}{5}-\dfrac{2}{5}\\\\a=\frac{1}{5}-\frac{2}{5}\\\\a=\dfrac{-1}{5}

So, the correct answer is Negative one-fifth. Hence, the correct option is (b).

5 0
3 years ago
Which one is > greater or < less than or = equal to?? 6+√8____√6+8​
brilliants [131]

\huge\boxed{\sqrt{6}+8\;is\;greater\;than \;6+\sqrt{8}  }

6+\sqrt[]{8} =8.82\\\sqrt{6}+8=10.44\\\\\boxed{Therefore, \sqrt{6}+8\;is\;greater\;than\;6+\sqrt{8}. }

4 0
3 years ago
You are supposed to distribute to only one term (the first).
Tom [10]

Answer:

true

Step-by-step explanation:

There is no distribution (unless you count substitution) in the second set of terms.

Edit: I think I get what your question is now. You are supposed to distribute to every term inside the parenthesis (false)

6 0
3 years ago
Read 2 more answers
Can someone please please please help me please
anzhelika [568]

#9 is answer A) yes,

#10 the answer is A) (n,j)

5 0
3 years ago
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