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Olenka [21]
3 years ago
6

What is four plus the number two basically mathematics

Mathematics
1 answer:
Oksi-84 [34.3K]3 years ago
6 0

Answer:

6

Step-by-step explanation:

4+2 = 6

Hope this helps

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Need answer ASAP thanks!
barxatty [35]

Answer:

assuming its an annual interest

Okay so 6 percent interest, the bank is paying you.
So with this it’s 6 percent of 1500 and add it to 1500.

You can always find 6 percent of 1500 and then add but here’s a short cut.
Your principle (beginning) balance is 1500.
That’s already 100 percent since thats yoru original value.
You then get added 6 percent interest.
We are jsut adding 6 percent to 100 percent so 106 percent.
Now we solve normally and you’d get the answer faster.

106 percent is 106/100 or 1 3/5 or 1.06

now we multiply
1500 * 1.06 = 1590

Your final balance would be 1590 after the 6 percent interest is added.

5 0
2 years ago
How is math used in making murals?(subject:Collage and Career readiness)
Stella [2.4K]
She 56/92 6-5=1 and 567-1=566
3 0
3 years ago
What is the missing operations to 4×9+7×9=(4+7)×
ICE Princess25 [194]

Answer:

9, if you're asking for only the number of it. But If it needs more then please elaborate.

Step-by-step explanation:

4×9+7×9=(4+7)×9

36+63=11*9

99=99

Order Of Operation Helps!

8 0
3 years ago
Read 2 more answers
Evaluate cube root of 5 multiplied by square root of 5 over cube root of 5 to the power of 5.
Andre45 [30]

Answer:

Option d)  5 to the power of negative 5 over 6 is correct.

\dfrac{\sqrt[3]{\bf 5} \times \sqrt{\bf 5}}{\sqrt[3]{\bf 5^{\bf 5}}}= 5^{\frac{\bf -5}{\bf 6}}

Above equation can be written as 5 to the power of negative 5 over 6.

ie, 5^\frac{\bf -5}{\bf 6}

Step-by-step explanation:

Given that cube root of 5 multiplied by square root of 5 over cube root of 5 to the power of 5.

It can be written as below

\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}

\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{1}{3}} \times 5^{\frac{1}{2}}}{5^{\frac{5}{3}}}

\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{1}{3}+\frac{1}{2}}}{5^{\frac{5}{3}}}

\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{2+3}{6}}}{5^{\frac{5}{3}}}

\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= 5^{\frac{5}{6}} \times 5^{\frac{-5}{3}}

\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= 5^{\frac{5-10}{6}}

\dfrac{\sqrt[3]{5} \times \sqrt{5}}{5^5}= 5^{\frac{-5}{6}}

Above equation can be written as 5 to the power of negative 5 over 6.

7 0
3 years ago
Can somebody help me please!!!
Effectus [21]

Answer:

Part A: 0

Part B: 0 or 1

Step-by-step explanation:

Part A: anything to the 0 power is 1 (x^0= x^y-y) if you solve 7^2 you get 49. 49^0= 1

Part B: Because it is already being raised to the power of 0, if x=0 it will remain equal to one (as explained above). Anything raised to the 1 power is itself, so 7^0= 1 and 1^1 =1

Hope this helped!

3 0
3 years ago
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