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3 years ago

Urgent Please Help Add The Fractions. 7 1/5 - ( - 3/5 )

Mathematics

2 answers:

Bezzdna [24]3 years ago

6 0

$\frac{74}{5}$

Komok [63]3 years ago

5 0

7 1/5 - (- 3/5) = 7 4/5 ( Or as a decimal 7.8)

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I need help ASAP!!!

jenyasd209 [6]

Answer:

t is no less than 5

Step-by-step explanation:

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4 years ago

How can the order of operations be used to simplify expressions? Create your own unique example showing order of operations.

Lostsunrise [7]

10(13+12)=x thats a good one

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4 years ago

Explain how you would round the numbers 33 and 89 to accelerate their sum

aleksandr82 [10.1K]

33 rounds to 30 and 89 rounds to 80. 80 x 30= 2400

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4 years ago

Read 2 more answers

Call an integer $n$ oddly powerful if there exist positive integers $a$ and $b$, where $b>1$, $b$ is odd, and $a^b = n$. How

azamat

Answer:

There are 16 oddly powerful integers less than 2010

Step-by-step explanation:

∵ b is an odd integer

∵ b > 1

∴ The first value of b is 3

∵ a is an integer

- We can use a = 1, 2, 3, ..........

∵ $a^{b}=n$

∵ n < 2010

- Let a = 1, 2, ............... 12 because 12³ is greatest integer < 2010

∵ 1³ = 1, 2³ = 8, 3³ = 27, 4³ = 64, 5³ = 125, 6³ = 216, 7³ = 343,

8³ = 512, 9³ = 729, 10³ = 1000, 11³ = 1331, 12³ = 1728

∴ There are 12 oddly powerful integers with b = 3

Now the second value of b is 5

$1^{5}=1$ but we took 1 before so we will start with 2

∵ $2^{5}=32$ , $3^{5}=243$ , $4^{5}=1024$

- $4^{5}$ is the greatest integer < 2010

∴ There are 3 oddly powerful integers with b = 5

Now the third value of b is 7

∵ $2^{7}=128$

- $2^{7}$ is the greatest integer < 2010

∴ There is 1 oddly powerful integers with b = 7

Now the fourth value of b is 9

∵ $2^{9}=512$

- $2^{9}$ is the greatest integer < 2010

- But we used 512 before

∴ There is no oddly powerful integers with b = 9

- 9 is the greatest value of b which makes $a^{b}$

∵ 12 + 3 + 1 = 16

∴ There are 16 oddly powerful integers less than 2010

5 0

3 years ago

What is the domain of the function?

ivann1987 [24]

For a radical with an even root, like 2 in this case, if the radicand turns to a negative value, the result is just an imaginary value, which is another way to say, there's no such a root, no solution per se.

for that, let's first check when the radicand turns to 0 first.

$\bf -8x+8=0\implies -8x=-8\implies x=\cfrac{-8}{-8}\implies \boxed{x=1}$

so, that happens when x = 1, -8(1) +8, is just -8+8 or 0, ok.

so.. if "x" goes a bit higher, like say 2, -8(2) + 8 --> -16 + 8 --> -16

you'd get a negative value, and the radical doesn't have a root for that.

so... the domain, or values "x" can safely take without making the square root expression a negative value, are 1 or below, for example, if we use say -5

-8(-5) + 8 --> 40+8 --> 48 <------ a positive value

thus, (-∞, 1]

8 0

3 years ago