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grin007 [14]
2 years ago
14

Y ³ ——— 10y + 4 i got 6/24 or just 4 is this correct

Mathematics
1 answer:
Helga [31]2 years ago
7 0

2 to the power of 3 is

2 x 2 x 2 = 8

10 to the power of 2 is

10 x 10 = 100

then add 4

104

8/104

I believe this is how you do it :) hope this helps

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3a-8a>-20 help I suck at math ​
vazorg [7]

Answer:

Step-by-step explanation:

3a-8a>-20

-5a>-20

a>-20/-5

a>4

5 0
2 years ago
What is the equivalent expression of y+2/5y
Nonamiya [84]
The answer Is going to be 24
8 0
3 years ago
What is 10 squared plus b squared equals 22 squared
klio [65]
I'm pretty sure you are asking for the value of b. Let's actually write it and then solve it:
10^2 + b^2 = 22^2

Solving for b yields:
b = \sqrt{22^2-10^2} = \sqrt{484-100} = \sqrt{384} = 19.59

So, b = 19.59. If you want the answer in simplified radical form, we can also do that:
\sqrt{384} =  \sqrt{16*24} = 4 \sqrt{24} = 4 \sqrt{4*6} = 8 \sqrt{6}

So, b = 8√6 = 19.59
4 0
3 years ago
show that thw roots of the equation (x-a)(x_b)=k^2 are always real if a,b and k are real. Please I really need help with this
VLD [36.1K]

Answer:

see explanation

Step-by-step explanation:

Check the value of the discriminant

Δ = b² - 4ac

• If b² - 4ac > 0 then roots are real

• If b² - 4ac = 0 roots are real and equal

• If b² - 4ac < 0 then roots are not real

given (x - a)(x - b) = k² ( expand factors )

x² - bx - ax - k² = 0 ( in standard form )

x² + x(- a - b) - k² = 0

with a = 1, b = (- a - b), c = -k²

b² - 4ac = (- a - b)² + 4k²

For a, b, k ∈ R then (- a - b)² ≥ 0 and 4k² ≥ 0

Hence roots of the equation are always real for a, b, k ∈ R


           

8 0
3 years ago
Explaining the Process
gladu [14]

Answer:

ok ?????????????????????????????

3 0
2 years ago
Read 2 more answers
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