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3 years ago

R^2+6r-27 factoring quadratics

Mathematics

1 answer:

Vanyuwa [196]3 years ago

5 0

Answer:

(r + 9)(r - 3)

Step-by-step explanation:

r² + 6r - 27 =

= r² + 9r - 3r - 27

= r(r + 9) - 3(r + 9)

= (r + 9)(r - 3)

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Which table represents the graph of a logarithmic function in the form y=log3x when b>1?

alex41 [277]

Answer:

The satisfied table of the given function $y = log_{b} (x)$

x 1/8 1/4 1/2 1 2

y -3 -2 -1 0 1

Step-by-step explanation:

Explanation :-

Given logarithmic function $y = log_{b} (x)$ if b >1

Given first table

put x = $\frac{1}{8}$ given b > 1 so we can choose b = 2

$y = log_{2} (\frac{1}{8} )$

$y = log_{2} (2^{-3} )$

we will apply logarithmic formula

log x ⁿ = n log (x)

$y = log_{2} (2^{-3} ) = -3 log_{2} (2) = -3 (1) = -3$

y = -3

ii)

put x = $\frac{1}{4}$ given b > 1 so we can choose b = 2

$y = log_{2} (\frac{1}{4} )$

$y = log_{2} (2^{-2} )$

we will apply logarithmic formula

log x ⁿ = n log (x)

$y = log_{2} (2^{-2} ) = -2 log_{2} (2) = -2 (1) = -2$

y = -2

iii)

put x = $\frac{1}{2}$ given b > 1 so we can choose b = 2

$y = log_{2} (\frac{1}{2} )$

$y = log_{2} (2^{-1} )$

we will apply logarithmic formula

log x ⁿ = n log (x)

$y = log_{2} (2^{-1} ) = -1 log_{2} (2) = - (1) = -1$

y = -1

iv)

put x = 1 given b > 1 so we can choose b = 2

$y = log_{2} (1 )$ = 0

y = 0

v)

put x = $2$ given b > 1 so we can choose b = 2

$y = log_{2} (2 )$

y = 1

Final answer:-

The satisfied table of the given function

x 1/8 1/4 1/2 1 2

y -3 -2 -1 0 1

8 0

3 years ago

Read 2 more answers

Which equation represents the linear relationships shown in the table?

saveliy_v [14]

Answer:

idk

Step-by-step explanation:

4 0

3 years ago

Hey.... sorry for that, are you still mad at me ?

Digiron [165]

Answer:

do u mean me???

8 0

2 years ago

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For 0 ≤ ϴ < 2π, how many solutions are there to tan(StartFraction theta Over 2 EndFraction) = sin(ϴ)? Note: Do not include va

Black_prince [1.1K]

Answer:

3 solutions:

$\theta={0, \frac{\pi}{2}, \frac{3\pi}{2}}$

Step-by-step explanation:

So, first of all, we need to figure the angles that cannot be included in our answers out. The only function in the equation that isn't defined for some angles is $tan(\frac{\theta}{2})$ so let's focus on that part of the equation first.

We know that:

$tan(\frac{\theta}{2})=\frac{sin(\frac{\theta}{2})}{cos(\frac{\theta}{2})}$

therefore:

$cos(\frac{\theta}{2})\neq0$

so we need to find the angles that will make the cos function equal to zero. So we get:

$cos(\frac{\theta}{2})=0$

$\frac{\theta}{2}=cos^{-1}(0)$

$\frac{\theta}{2}=\frac{\pi}{2}+\pi n$

$\theta=\pi+2\pi n$

we can now start plugging values in for n:

$\theta=\pi+2\pi (0)=\pi$

if we plugged any value greater than 0, we would end up with an angle that is greater than $2\pi$ so, that's the only angle we cannot include in our answer set, so: