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3 years ago

Log(8)a/2. Expand the logarithmic expression.

1 answer:

maks197457 [2]3 years ago

5 0

Log(8a/2) = log8 + log a - log 2

The simple rules here are: If logarithms (as in this case) have same basis, factors inside logarithm can be written as sum of logarithms or difference between logarithms (first if factors are multiplying and second if factors are dividing)

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The perimeter of a rectangle is 500 feet. The ratio of the base and height is 3:2. What is the measure of the height ?

nika2105 [10]

Answer:

height=100feet

Step-by-step explanation:

perimeter of a rectangle=p=500 feet

p=2a+2b equation 1

a=height

b=base

we have: b/a=3/2

so a=(2*b)/3

in equation 1 we have:

500=2*((2*b)/3)+2*b

so we have

500=(4*b)/3 + 2*b

500=(4b+6b)/3

500=(10b)/3

500*3=10b

b=(500*3)/10

b=150 feet =base

a=(2*150)/3=100 feet = height

8 0

3 years ago

Y < -2x + 4 what is the solution to the inequality?

Dafna11 [192]

Answer:

x<\frac{-y+4}{2}

Step-by-step explanation:

Hope this helps!!!

can i get brainliest plzzz thx

7 0

3 years ago

Use substitution to find the solution to the system of equations.

mote1985 [20]

Answer:

Ummm, so the answer is A. (-3,-2)

StepByStep Solution:

5x + 7(x+1) =-29

5x+7x+7=-29

12x=-36

x=-3

y=-3+1

y=-2

4 0

3 years ago

Read 2 more answers

Solve irrational equation pls

rusak2 [61]

$\hbox{Domain:}\\
x^2+x-2\geq0 \wedge x^2-4x+3\geq0 \wedge x^2-1\geq0\\
x^2-x+2x-2\geq0 \wedge x^2-x-3x+3\geq0 \wedge x^2\geq1\\
x(x-1)+2(x-1)\geq 0 \wedge x(x-1)-3(x-1)\geq0 \wedge (x\geq 1 \vee x\leq-1)\\
(x+2)(x-1)\geq0 \wedge (x-3)(x-1)\geq0\wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle1,\infty) \wedge x\in(-\infty,1\rangle \cup\langle3,\infty) \wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle3,\infty)$

$
\sqrt{x^2+x-2}+\sqrt{x^2-4x+3}=\sqrt{x^2-1}\\
x^2-1=x^2+x-2+2\sqrt{(x^2+x-2)(x^2-4x+3)}+x^2-4x+3\\
2\sqrt{(x^2+x-2)(x^2-4x+3)}=-x^2+3x-2\\
\sqrt{(x^2+x-2)(x^2-4x+3)}=\dfrac{-x^2+3x-2}{2}\\
(x^2+x-2)(x^2-4x+3)=\left(\dfrac{-x^2+3x-2}{2}\right)^2\\
(x+2)(x-1)(x-3)(x-1)=\left(\dfrac{-x^2+x+2x-2}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-x(x-1)+2(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-(x-2)(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\dfrac{(x-2)^2(x-1)^2}{4}\\
4(x+2)(x-3)(x-1)^2=(x-2)^2(x-1)^2\\$
$
4(x+2)(x-3)(x-1)^2-(x-2)^2(x-1)^2=0\\
(x-1)^2(4(x+2)(x-3)-(x-2)^2)=0\\
(x-1)^2(4(x^2-3x+2x-6)-(x^2-4x+4))=0\\
(x-1)^2(4x^2-4x-24-x^2+4x-4)=0\\
(x-1)^2(3x^2-28)=0\\
x-1=0 \vee 3x^2-28=0\\
x=1 \vee 3x^2=28\\
x=1 \vee x^2=\dfrac{28}{3}\\
x=1 \vee x=\sqrt{\dfrac{28}{3}} \vee x=-\sqrt{\dfrac{28}{3}}\\$

There's one more condition I forgot about
$-(x-2)(x-1)\geq0\\
x\in\langle1,2\rangle\\$

Finally
$x\in(-\infty,-2\rangle\cup\langle3,\infty) \wedge x\in\langle1,2\rangle \wedge x=\{1,\sqrt{\dfrac{28}{3}}, -\sqrt{\dfrac{28}{3}}\}\\
\boxed{\boxed{x=1}}$

3 0

3 years ago

376 feet above sea level

galina1969 [7]

Answer:

376 feet high standard

Step-by-step explanation:

above sea level its positive

3 0

3 years ago