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3 years ago

What affect does the "h" value have on the function f(x)=log(x)?

Mathematics

1 answer:

Neporo4naja [7]3 years ago

7 0

Y = a(x - h)² + k is the equation, so the "h" value moves the curve of the graph left or right.

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The value of x-y is negative, and x and y are both positive. What can you conclude about the values of x

Stolb23 [73]

Well for two positive numbers subtracted to equal a negative number, the first number must have a value that is less than the second number. (Ex. 3-4= -1) In this case, you can conclude that x is less than y! (x<y)

3 0

3 years ago

8) A pyramid has a rectangluar base with length 12 km and width of 5 km. If the volume is 200

SashulF [63]

Hi there!

$\large\boxed{h = 10km}$

We know that the equation for the volume of a pyramid is:

V = 1/3(bh)

We know that:

V = 200 km³

b = 12 × 5 = 60 km²

We can plug these into the equation:

200 = 1/3(60)(h)

Simplify:

200 = 20h

Divide both sides by 20:

200/20 = 20h/20

h = 10 km

6 0

3 years ago

Find the integral using substitution or a formula.

Nadusha1986 [10]

$\rm \int \dfrac{x^2+7}{x^2+2x+5}~dx$

Derivative of the denominator:
$\rm (x^2+2x+5)'=2x+2$

Hmm our numerator is 2x+7. Ok this let's us know that a simple u-substitution is NOT going to work. But let's apply some clever Algebra to the numerator splitting it up into two separate fractions. Split the +7 into +2 and +5.

$\rm \int \dfrac{x^2+2+5}{x^2+2x+5}~dx$

and then split the fraction,

$\rm \int \dfrac{x^2+2}{x^2+2x+5}~dx+\int\dfrac{5}{x^2+2x+5}~dx$

Based on our previous test, we know that a simple substitution will work for the first integral: $\rm \quad u=x^2+2x+5\qquad\to\qquad du=2x+2~dx$

So the first integral changes,

$\rm \int \dfrac{1}{u}~du+\int\dfrac{5}{x^2+2x+5}~dx$

integrating to a log,

$\rm ln|x^2+2x+5|+\int\dfrac{5}{x^2+2x+5}~dx$

Other one is a little tricky. We'll need to complete the square on the denominator. After that it will look very similar to our arctangent integral so perhaps we can just match it up to the identity.

$\rm x^2+2x+5=(x^2+2x+1)+4=(x+1)^2+2^2$

So we have this going on,

$\rm ln|x^2+2x+5|+\int\dfrac{5}{(x+1)^2+2^2}~dx$

Let's factor the 5 out of the intergral,
and the 4 from the denominator,

$\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\frac{(x+1)^2}{2^2}+1}~dx$

Bringing all that stuff together as a single square,

$\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(\dfrac{x+1}{2}\right)^2+1}~dx$

Making the substitution: $\rm \quad u=\dfrac{x+1}{2}\qquad\to\qquad 2du=dx$

giving us,

$\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(u\right)^2+1}~2du$

simplying a lil bit,

$\rm ln|x^2+2x+5|+\frac52\int\dfrac{1}{u^2+1}~du$

and hopefully from this point you recognize your arctangent integral,

$\rm ln|x^2+2x+5|+\frac52arctan(u)$

undo your substitution as a final step,
and include a constant of integration,

$\rm ln|x^2+2x+5|+\frac52arctan\left(\frac{x+1}{2}\right)+c$

Hope that helps!
Lemme know if any steps were too confusing.

8 0

3 years ago

Please help me solve this problem

algol [13]

1.afgx2
2.bfgx2

I hope it helps Im really not that smart