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4 years ago

How many solutions does the equation have?

Mathematics

2 answers:

I am Lyosha [343]4 years ago

8 0

C. 1 Would be your answer. Your welcome

Neporo4naja [7]4 years ago

7 0

Im going with C
hope this helps

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Math help guys how wouls i work this out

Andrei [34K]

8 > 7 + $\frac{x}{6}$ Subtract 7 from both sides

1 > $\frac{x}{6}$ Multiply both sides by 6

6 > x Flip it around so it's easier to read

x < 6

You can graph your answer by drawing an open circle at the 6 and coloring the line to the left. The circle should be open, because x is <em>less than 6</em>, not less than or equal to. You would color to the left to show that x can be anything less than 6.

4 0

3 years ago

HOW DO I SOLVE THIS PLEASE HELP

malfutka [58]

Answer:

x = 15

Step-by-step explanation:

We assume you want to find the value of x.

Know (or prove) that in this geometry, all of the right triangles are similar. That means the ratios of corresponding sides are proportional.

short side / hypotenuse = 9/x = x/25

x^2 = (9)(25) . . . . . . . . . . multiply by 25x ("cross multiply")

x = √((9)(25)) = (3)(5) . . . take the square root

x = 15

3 0

2 years ago

Recent research published by Frumin and colleagues (2011) in the journal Scienceaddresses whether females' tears have an effect

iogann1982 [59]

Answer: Option C. [0.95, 1.59].

Step-by-step explanation:

We know that the mean is:

M = 1.27

and the margin of error is:

e = 0.32

This means that the actual value can be at a maximum distance of 0.32 from the mean.

then the interval will be:

[M - e, M + e].= [1.27 - 0.32, 1.27 + 0.32].= [0.95, 1.59].

The correct option is C.

3 0

4 years ago

HELP FAST DUE IN 10 minutes WILL MARK BRAINLIEST SOMEONE PLEASE HELP

WARRIOR [948]

Answer:

your answer is correct -7,-2

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

"find the reduction formula for the integral" sin^n(18x)

dexar [7]

Let

$I(n,a)=\displaystyle\int\sin^nax\,\mathrm dx$
For demonstration on how to tackle this sort of problem, I'll only work through the case where $n$ is odd. We can write

$\displaystyle\int\sin^nax\,\mathrm dx=\int\sin^{n-2}ax\sin^2ax\,\mathrm dx=\int\sin^{n-2}ax(1-\cos^2ax)\,\mathrm dx$
$\implies I(n,a)=I(n-2,a)-\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx$

For the remaining integral, we can integrate by parts, taking

$u=\sin^{n-3}ax\implies\mathrm du=a(n-3)\sin^{n-4}ax\cos ax\,\mathrm dx$ $\mathrm dv=\sin ax\cos^2ax\,\mathrm dx\implies v=-\dfrac1{3a}\cos^3ax$

$\implies\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx=-\dfrac1{3a}\sin^{n-3}ax\cos^3ax+\dfrac{a(n-3)}{3a}\int\sin^{n-4}ax\cos^4ax\,\mathrm dx$

For this next integral, we rewrite the integrand

$\sin^{n-4}ax\cos^4ax=\sin^{n-4}ax(1-\sin^2ax)^2=\sin^{n-4}ax-2\sin^{n-2}ax+\sin^nax$
$\implies\displaystyle\int\sin^{n-4}ax\cos^4ax\,\mathrm dx=I(n-4,a)-2I(n-2,a)+I(n,a)$

So putting everything together, we found

$I(n,a)=I(n-2,a)-\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx$
$I(n,a)=I(n-2,a)-\left(-\dfrac1{3a}\sin^{n-3}ax\cos^3ax+\dfrac{n-3}3\displaystyle\int\sin^{n-4}ax\cos^4ax\,\mathrm dx\right)$
$I(n,a)=I(n-2,a)-\dfrac{n-3}3\bigg(I(n-4,a)-2I(n-2,a)+I(n,a)\bigg)+\dfrac1{3a}\sin^{n-3}ax\cos^3ax$
$\dfrac n3I(n,a)=\dfrac{2n-3}3I(n-2,a)-\dfrac{n-3}3I(n-4,a)+\dfrac1{3a}\sin^{n-3}ax\cos^3ax$

$\implies I(n,a)=\dfrac{2n-3}nI(n-2,a)-\dfrac{n-3}nI(n-4,a)+\dfrac1{na}\sin^{n-3}ax\cos^3ax$

7 0

3 years ago