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3 years ago

PLEASE HELP!!! I WILL GIVE BRIANLEY TO CORRECT ANSWER!!

Mathematics

1 answer:

Tatiana [17]3 years ago

7 0

Would it just be 45 songs?

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Please do this math for me or else i will steal your cookies

kompoz [17]

Answer:

Step-by-step explanation:

$-7x+16=58\\\\\mathrm{Subtract\:}16\mathrm{\:from\:both\:sides}\\-7x+16-16=58-16\\\\-7x=42\\\\\mathrm{Divide\:both\:sides\:by\:}-7\\\\\frac{-7x}{-7}=\frac{42}{-7}\\\\x =-6$

$-2x+15=-9\\\\\mathrm{Subtract\:}15\mathrm{\:from\:both\:sides}\\\\-2x+15-15=-9-15\\\\-2x=-24\\\\\mathrm{Divide\:both\:sides\:by\:}-2\\\\\frac{-2x}{-2}=\frac{-24}{-2}\\\\x=12\\$

$5x-4=36\\\\\mathrm{Add\:}4\mathrm{\:to\:both\:sides}\\\\5x-4+4=36+4\\\\5x=40\\\\\mathrm{Divide\:both\:sides\:by\:}5\\\\\frac{5x}{5}=\frac{40}{5}\\\\x=8$

$25-3x=88\\\\\mathrm{Subtract\:}25\mathrm{\:from\:both\:sides}\\\\25-3x-25=88-25\\\\\mathrm{Divide\:both\:sides\:by\:}-3\\\\\frac{-3x}{-3}=\frac{63}{-3}\\\\x=-21$

$-11=7-x\\\\\mathrm{Add\:}x\mathrm{\:to\:both\:sides}\\\\-11+x=7-x+x\\\\-11+x=7\\\\\mathrm{Add\:}11\mathrm{\:to\:both\:sides}\\\\-11+x+11=7+11\\\\x=18$

$65+15x=35\\\\\mathrm{Subtract\:}65\mathrm{\:from\:both\:sides}\\\\65+15x-65=35-65\\\\\mathrm{Divide\:both\:sides\:by\:}15\\\\\frac{15x}{15}=\frac{-30}{15}\\\\x=-2$

$\frac{1}{2} x-18=2\\\\\mathrm{Add\:}18\mathrm{\:to\:both\:sides}\\\\\frac{1}{2}x-18+18=2+18\\\\\frac{1}{2}x=20\\\\\mathrm{Multiply\:both\:sides\:by\:}2\\\\2\times\frac{1}{2}x=20\times \:2\\\\x=4$

$\frac{2}{3}x-10=-12 \\\\\mathrm{Add\:}10\mathrm{\:to\:both\:sides}\\\\\frac{2}{3}x-10+10=-12+10\\\\\frac{2}{3}x=-2\\\\Divide\:both\:sides\:by\: 2/3\\\\\frac{2}{3}x\div \frac{2}{3} =-2\div \frac{2}{3} \\\\\frac{2}{3} x\times \frac{3}{2} =-2\times\frac{3}{2} \\\\x =-3$

$6-\frac{1}{3}x=-1 \\\\\mathrm{Subtract\:}6\mathrm{\:from\:both\:sides}\\\\6-\frac{1}{3}x-6=-1-6\\\\-\frac{1}{3}x=-7\\\\\mathrm{Multiply\:both\:sides\:by\:}-3\\\\\left(-\frac{1}{3}x\right)\left(-3\right)=\left(-7\right)\left(-3\right)\\\\x=21$

10.

$4-9x=-14\\\\\mathrm{Subtract\:}4\mathrm{\:from\:both\:sides}\\\\4-9x-4=-14-4\\\\-9x=-18\\\\\mathrm{Divide\:both\:sides\:by\:}-9\\\\\frac{-9x}{-9}=\frac{-18}{-9}\\\\x=2$

11.

$11-x=29\\\\\mathrm{Subtract\:}11\mathrm{\:from\:both\:sides}\\\\-x=18\\\\\mathrm{Divide\:both\:sides\:by\:}-1\\\\\frac{-x}{-1}=\frac{18}{-1}\\\\x=-18$

12.

$-9-11x=68\\\\\mathrm{Add\:}9\mathrm{\:to\:both\:sides}\\\\-9-11x+9=68+9\\\\-11x=77\\\\\mathrm{Divide\:both\:sides\:by\:}-11\\\\\frac{-11x}{-11}=\frac{77}{-11}\\\\x=-7$

13.

$45+\frac{5}{6}x =50\\\\\mathrm{Subtract\:}45\mathrm{\:from\:both\:sides}\\\\45+\frac{5}{6}x-45=50-45\\\\\frac{5}{6}x=5\\\\\mathrm{Divide\:both\:sides\:by\:}5/6\\\\\frac{5}{6} x\div\frac{5}{6}=5\div\frac{5}{6}\\\\\frac{5}{6}x\times \frac{6}{5}= 5\times\frac{6}{5}\\\\ x=6$

14.

$-5x+17=-33\\\\\mathrm{Subtract\:}17\mathrm{\:from\:both\:sides}\\\\-5x+17-17=-33-17\\\\\mathrm{Divide\:both\:sides\:by\:}-5\\\\\frac{-5x}{-5}=\frac{-50}{-5}\\\\x=10$

15.

$95=-4+33x\\\\-4+33x=95\\\\\mathrm{Add\:}4\mathrm{\:to\:both\:sides}\\\\-4+33x+4=95+4\\\\\mathrm{Divide\:both\:sides\:by\:}33\\\\\frac{33x}{33}=\frac{99}{33}\\\\x=3$

3 0

3 years ago

Suppose h(x)=3x-2 and j(x) = ax +b. Find a relationship between a and b such that h(j(x)) = j(h(x))

sergejj [24]

Answer:

$\displaystyle a = \frac{1}{3} \text{ and } b = \frac{2}{3}$

Step-by-step explanation:

We can use the definition of inverse functions. Recall that if two functions, f and g are inverses, then:

$\displaystyle f(g(x)) = g(f(x)) = x$

So, we can let j be the inverse function of h.

Function h is given by:

$\displaystyle h(x) = y = 3x-2$

Find its inverse. Flip variables:

$x = 3y - 2$

Solve for y. Add:

$\displaystyle x + 2 = 3y$

Hence:

$\displaystyle h^{-1}(x) = j(x) = \frac{x+2}{3} = \frac{1}{3} x + \frac{2}{3}$

Therefore, a = 1/3 and b = 2/3.

We can verify our solution:

$\displaystyle \begin{aligned} h(j(x)) &= h\left( \frac{1}{3} x + \frac{2}{3}\right) \\ \\ &= 3\left(\frac{1}{3}x + \frac{2}{3}\right) -2 \\ \\ &= (x + 2) -2 \\ \\ &= x \end{aligned}$

And:

$\displaystyle \begin{aligned} j(h(x)) &= j\left(3x-2\right) \\ \\ &= \frac{1}{3}\left( 3x-2\right)+\frac{2}{3} \\ \\ &=\left( x- \frac{2}{3}\right) + \frac{2}{3} \\ \\ &= x \stackrel{\checkmark}{=} x\end{aligned}$