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3 years ago

Sederhanakanlah 5X²+7X-2X²+X

Mathematics

1 answer:

horsena [70]3 years ago

8 0

5X²+7X-2X²+X
5X² - 2X² +7X +X
3X² + 8X
X (3X + 8)

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I am stuck please help!

olga2289 [7]

Answer:

5x - 63 = 6

Step-by-step explanation:

Okay, first step, let's get all fractions having a common denominator, to do this, multiply 2/3 by 3 to get 6/9. Now, we have 5/9x-7=6/9. To make these into whole numbers, we can multiply the whole equation by the denominator (9). 5/9x multiplied by 9 is 5. 7 multiplied by 9 is 63. 6/9 multiplied by 9 is 6. Now we have 5x-63=6. If I were to simplify the equation, it would require decimals, so I'll leave it at this, but the equation simplified is x=13.8.

7 0

3 years ago

What is the rational number equivalent to 1.28? The 1.28 has a line above the 28

Svetach [21]

the line over the 28 means the 28 repeats forever.
1.282828.... and so on
let x be the rational number 1.28...
we can use this trick:
100*1.282828....= 128.282828...
(the decimal 28 part repeats) 100x = 128.28...
next: 100x - x = 128.282828... - 1.282828...
the .282828... part will be subtracted away
99x = 127
divide both sides by 99 to get
x= 127/99

8 0

3 years ago

Tickets to a concert cost $24 before a 5% fee is added.How much are the tickets with the fee?

vagabundo [1.1K]

Answer:

$25.20

Step-by-step explanation:

10%= 2.4 because you divide 24 by 10

So 5%= half 2.4= 1.2

24+1.2=25.20

5 0

3 years ago

Read 2 more answers

5^(-x)+7=2x+4 This was on plato

Setler79 [48]

Answer:

Below

I hope its not too complicated

$x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$

Step-by-step explanation:

$5^{\left(-x\right)}+7=2x+4\\\\\mathrm{Prepare}\:5^{\left(-x\right)}+7=2x+4\:\mathrm{for\:Lambert\:form}:\quad 1=\left(2x-3\right)e^{\ln \left(5\right)x}\\\\\mathrm{Rewrite\:the\:equation\:with\:}\\\left(x-\frac{3}{2}\right)\ln \left(5\right)=u\mathrm{\:and\:}x=\frac{u}{\ln \left(5\right)}+\frac{3}{2}\\\\1=\left(2\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)-3\right)e^{\ln \left(5\right)\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)}$

$Simplify\\\\\mathrm{Rewrite}\:1=\frac{2e^{u+\frac{3}{2}\ln \left(5\right)}u}{\ln \left(5\right)}\:\\\\\mathrm{in\:Lambert\:form}:\quad \frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}$

$\mathrm{Solve\:}\:\frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}:\quad u=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)\\\\\mathrm{Substitute\:back}\:u=\left(x-\frac{3}{2}\right)\ln \left(5\right),\:\mathrm{solve\:for}\:x$

$\mathrm{Solve\:}\:\left(x-\frac{3}{2}\right)\ln \left(5\right)=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right):\\\quad x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$

3 0

3 years ago

Pls help a brother out!

Leviafan [203]

Answer:

12 maybe the ANSWER.

hope it helps you

3 0

3 years ago

Read 2 more answers