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jeka57 [31]
3 years ago
6

Four times the sum of a number and ter is the same as one-hundred added to eight

Mathematics
1 answer:
AfilCa [17]3 years ago
7 0

Answer:

GIVE BRAINLIEST PLZ

I DONT NEED THNX

x = -15

Step-by-step explanation:

1) Four times sum of a number and 10

4(x + 10)

2) is equal to 100 added to 8 times the number

4(x+10) = 8x + 100

x + 10 = 2x + 25

-15 = x

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21. In 4 + In(4x - 15) = In(5x + 19)
Alexxx [7]

Answer:

x=-30;\quad \:I\ne \:0

Step-by-step explanation:

In\cdot \:4+In\left(4x-15\right)=In\left(5x+19\right)

\:4+In\left(4x-15\right):\quad -11nI+4nxI\\In\cdot \:4+In\left(4x-15\right)\\=4nI+nI\left(4x-15\right)\\\\\:In\left(4x-15\right):\quad 4nxI-15nI\\\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b-c\right)=ab-ac\\a=In,\:b=4x,\:c=15\\=In\cdot \:4x-In\cdot \:15\\=4nxI-15nI\\=In\cdot \:4+4nxI-15nI\\\mathrm{Simplify}\:In\cdot \:4+4nxI-15nI:\quad -11nI+4nxI\\In\cdot \:4+4nxI-15nI\\\mathrm{Group\:like\:terms}\\=4nI-15nI+4nxI\\\mathrm{Add\:similar\:elements:}\:4nI-15nI=-11nI\\=-11nI+4nxI\\

\mathrm{Expand\:}In\left(5x+19\right):\quad 5nxI+19nI\\In\left(5x+19\right)\\=nI\left(5x+19\right)\\\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b+c\right)=ab+ac\\a=In,\:b=5x,\:c=19\\=In\cdot \:5x+In\cdot \:19\\=5nxI+19nI\\\\-11nI+4nxI=5nxI+19nI\\\\\mathrm{Add\:}11nI\mathrm{\:to\:both\:sides}\\-11nI+4nxI+11nI=5nxI+19nI+11nI\\Simplify\\4nxI=5nxI+30nI\\\mathrm{Subtract\:}5nxI\mathrm{\:from\:both\:sides}\\4nxI-5nxI=5nxI+30nI-5nxI\\\mathrm{Simplify}\\-nxI=30nI\\

\mathrm{Divide\:both\:sides\:by\:}-nI;\quad \:I\ne \:0\\\frac{-nxI}{-nI}=\frac{30nI}{-nI};\quad \:I\ne \:0\\\mathrm{Simplify}\\\frac{-nxI}{-nI}=\frac{30nI}{-nI}\\\mathrm{Simplify\:}\frac{-nxI}{-nI}:\quad x\\\frac{-nxI}{-nI}\\\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{-b}=\frac{a}{b}\\=\frac{nxI}{nI}\\\mathrm{Cancel\:the\:common\:factor:}\:n\\=\frac{xI}{I}\\\mathrm{Cancel\:the\:common\:factor:}\:I\\=x\\\mathrm{Simplify\:}\frac{30nI}{-nI}:\quad -30\\\mathrm{Apply\:the\:fraction\:rule}:

\quad \frac{a}{-b}=-\frac{a}{b}\\\mathrm{Cancel\:the\:common\:factor:}\:n\\=\frac{30I}{I}\\\mathrm{Cancel\:the\:common\:factor:}\:I\\=-30\\x=-30;\quad \:I\ne \:0

3 0
3 years ago
I need an answer ASAP.
melisa1 [442]
6 positive tiles because there’s 6 negative so 6 positive will cancel it out
7 0
3 years ago
What is the range of the function?
levacccp [35]

Answer:

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Step-by-step explanation:

7 0
3 years ago
2/(x^2)=4/x+1<br>can any one solve this for me please
levacccp [35]
2/X^2=4/X+1                                   times by both denominators
2(x+1)=4(x^2)
2x+2=4x^2
4x^2-2x-2=0
(2x+1)(2x-2)=0
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5 0
3 years ago
F(1)=-12, f(n)=2f(n-1) <br> find first 4 terms
amid [387]
<h3>First four terms are:   -12, -24, -48, -96</h3>

============================================

Explanation:

The notation f(1) = -12 tells us that the first term is -12.

The equation f(n) = 2*f(n-1) indicates that we're multiplying the (n-1)th term by 2 to get the nth term. Basically we multiply each term by 2 to get the next term.

-12*2 = -24

-24*2 = -48

-48*2 = -96

So the first four terms are -12, -24, -48, -96.

-----------------------------------

Alternative Approach

f(1) = -12 .... first term

f(n) = 2*f(n-1) ... nth term recursive formula

f(2) = 2*f(2-1) ... plug in n = 2

f(2) = 2*f(1)

f(2) = 2*(-12) ... plug in f(1) = -12

f(2) = -24 ... second term

f(n) = 2*f(n-1) .... resetting for the next computation

f(3) = 2*f(3-1) ... plug in n = 3

f(3) = 2*f(2)

f(3) = 2*(-24)  ... plug in f(2) = -24

f(3) = -48 .... third term

f(n) = 2*f(n-1) ... one more reset

f(4) = 2*f(4-1) .... plug in n = 4

f(4) = 2*f(3)

f(4) = 2*(-48) ... plug in f(3) = -48

f(4) = -96 ... fourth term

We get the first four terms to be -12, -24, -48, -96, which was the same as earlier.

4 0
3 years ago
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