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3 years ago

8

-15 + 8x = x+3 Help me please!!

1 answer:

Amanda [17]3 years ago

4 0

Take the x and -15 over,

7x=18

then divide by 7,

x=18/7

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Let f(x) = 2x – 1, g(x) = 3x, and h(x) = x2 + 1. Compute the following.

bonufazy [111]

4.
h(f(x)=h(2x-1)=
(2x-1)^2+1=
4x²-4x+1+1=
4x²-4x+2

5.
f(f(x))=2(2x-1)-1=4x-2-1=4x-3

6.
f o g (x)=f(g(x))
h o g (x)=h(g(x))=
(3x)^2+1=
9x²+1

4 0

3 years ago

WARNING : NON SENSE ANSWER REPORT TO MODERATOR

Rzqust [24]

Answer:

$\frac{180}{147}$

Step-by-step explanation:

Simplify $(\frac{3}{-7} - \frac{11}{21} )$

$=> \frac{(-3 \times 3) - 11}{21}$

$=> \frac{-9 - 11}{21}$

$=> \frac{-20}{21}$

Find the additive inverse of $\frac{-20}{21}$ by using its property - <em>"Sum of a number & its additive inverse is always zero". </em>Assume that 'x' is an additive inverse of $\frac{-20}{21}$ .

$=> x + \frac{-20}{21} = 0$

$=> x = 0 - (-\frac{20}{21}) = \frac{20}{21}$

Simplify $(\frac{9}{5} \div \frac{7}{5} )$

$=> \frac{9}{5} \times \frac{1}{\frac{7}{5} }$

$=> \frac{9}{5} \times \frac{5}{7}$

$=> \frac{9}{7}$

Now, find the product of $\frac{9}{7}$ & $\frac{20}{21}$

$=> \frac{9}{7} \times \frac{20}{21}$

$=> \frac{180}{147}$

3 0

3 years ago

Helpppp please i’m struggling

IrinaK [193]

C is the correct answer

6 0

2 years ago

1,000,000,000 divided by 167?

tester [92]

Tge answer is 598,802.395 because if you put it into tge calculator you should get this answer

4 0

3 years ago

Read 2 more answers

Find the product of all real values of r for which 1/2x=r-x/7

Dahasolnce [82]

Answer:

$r = \±\sqrt{14$

$Product = -14$

Step-by-step explanation:

Given

$\frac{1}{2x} = \frac{r - x}{7}$

Required

Find all product of real values that satisfy the equation

$\frac{1}{2x} = \frac{r - x}{7}$

Cross multiply:

$2x(r - x) = 7 * 1$

$2xr - 2x^2 = 7$

Subtract 7 from both sides

$2xr - 2x^2 -7= 7 -7$

$2xr - 2x^2 -7= 0$

Reorder

$- 2x^2+ 2xr -7= 0$

Multiply through by -1

$2x^2 - 2xr +7= 0$

The above represents a quadratic equation and as such could take either of the following conditions.

(1) No real roots:

This possibility does not apply in this case as such, would not be considered.

(2) One real root

This is true if

$b^2 - 4ac = 0$

For a quadratic equation

$ax^2 + bx + c = 0$

By comparison with $2x^2 - 2xr +7= 0$

$a = 2$

$b = -2r$

$c =7$

Substitute these values in $b^2 - 4ac = 0$

$(-2r)^2 - 4 * 2 * 7 = 0$

$4r^2 - 56 = 0$

Add 56 to both sides

$4r^2 - 56 + 56= 0 + 56$

$4r^2 = 56$

Divide through by 4

$r^2 = 14$

Take square roots

$\sqrt{r^2} = \±\sqrt{14$

$r = \±\sqrt{14$

Hence, the possible values of r are:

$\sqrt{14$ or $-\sqrt{14$

and the product is:

$Product = \sqrt{14} * -\sqrt{14}$

$Product = -14$

8 0

2 years ago