All categories

4 years ago

Find x show steps too please will give brainliest

Mathematics

1 answer:

9966 [12]4 years ago

7 0

Hello from MrBillDoesMath!

Answer: x = 7/10

Discussion:

To eliminate fractions multiply each side by 10:

10 ( 9/5x - 2/5 ) = 10 * ( 8x +3)/10 or

10 * ( 9/5)x - 10 *(2/5) = 8x + 3 or

9 *2 x - 2*2 = 8x +3 or

18x - 4 = 8x + 3. Subtract 8x for each side:

18x - 8x - 4 = (8x - 8x) + 3 or

10x -4 = 0x + 3 or

10x -4 = 3 or

10x = 7

x = 7/10

Thank you,

MrB

You might be interested in

the ad giants baseball tearm has 20% of its 40 players that are left handed.how many left handed are their

dolphi86 [110]

8 are left handed bc 20% of its 40 is 0.2×40 =8

3 0

3 years ago

Identify the construction that the figure represents.

Aliun [14]

Angle bisectors is the correct ans

5 0

3 years ago

Read 2 more answers

-3.1x+7-7.4x-1.5x-6(x-3/2

nalin [4]

Answer:

-2(9x-8)

Step-by-step explanation:

We are to simplify the expression -3.1x+7-7.4x-1.5x-6(x-3/2)

Open the parenthesis

-3.1x+7-7.4x-1.5x-6(x-3/2)

= -3.1x+7-7.4x-1.5x-6x+6(3/2)

= -3.1x+7-7.4x-1.5x-6x+9

Collect like terms

= -3.1x-7.4x-1.5x-6x+7+9

= -12x-6x+16

= -18x+16

= -2(9x-8)

Hence the simplified form of the expression is -2(9x-8)

6 0

3 years ago

744 ÷ 40 in long division

wariber [46]

18.5 would be the answer and here’s my work

7 0

3 years ago

Find the absolute extrema of f(x) = e^{x^2+2x}f ( x ) = e x 2 + 2 x on the interval [-2,2][ − 2 , 2 ] first and then use the com

fredd [130]

$f(x)=e^{x^2+2x}\implies f'(x)=2(x+1)e^{x^2+2x}$

$f$ has critical points where the derivative is 0:

$2(x+1)e^{x^2+2x}=0\implies x+1=0\implies x=-1$

The second derivative is

$f''(x)=2e^{x^2+2x}+4(x+1)^2e^{x^2+2x}=2(2x^2+4x+3)e^{x^2+2x}$

and $f''(-1)=\frac2e>0$ , which indicates a local minimum at $x=-1$ with a value of $f(-1)=\frac1e$ .

At the endpoints of [-2, 2], we have $f(-2)=1$ and $f(2)=e^8$ , so that $f$ has an absolute minimum of $\frac1e$ and an absolute maximum of $e^8$ on [-2, 2].

So we have

$\dfrac1e\le f(x)\le e^8$

$\implies\displaystyle\int_{-2}^2\frac{\mathrm dx}e\le\int_{-2}^2f(x)\,\mathrm dx\le\int_{-2}^2e^8\,\mathrm dx$

$\implies\boxed{\displaystyle\frac4e\le\int_{-2}^2f(x)\,\mathrm dx\le4e^8}$

5 0

3 years ago