Simplify: 26 - 2(7 + 3^2) please answer with one of the options below:)

Mathematics

1 answer:

ad-work [718]2 years ago

7 0

-6 so option D is the answer

You might be interested in

What is the perimeter of this quadrilateral?<br> (5,5)<br> (2, 4)<br> (4, 1)<br> (6, 1)

lbvjy [14]

$~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_1}{5}~,~\stackrel{y_1}{5})\qquad B(\stackrel{x_2}{2}~,~\stackrel{y_2}{4}) ~\hfill AB=\sqrt{[ 2- 5]^2 + [ 4- 5]^2} \\\\\\ AB=\sqrt{(-3)^2+(-1)^2}\implies \boxed{AB=\sqrt{10}} \\\\[-0.35em] ~\dotfill\\\\ B(\stackrel{x_1}{2}~,~\stackrel{y_1}{4})\qquad C(\stackrel{x_2}{4}~,~\stackrel{y_2}{1}) ~\hfill BC=\sqrt{[ 4- 2]^2 + [ 1- 4]^2}$

$BC=\sqrt{2^2+(-3)^2}\implies \boxed{BC=\sqrt{13}} \\\\[-0.35em] ~\dotfill\\\\ C(\stackrel{x_1}{4}~,~\stackrel{y_1}{1})\qquad D(\stackrel{x_2}{6}~,~\stackrel{y_2}{1}) ~\hfill CD=\sqrt{[ 6- 4]^2 + [ 1- 1]^2} \\\\\\ CD=\sqrt{2^2+0^2}\implies \boxed{CD=2} \\\\[-0.35em] ~\dotfill\\\\ D(\stackrel{x_1}{6}~,~\stackrel{y_1}{1})\qquad A(\stackrel{x_2}{5}~,~\stackrel{y_2}{5}) ~\hfill DA=\sqrt{[ 5- 6]^2 + [ 5- 1]^2}$

$DA=\sqrt{(-1)^2+4^2}\implies \boxed{DA=\sqrt{17}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\Large Perimeter}}{\sqrt{10}~~ + ~~\sqrt{13}~~ + ~~2~~ + ~~\sqrt{17}}~~ \approx ~~ 12.89$

8 0

2 years ago

Round 2.794 to the tenths place

viva [34]

2.8

The tenths place is right after the decimal, so when we round 2.794 to the nearest tenths place, we get 2.8

Hope this helps! Have a good day :)

8 0

2 years ago

100 POINTS

likoan [24]

Answer: True because when you divide both sides by 2 that would be the right equation.

Step-by-step explanation:

3 0

3 years ago

Solve for x 4/x + 4/x^2-9= 3/x-3

Goshia [24]

$\dfrac{4}{x}+\dfrac{4}{x^2-9}=\dfrac{3}{x-3}\qquad\text{subtract}\ \dfrac{4}{x^2-9}\ \text{from both sides}\\\\\dfrac{4}{x}=\dfrac{3}{x-3}-\dfrac{4}{x^2-9}\qquad\text{use}\ a^2-b^2=(a-b)(a+b)\\\\\dfrac{4}{x}=\dfrac{3}{x-3}-\dfrac{4}{x^2-3^2}\\\\\dfrac{4}{x}=\dfrac{3}{x-3}-\dfrac{4}{(x-3)(x+3)}\\\\\dfrac{4}{x}=\dfrac{3(x+3)}{(x-3)(x+3)}-\dfrac{4}{(x-3)(x+3)}\qquad\text{use distributive property}\\\\\dfrac{4}{x}=\dfrac{3x+9-4}{(x-3)(x+3)}\\\\\dfrac{4}{x}=\dfrac{3x+5}{x^2-9}\qquad\text{cross multiply}$

$4(x^2-9)=x(3x+5)\qquad\text{use distributive property}\\\\4x^2-36=3x^2+5x\qquad\text{subtract}\ 3x^2\ \text{and}\ 5x\ \text{from both sides}\\\\x^2-5x-36=0\\\\x^2-9x+4x-36=0\\\\x(x-9)+4(x-9)=0\\\\(x-9)(x+4)=0\iff x-9=0\ \vee\ x+4=0\\\\\boxed{x=9\ \vee\ x=-4}$

8 0

3 years ago

What is the intercepting point in the equations y=7/3-1/3x and y=6x-11

GuDViN [60]

It's right here. (2.105,1.632)

4 0

3 years ago

Read 2 more answers