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3 years ago

What is the solution to the equation 2^(x+4) -12- 20? O x=0 O x=1 O x=2 O x=9

Mathematics

2 answers:

garik1379 [7]3 years ago

8 0

Use Photomath if helps a lot

masya89 [10]3 years ago

6 0

The answers is 2x+4−12−20

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−5(1+5)=−6−24 helppppppppppppppppppppp

GarryVolchara [31]

Answer:

-30=-30

Step-by-step explanation:

Always true.

8 0

3 years ago

Read 2 more answers

What is the perimeter of this quadrilateral? (5,5) (2, 4) (4, 1) (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

3 years ago

4) An arithmetic sequence has a 7th term of 54 and a 13th term of 94. Find the common difference

san4es73 [151]

Answer:

20/3

Step-by-step explanation:

Every nth term takes the form of a + n*d, where a is the first term.

So 7th term = (54 = a + 7d), 13th term = (94 = a + 13d).

equate them both:

94 - 13d = 54 - 7d

40 = 6d

d = 40/6

6 0

3 years ago

Bethany, Lauren, Amanda, and David all meet at a family reunion. They are comparing their ages. They have found that Lauren is 1

tekilochka [14]

Lauren =l
David=d
Bethany's =x
Amanda =y
l=x+13
d=y+11
lx=2y
(x+13)x=2y
 $2y=x^2+13x$
$y= \frac{1}{2}(x^2+13x)$
or
$y= \frac{x^2}{2}+ \frac{13}{2}x$
(x-20)(l-20)=d
(x-20)(x+13-20)=y+11 
 $y+11=(x-20)(x-7)$
$y+11=x^2-27x+140$
$y=x^2-27x+129$
so I believe its b
Hope this helps :)

6 0

3 years ago

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teacher has 27 students in her class she asked the students to form as many groups of 4 as possible how many students are not be

stepan [7]

If there are 27 students in the class and they have to form as many groups of 4 as possible, you can calculate this using the following step:

27 / 4 = 6.75 = 6 3/4

Result: 27 students can form 6 groups of 4 students and 3 students have to form their own group.

4 0

3 years ago