All categories

3 years ago

Question 3 of 10 Solve |x + 5|= 7.

Mathematics

2 answers:

quester [9]3 years ago

6 0

Answer:

x = 2 and -2

Step-by-step explanation:

|x + 5| = 7

x + 5 = 7

x = 7 - 5

x = 2

|x + 5| = 7

-x + 5 = 7

-x = 7 - 5

-x = 2

x = -2

x = 2 and -2

damaskus [11]3 years ago

4 0

Answer:

|x + 5|= 7

-7<(x+5)<7

-7<x+5<7

-7-5<x+5-5<7-5

-12 < x <2

Step-by-step explanation:

You might be interested in

How to work out the volume of a cone?

Romashka [77]

Volume of a cone = πr²h/3. π=3.14 R=radius. hope this helps you :)

4 0

3 years ago

Help me, please I need this

Alex17521 [72]

Answer:

B) y = -3x - 5

Hope this helps!

Please be kind to mark Brainliest.☺

7 0

3 years ago

Jenny’s mom gave her one dollar on the day she was born. She plans to triple the amount she gives her every day. How much will s

astraxan [27]

Answer:

19,683

Step-by-step explanation:

3 0

3 years ago

likoan [24]

It's easy to show that $7\tan(4x)$ is strictly increasing on $x\in\left[0,\frac\pi8\right]$ . This means

$M = \max \left\{7\tan(4x) \mid \dfrac\pi{16} \le x \le \dfrac\pi{12}\right\} = 7\tan(4x) \bigg|_{x=\pi/12} = 7\sqrt3$

and

$m = \min \left\{7\tan(4x) \mid \dfrac\pi{16} \le x \le \dfrac\pi{12}\right\} = 7\tan(4x) \bigg|_{x=\pi/16} = 7$

Then the integral is bounded by

$\displaystyle 7\left(\frac\pi{12} - \frac\pi{16}\right) \le \int_{\pi/16}^{\pi/12} 7\tan(4x) \, dx \le 7\sqrt3 \left(\frac\pi{12} - \frac\pi{16}\right)$

$\implies \displaystyle \boxed{\frac{7\pi}{48}} \le \int_{\pi/16}^{\pi/12} 7\tan(4x) \, dx \le \boxed{\frac{7\sqrt3\,\pi}{48}}$

7 0

2 years ago

Find tan if sin=1/4 and falls in Quadrant 1

borishaifa [10]

Ok, sin = prep / hyp

so, perpendicular = 1 , and hypo = 4
so, base = sq. root ( 16-1 ) = root 15

and tan = prep / base = 1 / root (15)

7 0

3 years ago