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4 years ago

What does 5-(-6) equal

Mathematics

2 answers:

Len [333]4 years ago

6 0

5-(-6) = 11
Subtracting a negative makes that negative positive so 5-(-6) is actually 5+6=11

allsm [11]4 years ago

3 0

Answer:

Step-by-step explanation:

- and - makes a plus

5+6=11

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Please help me solve this and please also show work so I can understand it BETTER PLEASEEE ITS DUE

ratelena [41]

Answer:

6. 8b²-5b-4

7. -3x²+9x-4

8. 3

9. 6a+15

Step-by-step explanation:

you will subtract/add the numbers with the same variables and exponent.

ex. (number 6) 7b²+ b²= 8b²

5b= 5b 8b²+5b-4

1-5= -4

(plus 5b because it didnt come out to a negative) if you have any questions, say in the comments and ill try my best to help

8 0

3 years ago

BRAINLIEST ANSWER WINS! A satellite is to be put into an elliptical orbit around a moon as shown below. A vertical ellipse is sh

bekas [8.4K]

Answer:

D. x²/1953² + y²/ 1466² = 1

Step-by-step explanation:

==>Given:

Radius of spherical moon = 1000km

Distance of satellite from moon surface = 953km to 466km

==>Required:

Derived equation of ellipse

==>Solution:

The formula for driving an equation of ellipse is given as:

x²/a² + y²/b² = 1

Where,

a = length of the semi-major axis, while,

b = length of the semi-major axis

Since we are told that the satellite distance to the surface of the moon varies from 953km to 466km, values of a and b is calculated by summing each length to the radius of the moon as follows:

a = radius of moon + the larger distance of the satellite = 1000+953 = 1,953km

b = radius of moon + the smaller distance of the satellite = 1000+466 = 1,466km

Thus, the equation of the ellipse would be:

x²/1953² + y²/ 1466² = 1

8 0

3 years ago

Find all solutions in the interval from [0,2pi)<br> 2cos(3x)= -sqrt{2}

algol [13]

Solutions of $2cos(3x)= -\sqrt{2}$ in the interval from [0,2pi) is $x =\frac{\pi}{12}$ and $x = \frac{23\pi}{12}$ .

<u>Step-by-step explanation:</u>

Find all solutions in the interval from [0,2pi)

$2cos(3x)= -\sqrt{2}$

⇒ $2cos(3x)= -\sqrt{2}$

⇒ $\frac{2cos(3x)}{2}= \frac{-\sqrt{2}}{2}$

⇒ $cos3x= \frac{-\sqrt{2}(\sqrt{2})}{2{\sqrt{2}}}$

⇒ $cos3x= \frac{-2}{2{\sqrt{2}}}$

⇒ $cos3x= \frac{-1}{{\sqrt{2}}}$

⇒ $cos^{-1}(cos3x)= cos^{-1}(\frac{-1}{{\sqrt{2}}})$

⇒ $3x=\pm \frac{\pi}{4}$

⇒ $x=\pm \frac{\pi}{12}$

Cosine General solution is :

$x = \pm cos^{-1}(y)+ 2k\pi$

⇒ $x = \pm \frac{\pi}{12}+ 2k\pi$ , k is any integer .

At k=0,

⇒ $x =\frac{\pi}{12}$ ,

At k=1,

⇒ $x = - \frac{\pi}{12}+ 2\pi$

⇒ $x = \frac{23\pi}{12}$

Therefore , Solutions of $2cos(3x)= -\sqrt{2}$ in the interval from [0,2pi) is $x =\frac{\pi}{12}$ and $x = \frac{23\pi}{12}$ .

5 0

3 years ago

Write each of the following expressions without using absolute value.

Stels [109]

Answer:

-3+7=4 is the answer if thats what you is looking for

Step-by-step explanation:

7 0

3 years ago

Can someone please help

Shalnov [3]

Answer:

I would say 0 if you add it all to be honest, but I’m not TOO aware- Hope this helps at least a bit!

Step-by-step explanation:

6 0

3 years ago