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2 years ago

Expand. If necessary, combine like terms (2+x)(2-x)

Mathematics

1 answer:

Anastaziya [24]2 years ago

5 0

Answer:

-x^2+4

Step-by-step explanation:

2*2=4

2(-x)+2(x)=0

x(-x)=-x^2

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What is x?????????????

Mekhanik [1.2K]

Answer:

BGC=180-90

=90

BGC=BGE+EGC

BGC=BGE+15

90=BGE+15

BGE=90-15

BGE=75

BGE=AGF

AGF=X,

X =75

8 0

3 years ago

The product of (x+2) (x-3) is

Aloiza [94]

Answer:

Step-by-step explanation:

$(x + 2)(x - 3)$

Multiply each term in the first parentheses by each term in the second parentheses ( FOIL)

$x(x - 3) + 2(x - 3)$

Calculate the product

${x}^{2} - 3x + 2x - 6$

Collect like terms

${x}^{2} - x - 6$

Hope this helps..

Best regards!

5 0

3 years ago

Read 2 more answers

Ng 20-21

Lisa [10]

you would spend $30

Step-by-step explanation:

if 20 calculator cost 120 then you would divide 20 by 120 which would give you 6 dollars per calculator then you would multiply that by 5 which would give you a $30 for five calculators.

4 0

2 years ago

Read 2 more answers

Gemma recently rode her bicycle to visit her friend who lives 6 miles away. On her way there, her average speed was 16 miles per

DochEvi [55]

first speed --- x mph

return speed -- x+16 mph

6/x + 6/(x+16) = 1

times each term by x(x+16)

6(x+16) + 6x = x(x+16)

x^2 + 4x - 96 = 0

(x-8)(x+12) = 0

x = 8 or x is a negative

her first speed was 8 mph

her return speed was 24 mph

check:

6/8 + 6/24 = 1 , that's good!

8 0

2 years ago

Find x<br><br><br><br><br> Thanks for the help in advance!

Marina86 [1]

I'm not sure if this is the easiest way of doing this, but it surely work.

Let the base of the triangle be AB, and let CH be the height. Just for reference, we have

$AH=2,\quad HB=6,\quad AC=x$

Moreover, let CH=y and BC=z

Now, AHC, CHB and ABC are all right triangles. If we write the pythagorean theorem for each of them, we have the following system

$\begin{cases}4+y^2=x^2\\36+y^2=z^2\\x^2+z^2=64\end{cases}$

If we solve the first two equations for y squared, we have

$y^2=x^2-4\\y^2=z^2-36$

And we can deduce

$z^2 = x^2+32$

So that the third equation becomes

$x^2+x^2+32=64 \iff 2x^2 = 32 \iff x^2=16 \iff x=4$

(we can't accept the negative root because negative lengths make no sense)

7 0

2 years ago