What kind of triangles does the Pythagorean theorem work for?
O A. Right
OB. Acute
O C. Obtuse
O D. All triangles

Answer:

g(x) = x^2 - 4   (Answer #3)

Step-by-step explanation:

The graph of the parent function f(x) = x^2 is a parabola with vertex at (0, 0).  That of g(x) is the same, except that the first graph has been translated down by 4 units.

When this happens, the original function f(x) = x^2 + 0 changes to

                                                                     g(x) = x^2 - 4   (Answer #3)

Answer:

(2,2)

Step-by-step explanation:

Answer:

-27 is your answer. If I'm right so,

Please mark me as brainliest. thanks!!!

Answer: 18

Step-by-step explanation:

Answer:

x(x+1)

Step-by-step explanation:

Answer:

#. x=12.75

Step-by-step explanation:

51/4

Answer:

-3-8=-11

Step-by-step explanation:

The missing value is -8 because if you add is to -3 then you will get -11 and -11 = -11

Answer:

a. 80 units

b. 482.843 units²

Step-by-step explanation:

First things first, we need to solve for the side of the octagon. We can do so by using Pythagorean Theorem: a² + b² = c² where c is the hypotenuse and a and b are the other two sides.

12² + b² = 13²

144 + b² = 169

b² = 25

b = 5

Now that we know half of the side of the octagon is 5, we know that one side of the octagon is 10.

a. To find perimeter we just add every side together (or since every side is the same size, we multiply by 8)

10 × 8 = 80

b. The formula for finding the area of an octagon is 2(1 + radical 2)a² where a is the length of a side. By plugging in to a calculator we get:

2(1 + radical 2)(10)² = 482.843

Answer:

Step-by-step explanation:

Disclaimer  it is up the student to verify all the geometry algabra and mathematical formulas and calculation .

cosФ = opp/hyp     sinФ = adj/hyp        hyp = 10      perimeter = a + b + c

for the 45° right triangle

a = b     so      2a² = hyp²                       perimeter 45 = a + a + hyp    

P45 = 2a + hyp         a = sin(45) (hyp)  = [1/2√2)] (hyp)        sin(45) = (√2)/2

P45 = 2([1/2√2)] (hyp) + hyp

P45  = hyp (√2 + 1)

for the 30° right triangle

a = hyp×sin(30)     b = hyp×cos(30)           perimeter 30 = a + b + hyp    

            sin(30) = 1/2       cos(30) = (√3)/2

a  = hyp(1/2)           b = hyp(1/2)(√3)

P30 =        a        +      b         + hyp

        =  hyp(1/2)  +  hyp(1/2)(√3) + hyp

        =  hyp[ 1/2  +  (1/2)√3  + 1]

        =  hyp[ 1/2  +  (1/2)√3  + 1]             factored out the hyp

        =  (1/2)hyp[ 1 + √3 + 2]                  factored out a 1/2

P30  =  (1/2)hyp[ 3 + √3]

which is greater  P45 or P30?   vso I set both perimeters equal to each other

              P45           vs             P30

      hyp (√2 + 1)     =    (1/2)hyp[ 3 + √3]            the hyp cancel

               (√2 + 1)     =    (1/2)[ 3 + √3]                    still tough to tell  needed

               1.414 + 1    =    (1/2)[ 3 + 1.732]                to go to the calsulator

                 2.414      vs    1/2( 4.732)

                 2.414      >     2.366        thus P45 > P30    < -----  Final Answer

Disclaimer  it is up the student to verify all the geometry algabra and mathematical calculation.

Answer:

what he said

Step-by-step explanation:

Answer: wouldn’t it be $9,000? Or $12,000?

Step-by-step explanation:

Answer:

0ption C is your answer

Step-by-step explanation:

by using Pythagoras law

x²=8²+15²

x²=289

x=√289=17

Step-by-step explanation:

So looking at this question, we have two variables (1 - a glass of milk; 2 - a snack bar) and two equations to use to solve for the two variables. Remember that for any number of variables, you need that same number or more equations to solve for the variables.

Let's say that a glass of milk is represented by g and a snack bar is represented by s.

So we essentially have two equations here:

4 glasses of milk + 3 snack bars = 88 carbs, or:

4g + 3s = 88

2 glasses of milk + 4 snack bars = 74 carbs, or:

2g + 4s = 74

Now that we have this, we can solve for our equations. There are two ways to do this:

SUBSTITUTION - solve for one variable, then substitute it in

Let's solve for g using the second equation. First, we'll isolate g by subtracting 4s from both sides.

2g = 74 - 4s

Next, we can divide the whole equation by 2 to solve for g.

g = 37 - 2s

Now, we can substitute this into the first equation for g and solve for s.

4g + 3s = 88

4(37 - 2s) + 3s = 88

148 - 8s + 3s = 88

148 - 5s = 88 (now subtract 88 from both sides and add 5s to both sides)

60 = 5s

s = 12

Now we can use this to solve for g in the original equation

g = 37 - 2s

g = 37 - 2(12) = 37 - 24 = 13

ELIMINATION - eliminate one variable to solve for the other, then use that to solve for the first

Let's look at our two equations

4g + 3s = 88

2g + 4s = 74

Let's eliminate the variable g first by multiplying the second question by -2.

4g + 3s = 88

-4g - 8s = -148

If we add these two equations together, we get

-5s = -60

Divide both sides by -5 and we get

s = 12

Now, we can plug s into either equation to solve for g as we did above.

the answer will be thirty three cm.

Answer:

b

Step-by-step explanation:

get y by itself in the second equation: y= 2x-3

substitute this into the y of the first equation:

2x+2x-3= 5

Answer:

36

Step-by-step explanation:

Answer:

C

Step-by-step explanation:

72° and 3x are adjacent angles and are supplementary , that is

72° + 3x = 180° ( subtract 72° from both sides )

3x = 108° ( divide both sides by 3 )

x = 36° → C

Answer:

1. Boys =61

2.=53

3.122

4.128

Answer:

d

Step-by-step explanation:

Answer:

4782969

Step-by-step explanation:

Answer:

[tex]\sqrt[7]{3^{2}}[/tex]

Step-by-step explanation:

[tex]3^{2/7}[/tex]

^ That is 3 to the 2/7ths power. We know that. Now it needs to be in radical form. That takes a bit of thinking.

If you have a power that is less than one (like 1/2) it is can be written as a root. So say you're given [tex]2^{1/2}[/tex]. That is the same exact thing as saying [tex]\sqrt{2}[/tex]. It's the same thing. You can say it either way.

In the same way, a squared power, like [tex]2^{2}[/tex], is the same as representing it like a fraction, like [tex]2^{2/1}[/tex]. They both mean the same thing.

So by pointing these out, a fraction power = the power / the root.

If that makes sense.

So, with this in mind, let's look at what we have again.

[tex]3^{2/7}[/tex]

So we can split it up to say that 3 is raised to a 2nd power times a 1/7th power.

So we know what a 3 to the 2nd power is: [tex]3^{2}[/tex].

And since we have a fraction that's also included, we now know that is a root. So we can represent that like this: [tex]\sqrt[7]{3}[/tex].

But those two answers need to be combined. We can do that like this: [tex]\sqrt[7]{3^{2}}[/tex]

Answer: [tex]\sqrt[7]{3^{2}}[/tex]

So there it is! I hope that helped you!!

Answer:

ΔCFE is a right triangle

ΔBEC is an acute triangle

ΔBFG is an equilateral triangle

ΔGAF is an obtuse isosceles triangle

Answer:

the 2nd one i think please be right

Answer:

Where are the graphs?

Step-by-step explanation:

Anyway use this method.

Use the vertical line test to determine whether or not a graph represents a function. If a vertical line is moved across the graph and, at any time, touches the graph at only one point, then the graph is a function. If the vertical line touches the graph at more than one point, then the graph is not a function.

Answer:

120 grams because 120 kilograms is a lot

:)