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2 years ago

Geometry help please!! 20 points

Mathematics

1 answer:

vekshin12 years ago

6 0

Answer:

Step-by-step explanation:

Opposite angles of a cyclic quadrilateral are supplementary, so (6x+72)+3x=180

$(6x+72)+3x=180\\9x+72=180\\9x=108\\x=12$

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0.2 Times 10^6 in scientific notatation

Ronch [10]

You could move the decimal to the right 6 times
0.2 x 10^6
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3 years ago

What are the like terms in this expression? 9a - 3b + 4ab

sergey [27]

There are no like terms

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3 years ago

Find h of the following cone

Reptile [31]

Sqrt55. Use the pythagorean theorem; 3^2 + x^2 = 8^2

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3 years ago

What is the surface area of this right triangular prism ? round your answer to the nearest tenth.

velikii [3]

Answer:

78 m^2 Unless I got some of the measurements wrong, kinda hard to see them.

Step-by-step explanation:

The surface area will be the sum of the area of each face. so lets see, there is the base (in this orientation) which is a 8x3 rectangle it looks like.

then two trianglular sides with both having a base of 8 and height of 3

then two "roof" pieceswhich are rectangles again, one side of 3 and the other 5.

Now we add each face up.

rectangular base = 3x8 = 24

2 triangular sides = 2(.5*8*3) = 24

2 rectangular roof pieces = 2(5*3) = 30

Now add it all up

24+24+30 = 78

5 0

3 years ago

Read 2 more answers

Find the zeros of the function f(x) = x2 + 5x + 6. A) y = 6 because the graph crosses the y-axis at 6. B) y = -0.25 because that

Fofino [41]

Answer:

D.)

Step-by-step explanation:

The zero's are referencing when y=0, note that when y=0 they are talking about the x-intercepts. You can graph the function and see when the graph crosses the x-axis or solve for the x-values. I will solve it via factoring and so:

$f(x)=x^2+5x+6$

Multiply the outer coefficients, in this case 1 and 6, and 1×6=6. Now let's think about all the factors of 6 we have: 6×1 and 2×3. Now is there a way that if we use any of these factors and add/subtract them they will return the middle term 5? Actually we can say 6-1=5 and 2+3=5. Let's try both.

First let's use 6 and -1 and so:

$x^2+5x+6\\\\x^2+6x-x+6\\\\x(x+6)-1(x-6)$

Notice how we have (x+6) and (x-6), these factors do not match so this is incorrect.

Now let's try 2 and 3 and so:

$x^2+5x+6\\\\x^2+3x+2x+6\\\\x(x+3)+2(x+3)\\\\(x+2)(x+3)$

Notice how the factors (x+3) matched up so this is a factor and so is (x+2), now to solve for the zero's let's make f(x)=0 and solve each factor separately:

Case 1:

$f(x)=x+2\\\\0=x+2\\\\x=-2$

Case 2:

$f(x)=x+3\\\\0=x+3\\\\x=-3$

So your zero's are when x=-2 and x=-3.

D.) x=-3 and x=-2 because the graph crosses the x-axis at -3 and -2.

~~~Brainliest Appreciated~~~

8 0

3 years ago