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Gwar [14]
3 years ago
14

6x4/12+72/8-9 using pemdas

Mathematics
2 answers:
viktelen [127]3 years ago
4 0

Answer:

the answer is 2

Step-by-step explanation:

You do multiplication and division left to right first and then you add the two answers

V125BC [204]3 years ago
3 0
Here ya go ^v^ try to use math.way
I hope this helps
Ya have a wonderful day ^v^

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. How many faces does a polyhedron with<br> 20 vertices and 30 edges have?
Alisiya [41]

Answer:

12

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
What are the zero(s) of the function f(x)=<img src="https://tex.z-dn.net/?f=%5Cfrac%7B4x%5E%7B2%7D-36x%20%7D%7Bx-9%7D" id="TexFo
Contact [7]

Answer:

D

Step-by-step explanation:

The function will be zero when the numerator is equal to 0. So, set the numerator equal to 0 and solve:

0 = 4 {x}^{2}  - 36x

I will use the quadratic formula:

\frac{ 36  +  \sqrt{ {36}^{2}  - 4(4)(0)} }{2(4)}  \\  \frac{  36 + 36}{8}  = 9

Therefore, x = 9 is a 0. Let's check with subtracted root now:

\frac{ 36   -    \sqrt{ {36}^{2}  - 4(4)(0)} }{2(4)}  \\  \frac{ 36  -  36}{8}  =  0

It appears 0 is also a root.

Therefore, the answer is D.

7 0
3 years ago
Need some help with these problems please.
goldenfox [79]

Answer:

See below.

Step-by-step explanation:

a.  

Divide the leading terms:

6x^4 / 2x^2 = 3x^2

3x2 is a parabola so the long run is

as x ---> +/- infinity r(x) ----> + infinity. (answer).

b.

10,000x^3 / 50 x^3

= 200.

So r(x) as a horizontal asymptote at y ( r(x)) = 200.  (answer).

3 0
3 years ago
What is HL???<br><br> HL=__<br><br> Image Below
Advocard [28]

Consider a trapezoid GJKF with:

GH=JH (hypothesis)

FL=KL (hypothesis)

so HL is the median of trapezoid GJKF

so: HL=1/2(2.5+1.2)=1.85

4 0
2 years ago
Q. A room measure 3m 55cm by 2m 25 cm. What is the perimeter of the room?
Bond [772]

Answer:

100cm = 1m

Thus, one more step should be added to your working to arrive at the final answer:

Perimeter

= 10m 160cm

= 11m 60cm

This is because 160cm= 100cm +60cm, which is also equivalent to 1m 60cm.

Alternative working:

Convert all the units to cm first.

\boxed{1 \: m= 100 \: cm}

Length

= 5m 55cm

= 300cm +55cm

= 355cm

Breadth

= 2m 25cm

= 200cm +25cm

= 225cm

Perimeter

= 2(length +breadth)

= 2(355 cm +225 cm)

= 2(580 cm)

= 1160 cm

= 11m 60cm

From the second last step to the last step, you could divide 1160 by 100 to find how many meters are there. You would get a quotient of 11 and a remainder of 60. Since this 60 cannot be changed into meters as a whole number, we can leave it as 11m 60cm. Otherwise, it is also correct to leave it as 1160cm or 11.6m unless otherwise stated.

However, 10m 160cm is not preferred since the conversion of cm to m is done partially, as the 160 cm can still be further simplified.

8 0
2 years ago
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