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MissTica
3 years ago
13

1 1/6 as a whole number

Mathematics
2 answers:
MAXImum [283]3 years ago
6 0
1 and 5/6 is the answer
dsp733 years ago
3 0
5/1 is correct i think

You might be interested in
How would I solve for the parabolic equation with these two given points. Please answer ASAP!!
monitta
So

y=ax^2+bx+c
(x,y)
sub the points and solve

(4.28,6.48)
6.48=a(4.28)^2+b(4.28)+c

(12.61,15.04)
15.04=a(12.61)^2+b(12.61)+c
well, for 3 variables, we need equations and therefor 3 points

maybe we are supposed to assume it starts at (0,0)
so then
0=a(0)^2+b(0)+c
0=c
so then
6.48=a(4.28)^2+b(4.28)
15.04=a(12.61)^2+b(12.61)

solve for a by subsitution
first equation, minut a(4.28)^2 from both sides
6.48-a(4.28)^2=b(4.28)
divide both sides by 4.28
(6.48/4.28)-4.28a=b
sub that for b in other equation

15.04=a(12.61)^2+b(12.61)
15.04=a(12.61)^2+((6.48/4.28)-4.28a)(12.61)
expand
15.04 =a(12.61)^2+(81.7128/4.28)-53.9708a
minus (81.7128/4.28) both sides
15.04-(81.7128/4.28)=a(12.61)^2-53.9708a
15.04-(81.7128/4.28)=a((12.61)^2-53.9708)
(15.04-(81.7128/4.28))/(((12.61)^2-53.9708))=a
that's the exact value of a
to find b, subsitute to get
(6.48/4.28)-4.28((15.04-(81.7128/4.28))/(((12.61)^2-53.9708)))=b

if we aprox
a≈-0.038573167896199
b≈1.6791118501845
so then the equation is
y=-0.038573167896199x²+1.6791118501845x
7 0
3 years ago
Point F' corresponds to<br> point D.<br> point E.<br> point F.<br> opoint E.
Arada [10]

refer to the attachment

6 0
2 years ago
Read 2 more answers
Using the rest of my points to ask this, please answer quickly :)
TEA [102]

Answer:

C

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
Evaluate lim x=π/4(3x - tanx)​
Karo-lina-s [1.5K]

Answer:

 \lim_{x \to\frac{\pi }{4} } (3x-tanx) =  3(\frac{\pi }{4}  )

Step-by-step explanation:

<u><em>Explanation:-</em></u>

Given that

         \lim_{x \to\frac{\pi }{4} } (3x-tanx)

  =  \lim_{x \to\frac{\pi }{4} } (3x) -  \lim_{x \to\frac{\pi }{4} }  tanx)

 = 3(\frac{\pi }{4} )+tan(\frac{\pi }{4} )

=  3(\frac{\pi }{4} )+ 1 )

=  3(\frac{\pi }{4}  )

5 0
2 years ago
How many different rectangles can you draw, that has an area of 28 cm sq. ? pls help me
jolli1 [7]

Answer:

You can draw three rectangles:

1- 28 * 1.

2- 14*2

3- 7*4

Step-by-step explanation:

We have a total area of 28. If we use only integer numbers, we can find all the divisors of 28. The possible rectangles will be organized with them, taking into account that the product of them, which means the area should be 28.

28 = 1*2*2*7

We can organize them as follows:

R1: 28 = 1*28

R2: 28 = 2*14

R3: 28 = 4*7

Finally, we can conclude that there are only three possibilities

1- 28 by 1.

2- 14 by 2

3- 7 by4

The perimeters will be:

Perimeter 1 = 2x1 + 2x28 = 58

Perimeter 2 = 2x2 + 2x14 = 32

Perimeter 3 = 2x4+2x7 = 22

3 0
3 years ago
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