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stiks02 [169]
3 years ago
11

Whats 16/40 as a mixed fraction?

Mathematics
1 answer:
Rainbow [258]3 years ago
8 0
You can't do that in less the 40 is over 16
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6 2/7 * 3 3/12= ? Anyways hi! I'm bo.red lol
Georgia [21]

Answer:

20  and 3/7

Step-by-step explanation:

8 0
2 years ago
Translate.
xxMikexx [17]

Answer:

b.

Step-by-step explanation:

The quantity x minus 2y is (x - 2y),   then we add  3.

4 0
3 years ago
Y ∝ <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx%7D" id="TexFormula1" title="\frac{1}{x}" alt="\frac{1}{x}" align="absmi
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4 0
2 years ago
1.Graph f(x) = -1.5x +6<br> 2.Graph f(x) = -1/2x-5
nalin [4]

Ans(1):

Given equation is f(x)=-1.5x+6

we can plug any number like x=0 and x=2 to find the f(x) also called y-value

plug x=0

f(x)=-1.5x+6 =-1.5*0+6 =0+6 =6

Hence first point is (0,6)

plug x=2

f(x)=-1.5x+6 =-1.5*2+6 =-3+6 =3

Hence first point is (2,3)

now we can graph both points then join them to get final graph of f(x)=-1.5x+6

---------------------

Ans(2):

We can repeat exactly same process for f(x) = -1/2x-5.

So the final graph will look like attached picture:


5 0
3 years ago
A hyperbola centered at the origin has verticies at (add or subtract square root of 61,0 and foci at (add or subtract square roo
deff fn [24]

Answer:

\frac{x^2}{61}-\frac{y^2}{37}  =1

Step-by-step explanation:

The standard equation of a hyperbola is given by:

\frac{(x-h)^2}{a^2} -\frac{(y-k)^2}{b^2} =1

where (h, k) is the center, the vertex is at (h ± a, k), the foci is at (h ± c, k) and c² = a² + b²

Since the hyperbola is centered at the origin, hence (h, k) = (0, 0)

The vertices is (h ± a, k) = (±√61, 0). Therefore a = √61

The foci is (h ± c, k) = (±√98, 0). Therefore c = √98

Hence:

c² = a² + b²

(√98)² = (√61)² + b²

98 = 61 + b²

b² = 37

b = √37

Hence the equation of the hyperbola is:

\frac{x^2}{61}-\frac{y^2}{37}  =1

6 0
3 years ago
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