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pentagon [3]
3 years ago
14

A potter can make 24 vases in 8 days. If the potter works 6 hours each day, how long does it take to make one vase?

Mathematics
1 answer:
NikAS [45]3 years ago
7 0
Answer: 2
Explaintion/
24/8 =3
6/3=2
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8.66

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10z + 5

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Find the absolute maximum and minimum values of the following function in the closed region bounded by the triangle with vertice
alukav5142 [94]

f(x,y)=2x^2-8x+y^2-8y+2

f_x(x,y)=4x-8=0\implies x=2

f_y(x,y)=2y-8=0\implies y=4

There is one critical point at (2, 4), but this point happens to fall on one of the boundaries of the region. We'll get to that point in a moment.

Along the boundary x=0, we have

f(0,y)=y^2-8y+2=(y-4)^2-14

which attains a maximum value of

f(0,4)=-14

Along y=44, we have

f(x,44)=2x^2-8x+1586=2(x-2)^2+1578

which attains a maximum of

f(2,44)=1578

Along y=2x, we have

f(x,2x)=6x^2-24x+2=6(x-2)^2-22

which attains a maximum of

f(2,4)=-22

So over the given region, the absolute maximum of f(x,y) is 1578 at (2, 44).

8 0
3 years ago
(3x2 + 10x + 3) by (x + 3)​
Nata [24]
Explanation: 3 x 2 + 10 x + 3 We can Split the Middle Term of this expression to factorise it. In this technique, if we have to factorise an expression like a x 2 + b x + c , we need to think of 2 numbers such that: N 1 ⋅ N 2 = a ⋅ c = 3 ⋅ 3 = 9 and, N 1 + N 2 = b = 10 After trying out a few numbers we get: N 1 = 9 and N 2 = 1 9 ⋅ 1 = 9 , and 9 + ( 1 ) = 10 3 x 2 + 10 x + 3 = 3 x 2 + 9 x + 1 x + 3 = 3 x ( x + 3 ) + 1 ( x + 3 ) ( 3 x + 1 ) ( x + 3 ) is the factorised form for the expression.

c
, we need to think of 2 numbers such that:
N
1
⋅
N
2
=
a
⋅
c
=
3
⋅
3
=
9

and,
N
1
+
N
2
=
b
=
10
After trying out a few numbers we get:
N
1
=
9
and
N
2
=
1

9
⋅
1
=
9
, and
9
+
(
1
)
=
10
3
x
2
+
10
x
+
3
=
3
x
2
+
9
x
+
1
x
+
3

=
3
x
(
x
+
3
)
+
1
(
x
+
3
)
(
3
x
+
1
)
(
x
+
3
)
is the factorised form for the expression.
6 0
2 years ago
The perimeters of the square and the triangle shown below are equal. What is the value of x?
Rudiy27
Perimeter is the sum of all the sides. So we can set up an equation:

\sf 3x+1+3x+1+3x+1+3x+1=2x+8+x+8+4x+2

Now solve for 'x', combine like terms:

When it comes to terms with variables it's just like normal addition but we keep the variable:

\sf 3x+3x+3x+3x=12x
\sf 2x+x+4x=7x

So we have:

\sf 12x+1+1+1+1=7x+8+8+2

Add:

\sf 12x+4=7x+18

Subtract 7x to both sides:

\sf 5x+4=18

Subtract 4 to both sides:

\sf 5x=14

Divide 5 to both sides:

\boxed{\sf x=2.8}
8 0
3 years ago
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