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Assoli18 [71]
3 years ago
8

Find three positive numbers that sum to 150 with the largest possible product of the three. (enter your answers as a comma-separ

ated list.)
Mathematics
1 answer:
liq [111]3 years ago
3 0
Largest possible number = 150 ÷ 3 = 50

Sum = 50 + 50 + 50 = 150
Product = 50 x 50 x 50 = 125000

Answer: The three numbers are 50, 50 and 50.
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Solve:<br>.<br>.<br>.<br><br>PLS DON SPAM
MrRissso [65]
4
)
10(x+6)+8(x−3)=5(5x−4)

2 Expand.
10
x
+
60
+
8
x
−
24
=
25
x
−
20
10x+60+8x−24=25x−20

3 Simplify
10
x
+
60
+
8
x
−
24
10x+60+8x−24 to
18
x
+
36
18x+36.
18
x
+
36
=
25
x
−
20
18x+36=25x−20

4 Subtract
18
x
18x from both sides.
36
=
25
x
−
20
−
18
x
36=25x−20−18x

5 Simplify
25
x
−
20
−
18
x
25x−20−18x to
7
x
−
20
7x−20.
36
=
7
x
−
20
36=7x−20

6 Add
20
20 to both sides.
36
+
20
=
7
x
36+20=7x

7 Simplify
36
+
20
36+20 to
56
56.
56
=
7
x
56=7x

8 Divide both sides by
7
7.
56
7
=
x
7
56
​
=x

9 Simplify
56
7
7
56
​
to
8
8.
8
=
x
8=x

10 Switch sides.
x
=
8
x=8


Your answer is

X=8
7 0
3 years ago
Which terms of the AP: 213, 207, 201… is its last positive term?
riadik2000 [5.3K]

Answer: the last pos term is 3

Details:

213

207

201

195

189

183

177

171

165

159

153

147

141

135

129

123

117

111

105

99

93

87

81

75

69

63

57

51

45

39

33

27

21

15

9

3

-3


5 0
3 years ago
Suppose h(x)=3x-2 and j(x) = ax +b. Find a relationship between a and b such that h(j(x)) = j(h(x))
sergejj [24]

Answer:

\displaystyle a = \frac{1}{3} \text{ and } b = \frac{2}{3}

Step-by-step explanation:

We can use the definition of inverse functions. Recall that if two functions, <em>f</em> and <em>g</em> are inverses, then:

\displaystyle f(g(x)) = g(f(x)) = x

So, we can let <em>j</em> be the inverse function of <em>h</em>.

Function <em>h</em> is given by:

\displaystyle h(x) = y = 3x-2

Find its inverse. Flip variables:

x = 3y - 2

Solve for <em>y. </em>Add:

\displaystyle x + 2 = 3y

Hence:

\displaystyle h^{-1}(x) = j(x) = \frac{x+2}{3} = \frac{1}{3} x + \frac{2}{3}

Therefore, <em>a</em> = 1/3 and <em>b</em> = 2/3.

We can verify our solution:

\displaystyle \begin{aligned} h(j(x)) &= h\left( \frac{1}{3} x + \frac{2}{3}\right) \\ \\ &= 3\left(\frac{1}{3}x + \frac{2}{3}\right) -2 \\ \\ &= (x + 2) -2 \\ \\ &= x \end{aligned}

And:

\displaystyle \begin{aligned} j(h(x)) &= j\left(3x-2\right) \\ \\ &= \frac{1}{3}\left( 3x-2\right)+\frac{2}{3} \\ \\ &=\left( x- \frac{2}{3}\right) + \frac{2}{3} \\ \\ &= x  \stackrel{\checkmark}{=} x\end{aligned}

3 0
2 years ago
Triangle ABC and triangle PQR each have two sides whose lengths are 7 and an angle whose measure is 40. Are the triangles congru
Elena L [17]

No

Step-by-step explanation:

they could be but it doesnt say where the angle is so there isn't enough information to tell if they are congruent. answer is no

8 0
3 years ago
Find the vertex: -4x² + 16x - 7​
guapka [62]

Find the vertex: -4x^2 + 16x - 7​

Vertex = ( x, f(x)).

x = -b/2a

x = -16/2(-4)

x = -16/-8

x = 2

f(x) = -4x^2 + 16x - 7

Let x = 2

f(2) = -4(2)^2 + 16(2) - 7

f(2) = -4(4) + 32 - 7

f(2) = -16 + 32 - 7

f(2) = 16 - 7

f(2) = 9

Vertex = (2, 9)

7 0
3 years ago
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