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3 years ago

I don't understand how to begin this problem

Mathematics

1 answer:

Ronch [10]3 years ago

7 0

I would first add 5 and 2 together, then multiply that by 3. Finally you would subtract the 1. After solving for what is inside the parentheses you would multiply it by 2 for your final answer.

2•(3(5+2)-1)
2•(3(7)-1)
2•(21-1)
2•(20)
40

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A party mix is made by adding nuts that sell for $2.50 per kg to a cereal mixture that sells for $1 per kg. How much of each sho

sesenic [268]

Answer:

The amount of nuts is 16 kg and cereal mixture is 44 kg to obtain 60 kg of a mix that will sell for $1.40 per kg

Step-by-step explanation:

Let

x=nuts at $2.50 per kg

y=cereal at $1 per kg

x+y=60 kg (1)

2.50x+y=1.40(60) (2)

From (1)

x=60-y

Substitute x=60-y into (2)

2.50x+y=1.40(60)

2.50(60-y)+y=84

150-2.5y+y=84

-1.5y=84-150

-1.5y=-66

Divide both sides by -1.5

y=44 kg

Substitute y=44 into (1)

x+y=60

x+44=60

x=60-44

=16

x=16 kg

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3 years ago

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Advocard [28]

U need to attach an image

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2 years ago

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lesantik [10]

Answer:

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For function of g,g(0)=1 and g(1)=5 assuming g is a linear function what is g(3)<br>

lawyer [7]

Answer:

$g(3) = 13$

Step-by-step explanation:

Given

$g(0) = 1$

$g(1) = 5$

Required

Find g(3)

Since g(x) is a linear function, then:

$g(x) = mx + c$

For: $g(0) = 1$

$g(x) = mx + c$

$g(0) = m*0 + c$

$1 = 0 + c$

$1 = c$

$c = 1$

For: $g(1) = 5$

$g(x) = mx + c$

$g(1) = m * 1 + c$

$5 = m * 1 + c$

$5 = m + c$

Substitute $c = 1$

$5 = m + 1$

$m = 5 - 1$

$m =4$

So, the function is:

$g(x) = mx + c$

$g(x) = 4x + 1$

Calculate g(3)

$g(3) = 4*3 + 1$

$g(3) = 12 + 1$

$g(3) = 13$

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3 years ago

Which of the following describes how to translate the graph y = |x| to obtain the graph of y = |x - 6|?. . . shift 6 units up. s

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Answer:

Shift 6 units right.

Step-by-step explanation:

Easiest way to answer this question is to plot the two graphs:

Let's plot these graphs and analyse each graph.

Lets take the graph of $y=|x|$ that is attached.

We can clearly see that graph intersects x axis at x=0 and then essentially looks like y=x is mirrored by the y axis.

Now when we look at the $y=|x-6|$ that is attached,

We can see that graph intersects x axis at x=6 and is a composite graph of $y=|x-6|$ and its mirrored graph by y=6.

We can clearly see that graph has moved 6 units to the right.

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