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Angelina_Jolie [31]
3 years ago
12

PLEASE HELP!!!!!!!!!!!!

Mathematics
1 answer:
katovenus [111]3 years ago
5 0
Let me ask my brother he's smart
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A company needs to purchase several new machines to meet its future production needs. It can purchase three different types of m
katovenus [111]

Answer:657

Step-by-step explanation:

3 0
3 years ago
25 POINTS AND BRAINLIEST IF ANSWERED CORRECTLY!
vesna_86 [32]
X=-1 hope this helped!
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3 years ago
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When 2(3/5 x + 2/3 y - 1/4 x -1 1/2 y + 3 )is simplified, what is the resulting expression?
vovangra [49]

Answer:

Simplifying  the expression: 2(\frac{3}{5}x+\frac{2}{3}y-\frac{1}{4}x-1\frac{1}{2}y +3) we get \mathbf{42x-100y-360}

Step-by-step explanation:

We need to simplify the expression: 2(\frac{3}{5}x+\frac{2}{3}y-\frac{1}{4}x-1\frac{1}{2}y +3)

First we will solve terms inside the bracket

2(\frac{3}{5}x+\frac{2}{3}y-\frac{1}{4}x-1\frac{1}{2}y +3)

Converting mixed fraction 1\frac{1}{2} y into improper fraction, we get: \frac{3}{2}y

Replacing the term:

2(\frac{3}{5}x+\frac{2}{3}y-\frac{1}{4}x-\frac{3}{2}y +3)

Now, taking LCM of: 5,3,4,2 we get 60

Now multiply 60 with each term inside the bracket

2(\frac{3}{5}x\times60+\frac{2}{3}y\times60-\frac{1}{4}x\times60-\frac{3}{2}y\times60 +3\times60)\\2(3x\times12+2y\times20-1x\times15-3x\times30-3\times60)\\2(36x+40y-15x-90y-180)

Now, combine like terms

2(36x-15x-90y+40y-180)\\2(21x-50y-180)

Now, multiply all terms with 2

42x-100y-360

So, Simplifying  the expression: 2(\frac{3}{5}x+\frac{2}{3}y-\frac{1}{4}x-1\frac{1}{2}y +3) we get \mathbf{42x-100y-360}

3 0
3 years ago
7-9: Practice Buddy: Independent Practice; Problem Solving please help!
strojnjashka [21]

Answer:

4/3/8

Step-by-step explanation:

12/1/2-8/8/1

=25/2-65/8

=100-65/8

=35/8

=4/3/8

3 0
3 years ago
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Which function is undefined for x = 0? y=3√x-2 y=√x-2 y=3√x+2 y=√x=2
Andre45 [30]

For this case, we must indicate which of the given functions is not defined forx = 0

By definition, we know that:

f (x) = \sqrt {x} has a domain from 0 to infinity.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. For it to be defined, the term within the root must be positive.

Thus, we observe that:

y = \sqrt {x-2} is not defined, the term inside the root is negative when x = 0.

While y = \sqrt {x + 2} if it is defined for x = 0.

f(x)=\sqrt[3]{x}, your domain is given by all real numbers.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. In the same way, its domain will be given by the real numbers, independently of the sign of the term inside the root.

So, we have:

y = \sqrt [3] {x-2} with x = 0: y = \sqrt [3] {- 2} is defined.

y = \sqrt [3] {x + 2}with x = 0: y = \sqrt [3] {2}in the same way is defined.

Answer:

y = \sqrt {x-2}

Option b


6 0
3 years ago
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