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algol13
3 years ago
7

Find the value of each variable x= y=

Mathematics
1 answer:
mamaluj [8]3 years ago
4 0
Can’t see it sorry jk the answer is x=567 y=766
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1. slope need help ASAP
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The slope is 5/1 and the y-intercept is -3
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3 years ago
The child’s physical density is being measured by the displacement method. A child of 50 pounds is placed in a tub filled
Mariana [72]

ANSWER

1.05 g/cm³

EXPLANTION

this is a multi step process and is really hard to type on the computer but this is the answer.

6 0
3 years ago
Please Help <br> Write a system of linear Inequalities to represent each graph.
docker41 [41]

Answer:

\left\{\begin{array}{l}5x-y-7\le 0\\2x+3y-9\ge 0\end{array}\right.

Step-by-step explanation:

1. Blue region. The boundary blue line passes through the points (0,-7) and (3,8), then its equation is

\dfrac{x-0}{3-0}=\dfrac{y+7}{8+7},\\ \\15x=3(y+7),\\ \\5x=y+7,\\ \\5x-y-7=0.

This line is solid, so the sign for the inequality should be with the notion "or equal to".

From the diagram you can see that the origin belongs to the blue region, then its coordinates satisfy the inequality. Thus,

5x-y-7\le 0.

2. Green region. The boundary green line passes through the points (6,-1) and (0,3), then its equation is

\dfrac{x-0}{6-0}=\dfrac{y-3}{-1-3},\\ \\-4x=6(y-3),\\ \\-2x=3y-9,\\ \\2x+3y-9=0.

This line is solid, so the sign for the inequality also should be with the notion "or equal to".

From the diagram you can see that the origin doesn't belong to the blue region, then its coordinates satisfy the inequality. Thus,

2x+3y-9\ge 0.

3. The system of linear inequalities that represents these graphs is

\left\{\begin{array}{l}5x-y-7\le 0\\2x+3y-9\ge 0\end{array}\right.

6 0
3 years ago
Apples are on sale 5 dor 3.15. At this rate, what is the cost of one apple?
soldi70 [24.7K]

63 cents per apple because 3.15 divided by 5 is 0.63

hope this helps

3 0
3 years ago
Read 2 more answers
(-2ab^-5)(4a^-2b)^-2
enyata [817]

Answer:

  -a^5/(8b^7)

Step-by-step explanation:

The applicable rules of exponents are ...

  (a^b)(a^c) = a^(b+c)

  (a^b)^c = a^(bc)

  a^-b = 1/a^b

__

  (-2ab^{-5})(4a^{-2}b)^{-2}=(-2ab^{-5})(4^{-2}a^{-2(-2)}b^{-2})\\\\=\dfrac{-2}{16}a^{1+4}b^{-5-2}=\boxed{-\dfrac{a^5}{8b^7}}

4 0
3 years ago
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