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notka56 [123]
3 years ago
5

Using the graph below, identify 2 solutions to the inequality y<-3/2x+2

Mathematics
1 answer:
lora16 [44]3 years ago
3 0
Any point in that shaded area  is a solution
(0,0) is a solution
(-1,-1) is a solution
(-3,3) is a solution
etc
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What is the equation of a line parallel to the x-axis that passes though the point (4,5)? What is the slope of this line
nignag [31]

Answer:

y = 5

slope = 0

Step-by-step explanation:

5 0
3 years ago
What does -3x + 7 + -6x t 9 equal
liberstina [14]

Answer:

-50x

Step-by-step explanation:

-6x times 9 which is -54x

-54x-3x = -57x

-54x+7

8 0
3 years ago
3(x-2)+2(x+1)=-14<br> plz show work
Rom4ik [11]
Hello!

First, let's write the problem.
3\left(x-2\right)+2\left(x+1\right)=-14
Apply the distributive property on the left side of the equation.
=3x-6+2x+2
Add like terms.
=5x-4
Let's plug that in into the original equation.
5x-4=-14
Add 4 to both sides.
5x-4+4=-14+4
5x=-10
Divide both sides by 5.
\frac{5x}{5}=\frac{-10}{5}

Our final answer would be,
x=-2

You can feel free to let me know if you have any questions regarding this!
Thanks!

- TetraFish
5 0
3 years ago
Read 2 more answers
(Congruent polygons)
olasank [31]
1. is line FG
2. is line IH
3. is angle J
4. is angle H
5. is 12 cm.
6. is 4 cm.
7. is 135
8. is 150
9. sorry can't help you on this one.
10. is 150+90+90=330
7 0
3 years ago
How many pounds of a 15% copper alloy must be mixed with 700lb of a 30% copper alloy to maybe a 25.5% copper alloy
lisov135 [29]

Answer:

300\; \rm lb.

Step-by-step explanation:

Let x represent the mass (in pounds) of that 15\% copper alloy required, such that the final mixture would contain 25.5\% copper by mass.

Consider: if x pounds of that 15\% copper alloy is mixed with 700 pounds that 30\% copper alloy, what would be the mass of copper in the mixture?

  • Mass of copper in x pounds of that 15\% copper alloy: (0.15\, x)\; \rm lb.
  • Mass of copper in 700 pounds of that 30\% copper alloy: 700 \times 0.30 = 210\; \rm lb.

Therefore, the mixture would contain (210 + 0.15\, x) \; \rm lb of copper.

The mass of that mixture would be (700 + x)\; \rm lb. The mass fraction of copper in that mixture would be:

\displaystyle \frac{(210 + 0.15\, x)\; \rm lb}{(700 + x)\; \rm lb} \times 100\%.

This ratio is supposed to be equal to 25.5\%. These two pieces of equations combine to give an equation about x:

\displaystyle \frac{(210 + 0.15\, x)\; \rm lb}{(700 + x)\; \rm lb} \times 100\% = 25.5\%.

\displaystyle \frac{210 + 0.15\, x}{700 + x} = 0.255.

Simplify and solve for x:

210 + 0.15\, x= 0.255\, (700 + x).

(0.255 - 0.15)\, x= 210 - 0.255 \times 700.

\displaystyle x = \frac{210 - 0.255 \times 700}{0.255 - 0.15} = 300.

Therefore, 300\; \rm lb of that 15\% alloy would be required.

4 0
3 years ago
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