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Aleks04 [339]
2 years ago
12

Exam help !!

Mathematics
1 answer:
omeli [17]2 years ago
3 0
M= 1/3
Remember m is the slope so to find the slope use (-1,-2) (-7,-4)
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If the segment OP is dilated by a scale factor r=2, what is the length of segment OP?
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The answer is 4 I think
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Part A: Identify the point that is an x-intercept in order from left to right (1st)
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3 years ago
Help me pls its due soon
RoseWind [281]

Answer:

do you mean in simplist form?? if so...

Step-by-step explanation:

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7 0
3 years ago
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Graph each circle given below right the center and radius of each circle
expeople1 [14]

Answer:

Center: (-3,-2)

Radius: √6

The graph is attached.

Step-by-step explanation:

The equation of the circle has the form:

(x -h)^{2}+(y-k)^{2}=r^{2}

Where (h,k) is the  point of the center of the circle and r is the radius of the circle.

The equation given in the problem is

(x +3)^{2}+(y+2)^{2}=6

 Therefore:

h=-3

k=-2

The center is at (-3,-2)

And the radius is:

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Then, you can graph it has you can see in the image attached.

5 0
3 years ago
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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
2 years ago
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