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Doss [256]
3 years ago
15

PLEASE HELP ASAP

Mathematics
1 answer:
zaharov [31]3 years ago
4 0
Wouldn’t it be 0, as that’s when their values are both equal to each other, making the equation true?
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Before a basketball game a referee noticed that the ball seemed under inflated. She dropped it from 6 feet and measured the firs
deff fn [24]

Answer: a) H = h( 0.5 )^n

b) H = 1.125inches

Step-by-step explanation:

Let H = height of the ball

n = number of time the ball bounces

h = initial height.

The exponential function to model the height of the ball will be:

H = h( 1 - 0.5)^n

H = h( 0.5 )^n

It's minus because the height of the ball is decreasing.

h = 36 inches

n = 5

H = 36( 1 - 0.5 ) ^5

H = 36( 0.5 )^5

H = 36 × 0.03125

H = 1.125inches

7 0
3 years ago
What is absolute value?
pochemuha

Answer:

Absolute value means the distance between a number and 0.

Examples:

The absolute value of -7 is 7.

The absolute value of 5 is 5.

Key Concepts:

The absolute value sign is |x|, where x is any real number.

The absolute value removes any negative sign in front of a number.

8 0
3 years ago
Read 2 more answers
Can someone please help?? what’s the measure of the vertex angle
maw [93]

Answer:

120 deg

Step-by-step explanation:

There is an angle symbol in the equilateral triangle and another symbol in half of the vertex angle. Those two angles are congruent.

The measure of the vertex angle is 2 * 60 deg = 120 deg

Answer: 120 deg

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
T-7=19 what is the answer and how do you do it
jeka57 [31]
T=26
you just add 7 to the other side with 19 which gives you the answer
7 0
3 years ago
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