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Sliva [168]
3 years ago
8

2⋅(36⋅3−5) . evaluate

Mathematics
2 answers:
Rasek [7]3 years ago
8 0

Answer:

206

2 \times  |108 - 5|=\\ 2 \times 103  = \\ 206

Answer: 206

lesya [120]3 years ago
6 0

Answer:

Answer is = 6

Step-by-step explanation:

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The difference of 3 times a number and 15 is no less than the square of the number.<br><br>​
Citrus2011 [14]

Answer:

Answer: 3x-15 greater than/= x^2

Step-by-step explanation:

6 0
4 years ago
Here is for 21 points
Andre45 [30]

Answer:

18

Step-by-step explanation:

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6 0
3 years ago
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What would the second point be with point (-2,-5) to get a line with the slope of -1
choli [55]

Answer:

A second point would be (-1, -6)

Step-by-step explanation:

Because the slope is -1, we know that for every number that x increases, we change y the amount of the slope. So, if we increase x by 1, we change y by -1.

x value

-2 + 1 = -1

y value

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3 0
3 years ago
A bottle rocket is shot off a bridge into the stream below. The height of the rocket (in feet) above
nalin [4]

The correct model of the height of rocket above water is;

h(t) = -16t² + 96t + 112

Answer:

time to reach max height = 3 seconds

h_max = 256 ft

Time to hit the water = 7 seconds

Step-by-step explanation:

We are given height of water above rocket;

h(t) = -16t² + 96t + 112

From labeling quadratic equations, we know that from the equation given, we have;

a = -16 and b = 96 and c = 112

To find the time to reach maximum height, we will use the vertex formula which is; -b/2a

t_max = -96/(2 × -16)

t_max = 3 seconds

Thus, maximum height will be at t = 3 secs

Thus;

h_max = h(3) = -16(3)² + 96(3) + 112

h_max = -144 + 288 + 112

h_max = 256 ft

Time for it to hit the water means that height is zero.

Thus;

-16t² + 96t + 112 = 0

From online quadratic formula, we have;

t = 7 seconds

7 0
3 years ago
If f(x)=x-6 and g(x)=x^3, find G(f(x)) a)x^3(x-7) b)x^3-7 c)x^3+x-7 d)(x-7)^3
nirvana33 [79]

Answer:

(x-6)^3

Step-by-step explanation:

f(x)=x-6 \\\\g(x)=x^3 \\\\g(f(x))=(x-6)^3

Hope this helps!

4 0
3 years ago
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