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Delvig [45]
3 years ago
7

(3 to the power of 2) + (5 − 2) ⋅ 4 − 6 over 3

Mathematics
2 answers:
Oduvanchick [21]3 years ago
4 0
Do parentheses first: 9+7x4-6/3 then multiplication/division: 9+28-2 then addition/subtraction: =35
matrenka [14]3 years ago
3 0

(3 to the power of 2) + (5 − 2) ⋅ 4 − 6 over 3 =19

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Last answer is the correct.
Distance between each hour is the same-13 miles. Good luck.
6 0
3 years ago
Which is the hypothesis of this statement?
zubka84 [21]
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The hypothesis is the part where the word <em>if </em>can be found. So, the hypothesis here has to be C. a and b are negative.
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2 years ago
Question in photo too lazy to solve lol​
Tamiku [17]

The value of 3centimeters to kilometers is 0,00003 km

<h3>Converting centimeters to meters</h3>

In order to convert centimeters to kilometers, we will use the conversion factor below.

100cm = 1m

1000m = 1km

Given 3cm

3cm = 3/100 m

3cm = 0.03m

Convert to km

Since 1000m = 1km

0.03m = 0.03/1000

0.03m = 0.00003km

Therefore the value of 3centimeters to kilometers is 0,00003 km

Learn more on cm to km here: brainly.com/question/1549097

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3 0
1 year ago
Angle 4 above has a measurement of 160 degrees. Angle 1 has a label
Elena-2011 [213]

Answer: A

Step-by-step explanation:

5 0
2 years ago
What are the approximate values of the minimum and maximum points of f(x) = x5 − 10x3 + 9x on [-3,3]?
nika2105 [10]

Answer:

Minimum : -37 at x=2.4 and

Maximum = 37 at x=-2.4.

Step-by-step explanation:

Given:

f(x)=x^5-10x^3+9x; [-3,3]

Explanation:

In order to find minimum/maximum of a function, we need to find the first derivative of the function and then set it equal to 0 to get critical points.

Therefore,

f'(x)=5x^4-30x^2+9

Setting derivative equal to 0, we get

5x^4-30x^2+9=0

On applying quadratic formula, we get

x=2.4, -2.4, -0.7, 0.7.

So, those are critical points of the given function.

Plugging the values x=2.4, -2.4, -0.7, 0.7, -3 and 3 in above function, we get

f(2.4)=(2.4)^5-10(2.4)^3+9(2.4)= -37.01376   : Minimum.

f(-2.4)=(-2.4)^5-10(-2.4)^3+9(-2.4)= 37.01376 : Maximum.

f(0.7)=(0.7)^5-10(0.7)^3+9(0.7) = 3.03807

f(-0.7)=(-0.7)^5-10(-0.7)^3+9(-0.7) = -3.03807

f(-3)=(-3)^5-10(-3)^3+9(-3) =0

f(3)=(3)^5-10(3)^3+9(3) =0

Therefore the approximate values of the minimum and maximum points of f(x) = x^5- 10x^3+ 9x on [-3,3] are:

Minimum : -37 at x=2.4 and

Maximum = 37 at x=-2.4.


7 0
3 years ago
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