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kenny6666 [7]
3 years ago
11

Solve for x. Round to the nearest tenth, if necessary.

Mathematics
1 answer:
Illusion [34]3 years ago
7 0
Cos(o)= ads
CoS61X
5.5=* (Cos (60)
11.344
-
X= 5.5

COSC61)
The answer is in the picture
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The length of a shadow of a building is
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H^2=x^2+y^2

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y^2=35^2-29^2

y^2=384

y=√384

y≈19.60 m  (to the nearest hundredth of a meter, or nearest centimeter)

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Justin weighs 15 pounds less than Greg weighs. Half of Greg's weight is 75 pounds less than Justin's weight. how much does each
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Greg is 180 and justin is 165
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5.074 rounded to the nearest hundredth is
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Read 2 more answers
Find f(x)=x^2-x^3,find f(-10)
Elanso [62]

Answer:

Value of f(-10) is 1100

Step-by-step explanation:

Given that f(x) = x² - x³

Here we need to find f(-10).

 f(-10) =  (-10)² - (-10)³

 f(-10) = 100 - (-1000)

 f(-10) = 100 + 1000

 f(-10) = 1100

Value of f(-10) is 1100

3 0
2 years ago
find three consecutive odd integers such that the sum of the smallest number and middle number is 27 less than 3 times the large
katrin2010 [14]

A generic odd number can be written as

2k+1,\quad k \in \mathbb{Z}

Since there is an odd number every two numbers, three consecutive odd numbers will be

2k+1,\quad 2k+3,\quad 2k+5

Now let's make up the equations: the sum of the first two is

(2k+1)+(2k+3)

And 27 less than 3 times the largest is

3(2k+5)-27

These two must be the same, so we have

(2k+1)+(2k+3)=3(2k+5)-27 \iff 4k+4 = 6k+30-27 \iff 4k+4=6k+3

Subtracting 4k and 3 from both sides gives

1=2k \iff k=\dfrac{1}{2}

Which means that the problem has no solution.

To confirm this hypothesis, we can observe that, on the left hand side, we have the sum of two odd numbers, which is even

On the right hand side, we have an odd number, multiplied by 3 (still odd), take away 27 (still odd).

So, the left hand side is even, and the right hand side is odd. They can't be the same number.

7 0
4 years ago
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