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alexira [117]
3 years ago
8

I need lots of help with 15 18 19 20 pls pls pls

Mathematics
1 answer:
Katena32 [7]3 years ago
5 0

Answer:

15) x = 112

18) x = 120

19) x = 31

20) x = 65

Step-by-step explanation:

I hope that is useful for you :)

You might be interested in
45.6/109.2 = x/115<br><br> Is x, 48.02?
MAXImum [283]
Yes correct x is 48.02 here is why.

simplify 45.6/109.2 to 0.417582

multiply both sides by 115

simplify 0.417582 x 115 to 48.021978

switch sides

Answer: x = 48.021978
4 0
3 years ago
a storage room measures 20 feet by 15 feet. the floor is covered by tiles that cost $7.50 per square foot. what will be the cost
Sidana [21]

Answer: $2250

Step-by-step explanation:

<u>Given information</u>

Dimension of the room = 20 feet × 15 feet

Cost of tiles = $7.50 / ft²

<u>Derived formula from the given information</u>

Total cost = Floor area × Cost of tiles

<u>Substitute values into the formula</u>

Total cost = Floor area × Cost of tiles

Total cost = (20 × 15) × (7.5)

<u>Simplify by multiplication</u>

Total cost = 300 × (7.5)

\Large\boxed{Total~cost~=~2250~Dollars}

Hope this helps!! :)

Please let me know if you have any questions

4 0
2 years ago
Answer choices:
Mamont248 [21]

Answer:

(x+8,y-6)

Step-by-step explanation:

it's the number of values that x and y differ

7 0
3 years ago
Solve the system by substitution.<br> y = -3x +4<br> y = 2
erma4kov [3.2K]

Answer:

(2/3,2)

Step-by-step explanation:

3*2/3+4

simplify  2=2

5 0
2 years ago
the value of c guaranteed to exist by the Mean Value Theorem for V(x) = x² in the interval [0, 3] is...? a) 1 b) 2 c) 3/2 d) 1/2
lilavasa [31]

Answer:  c) \dfrac{3}{2} .

Step-by-step explanation:

Mean value theorem : If f(x) is defined and continuous on the interval [a,b] and differentiable on (a,b), then there is at least one number c in the interval (a,b) (that is a < c < b) such that

\begin{displaymath}f'(c) = \frac{f(b) - f(a)}{b-a} \cdot\end{displaymath}

Given function : f(x) = x^2

Interval : [0,3]

Then, by the mean value theorem, there is at least one number c in the interval (0,3) such that

f'(c)=\dfrac{f(3)-f(0)}{3-0}\\\\=\dfrac{3^2-0^2}{3}=\dfrac{9}{3}\\\\=3

\Rightarrow\ f'(c)=3\ \ \ ...(i)

Since f'(x)=2x

then, at x=c, f'(c)=2c\ \ \ ...(ii)

From (i) and (ii), we have

2c=3\\\\\Rightarrow\ c=\dfrac{3}{2}

Hence, the correct option is c) \dfrac{3}{2} .

4 0
3 years ago
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