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aivan3 [116]
3 years ago
12

Please answer and fast ​

Mathematics
1 answer:
valentinak56 [21]3 years ago
3 0
18- H
19-C

Explanation:
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PLEASE HELP ASAP, GIVING BRAINLIEST
Svet_ta [14]

Answer:

(-6, 11)

(3, 5)

(15, -3)

Step-by-step explanation:

i just used desmos to help me get the answer :) have a nice day

4 0
3 years ago
Sin6x-sin4x<br><br> Express the given sum or difference as a product of sines or cosines.
myrzilka [38]
It will be 24 because 6 times 4 is 24.

7 0
4 years ago
A Construction contractor used the equation 11.52 = (1.44)8 to calculate how much 8 boxes of nails would cost him . How much wou
k0ka [10]

Answer:

The 2 boxes of nails will cost him $ 2.88.

Step-by-step explanation:

Cost of a given substance is the amount of money needed to acquire it. <u>The total cost involved in the purchase of a given number of units is equal to the cost of one unit multiplied by the total number of units. </u>

Total cost = Cost of 1 unit × Total number of units          ....equation 1

Given: Boxes of nails to be bought = 8

Price of 2 boxes of nails = ?

Assuming the money to be in dollars ($).

The given equation used to calculate the cost of 8 units or boxes of nails:

11.52 = (1.44)8                                                                        ....equation 2

By comparing equations (1) and (2), we get

The cost of 1 box of nails = $ 1.44

and the total cost of 8 boxes of nails = $ 11.52

∴ <u>The total cost of 2 boxes of nails</u> = cost of 1 box of nails × 2 = $ 1.44 × 2 =  $ 2.88

<u>Therefore, the 2 boxes of nails will cost him $ 2.88.</u>

4 0
3 years ago
Fggggggggggggggggggggb
zavuch27 [327]

Answer:

herbert can translate that for you

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
If the ball leaves her hand 6 feet above the ground at 65 feet/second: a) what is the maximum height of the ball
FinnZ [79.3K]

~~~~~~\textit{initial velocity in feet} \\\\ h(t) = -16t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}&65\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&6\\ \qquad \textit{of the object}\\ h=\textit{object's height}&\\ \qquad \textit{at "t" seconds} \end{cases} \\\\\\ h(t)=-16t^2+65t+6

Check the picture below, the maximum occurs at the vertex of the parabolic path, so

\textit{vertex of a vertical parabola, using coefficients} \\\\ h(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+65}t\stackrel{\stackrel{c}{\downarrow }}{+6} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)

\left(-\cfrac{ 65}{2(-16)}~~~~ ,~~~~ 6-\cfrac{ (65)^2}{4(-16)}\right)\implies \left(\cfrac{65}{32}~~,~~6+\cfrac{4225}{32} \right) \\\\\\ \left(\cfrac{65}{32}~~,~~\cfrac{4609}{32} \right) \implies \stackrel{maximum~height}{(2.03125~~,~~\stackrel{\downarrow }{72.015625})}

6 0
2 years ago
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