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sineoko [7]
3 years ago
15

Does 9, 12, 15 belong to a right traingle?

Mathematics
1 answer:
Leokris [45]3 years ago
7 0
Is there a picture? or something

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Can anyone help me??
Tasya [4]

Your in luck! I was just learning these percent things today 'xD but Not in luck at the same time. I will answer the 2 questions that's the most I can do.

1. $17.50 with a 7% sales tax

Answer: 17.50 times .07 equals 1.23  

              17.50 plus 1.23 equals 18.73


              The answer is $18.73

2. 21.95 with a 4.25% sales tax

Answer: 21.95+4.25=26.20


Hope this helps!! :D this took me a while to figure out and used a bit of my time. So I hope the time was worth it! :)

5 0
3 years ago
- 40 = -5r (algebra)
Gekata [30.6K]

Answer:

r=8

hope this helps

have a good day :)

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Use quadratic regression to find
vovangra [49]

Answer:

y = 6x² -13x -5

Step-by-step explanation:

(1,-12) (2,-7) (5,80)

y = ax² + bx + c

-12 = a*1² + b*1 + c = a+b+c    (1)

-7 = 4a + 2b + c    (2)

80 = 25a + 5b + c   (3)

(2)-(1): 3a + b = 5    (4)

(3)-(2): 21a + 3b = 87   (5)

(4)*3:     9a + 3b = 15    (6)

(5)-(6):  12a = 72      a = 6

(4)                             b = -13

(2)                             c = -5

4 0
3 years ago
At a point on the ground 46 feet from the foot of a tree, the angle of elevation of the top of the tree is 68 degrees. What is t
Mandarinka [93]

Answer:

114 ft

Step-by-step explanation:

Imagine or construct a right triangle with the 46 ft leg lying on the ground.  This is the "adjacent side" of the triangle; it lies immediately adjacent to the 68 degree angle.  The side opposite this angle is h, the height of the tree.

The tangent function includes angle, opp side and adj side:

tan 68 degrees = opp / adj = h / (46 ft), and so:

(46 ft)*tan (68 degrees) = opp = h

Then the height of the tree is h = (46 ft)(2.47) = 114 ft

6 0
3 years ago
Given the position of the particle, what the position(s) of the particle when it’s at rest
choli [55]

The position function of a particle is given by:

X\mleft(t\mright)=\frac{2}{3}t^3-\frac{9}{2}t^2-18t

The velocity function is the derivative of the position:

\begin{gathered} V(t)=\frac{2}{3}(3t^2)-\frac{9}{2}(2t)-18 \\ \text{Simplifying:} \\ V(t)=2t^2-9t-18 \end{gathered}

The particle will be at rest when the velocity is 0, thus we solve the equation:

2t^2-9t-18=0

The coefficients of this equation are: a = 2, b = -9, c = -18

Solve by using the formula:

t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}

Substituting:

\begin{gathered} t=\frac{9\pm\sqrt[]{81-4(2)(-18)}}{2(2)} \\ t=\frac{9\pm\sqrt[]{81+144}}{4} \\ t=\frac{9\pm\sqrt[]{225}}{4} \\ t=\frac{9\pm15}{4} \end{gathered}

We have two possible answers:

\begin{gathered} t=\frac{9+15}{4}=6 \\ t=\frac{9-15}{4}=-\frac{3}{2} \end{gathered}

We only accept the positive answer because the time cannot be negative.

Now calculate the position for t = 6:

undefined

6 0
1 year ago
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