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blagie [28]
3 years ago
14

Nicole has 15 nickels and dimes. If the value of her coins is 1.20, how many of each coin does she have?

Mathematics
2 answers:
otez555 [7]3 years ago
6 0

Answer:

9 Dimes & 6 Nickels

Step-by-step explanation:

9 × 10 = 90 cents

6 × 5 = 30 cents

90 + 30 = 120 which equals $1.20

Dmitry [639]3 years ago
5 0
She has 9 dimes and 6 nickels making up that value.
9*10=90
6*5=30
90+30=120
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Find the perimeter of a quadrilateral with vertices at R (-2, 1), S (-5, 5), T (2, 5), U (5, 1). Round your answer to the neares
Nezavi [6.7K]
Check the picture below.

now, you can pretty much count the units off the grid for the segments ST and RU, so each is 7 units long, and are parallel, meaning that the other two segments are also parallel, and therefore the same length each.

so we can just find the length for hmmmm say SR, since SR = TU, TU is the same length,

\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
S(\stackrel{x_1}{-2}~,~\stackrel{y_1}{1})\qquad 
R(\stackrel{x_2}{-5}~,~\stackrel{y_2}{5})\qquad \qquad 
%  distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
SR=\sqrt{[-5-(-2)]^2+[5-1]^2}\implies SR=\sqrt{(-5+2)^2+(5-1)^2}
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SR=\sqrt{(-3)^2+4^2}\implies SR=\sqrt{25}\implies SR=5

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6 0
3 years ago
Find the roots of h(t) = (139kt)^2 − 69t + 80
Sonbull [250]

Answer:

The positive value of k will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80, roots are those values of t so that h(t) = 0. That is:

19321\cdot k^{2}\cdot t^{2}-69\cdot t + 80=0 (1)

Roots are determined analytically by the Quadratic Formula:

t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}

t = \frac{69}{38642} \pm \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }

The smaller root is t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }, and the larger root is t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }.

h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80 has one real root when \frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0. Then, we solve the discriminant for k:

\frac{80\cdot k^{2}}{19321} = \frac{4761}{1493204164}

k \approx \pm 0.028

The positive value of k will result in exactly one real root is approximately 0.028.

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2 years ago
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Luba_88 [7]
Test contains 10 three-point questions and 14 five-point questions.
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