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Ostrovityanka [42]
3 years ago
13

Solve T = 7c-2d for d.

Mathematics
1 answer:
stepladder [879]3 years ago
4 0

Answer:

d =  \frac{t - 7c}{ - 2}

Hope this helps at all. :)

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The sum of one-half of d and one-third of p, minus 8.<br><br> Translate as algebraic expression.
Alinara [238K]

Answer:

1\frac{1}{2} + 1\frac{1}{3}p - 8

Step-by-step explanation:

The sum of one-half of d and one-third of p, minus 8;

i.e

1\frac{1}{2} + 1\frac{1}{3}p - 8

Putting it into improper fraction becomes:

\frac{3}{2}d + \frac{4}{3}p - 8

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3 years ago
Adult male heights are normally distributed with a mean of 70 inches and a standard deviation of 3 inches. The average basketbal
Marizza181 [45]
Z = (x - mean) / SD = (79 - 70) / 3 = 3 
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4) You can buy 3 apples at the Quick Market for $1.20. You can buy 5 of the
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Answer: Stop and Save is better

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you can get 3 apple at Quick market for $1.20, but you can buy 5 at Stop and Save for only 5 cents more, plus your getting two more apples.

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3 years ago
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Complete the square to solve the equation below
frutty [35]
x^2+x= \frac{7}{4}

Take one-half of the coefficient of x and square it, then add it to both sides.

x^2+x+ (\frac{1}{4}) =\frac{7}{4}+ (\frac{1}{4})

x^2+x+(\frac{1}{4})= 2

\left(x+\frac{1}{2}\right)^2=2

x+\frac{1}{2}=\pm \sqrt{2}
-----------------------------------------------------------------
x+\frac{1}{2}=- \sqrt{2}
x=-\frac{1}{2}- \sqrt{2} <=

AND

x+\frac{1}{2}= \sqrt{2}
x= \sqrt{2}-\frac{1}{2} <=
5 0
3 years ago
Can someone please help me with this!?
Yuki888 [10]

Answer:

Applying cosine theorem:

AB = sqrt(AC^2 + BC^2 - 2 x AB x BC x cosC)

     = sqrt( 72^2 + 135^2 - 2 x 72 x 135 x cos(119deg))

     = 181

Applying sine theorem:

BC/sinA = AB/sinC

=> sinA = BC x sinC/AB

=> A = arcsin(BC x sinC/AB)

        = arcsin( 135 x sin(119deg)/181)

        = 40.7 deg

Hope this helps!

:)

7 0
4 years ago
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