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Digiron [165]
3 years ago
6

The number 137.5 is 125% of what number

Mathematics
2 answers:
murzikaleks [220]3 years ago
8 0
125% of x =137,5

125/100 * x= 137.5

5/4 *x=137,5

5x/4= 137,5

x=4*137,5/5

x=4*27,5

x=110
ipn [44]3 years ago
7 0
125% = 1.25

x * 1.25 = 137.5

x = 110

137.5 is 125% of 110
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Dots sells t-shirts ($2) and shorts ($4). In April total sales were $456. People bought 4 times as many t-shirts as shorts. How
Doss [256]

Answer:

Dots sold 152 t-shirts and 38 shorts sold.

Step-by-step explanation:

Let t = t-shirts and s = shorts.

We know that people bought 4 times as many t-shorts as shorts, which can be set to the equation t = 4s

Since t-shirts cost $2, shorts cost $4, and the total sales was $456, we can set the equation 2t +4s = 456

So our two equations are:

t = 4s

2t + 4s = 456

Substitute t from the 1st equation into the 2nd:

2(4s) + 4s = 456

8s + 4s = 456

12s = 456

s = 38

There were 38 shorts sold.

Since we have the number of shorts, we can plug that back into the 1st equation to find the number of t-shirts sold:

t = 4s

t = 4(38)

t = 152

There were 152 t-shirts sold.

4 0
4 years ago
At 7 a.m., the temperature was 48°F. By noon the temperature was 63°F. By how many degrees did the temperature increase?​
vfiekz [6]

Answer: 15 degrees

Step-by-step explanation:

6 0
4 years ago
Describe the steps you can use to find an average of fractional amounts.
oksano4ka [1.4K]
Add up the fractional amounts and divide them by however many amounts there are..........

eg.

Step 1:Add the fractions
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Step 2: Divide the sum by the number of numbers in the set
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5 0
4 years ago
the temperature of a cup of coffee obeys newton's law of cooling. The initial temperature of the coffee is 150F and 1 minute lat
natka813 [3]

Answer:

Newton's law of cooling says that:

T(t) = Tₐ + (T₀ - Tₐ)*e^(k*t)

or:

\frac{dT}{dt} = -k*(T - T_a)

in the differential form.

where:

T is the temperature as a function of time

Tₐ  is the ambient temperature, in this case, 70F

T₀ is the initial temperature of the object, in this case, 150F

k is a constant, and we want to find the value of k.

Then our equation is:

T = 70F + (150F - 70F)*e^(k*t)

Now we also know that after a minute, or 60 seconds, the temperature was 135F

then:

135F = 70F + (150F - 70F)*e^(k*60s)

We can solve this for k:

135F = 70F + 80F*e^(k*60s)

135F - 70F = 80F*e^(k*60s)

65F =  80F*e^(k*60s)

(65/80) = e^(k*60s)

Now we can apply the Ln(x) function to both sides to get:

Ln(65/80) = Ln(e^(k*60s))

Ln(65/80) = k*60s

Ln(65/80)/60s = k = -0.0035 s^-1

Then the differential equation is:

\frac{dT}{dt} = -0.0035 s^-1*(T - 70F)

8 0
3 years ago
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