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2 years ago

12) You can exchange $5 US dollars for L123 Honduran Lempiras. How many dollars can you exchange L2,300 for? ----

Mathematics

1 answer:

Anit [1.1K]2 years ago

6 0

Step-by-step explanation:

$5 US dollar = L123 Honduran Lempiras

L2300Honduran Lempiras = x ( Let $ be x)

By using chain rule, we get

5×2300 = 123×x

or, 11500 = 123x

or, x = 11500÷123

hence, x = 93.49593496

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What is the math problem for 9,288 divided by 43

S_A_V [24]

Answer:

9,288 ÷ 43 = 216

Step-by-step explanation:

9,288 ÷ 43

First divide 9,288 by 43 we get quotiest as 216 and remainder 0

= 216

7 0

4 years ago

(x÷(y÷z))÷((x÷y)÷z)<br>also the variables are not specified.

Bumek [7]

The answer is: z² .
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Given: <span>(x÷(y÷z))÷((x÷y)÷z) ; without any specified values for the variables;
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we shall simplify.
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We have:
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</span>(x÷(y÷z)) / ((x÷y)÷z) .
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Start with the first term; or, "numerator": (x÷(y÷z)) ;
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x ÷ (y / z) = (x / 1) * (z / y) = (x * z) / (1 *y) = [(xz) / y ]
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Then, take the second term; or "denominator":
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((x ÷ y) ÷z ) = (x / y) / z = (x / y) * (1 / z) = (x *1) / (y *z) = [x / (zy)]
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So (x÷(y÷z)) / ((x÷y)÷z) = (x÷(y÷z)) ÷ ((x÷y)÷z) =

[(xz) / y ] ÷ [x / (zy)] = [(xz) / y ] / [x / (zy)] =
[(xz) / y ] * [(zy) / x] ;
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The 2 (two) z's "cancel out" to "1" ; and
The 2 (two) y's = "cancel out" to "1" ;
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And we are left with: z * z = z² . The answer is: z² .
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4 0

4 years ago

A graphing calculator is recommended. A crystal growth furnace is used in research to determine how best to manufacture crystals

Sophie [7]

Answer:

a) $w_1 = \frac{-2.156 -\sqrt{2.156^2 -4(0.1)(-183)}}{2*0.1}=-54.896$

$w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-183)}}{2*0.1}=33.336$

And since the power can't be negative then the solution would be w = 33.34 watts.

b) $202= 0.1w^2 +2.156 w +20$

$0.1w^2 +2.156 w-182=$

$w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-182)}}{2*0.1}=33.222$

$204= 0.1w^2 +2.156 w +20$

$0.1w^2 +2.156 w-184=$

$w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-184)}}{2*0.1}=33.449$

So then the range of voltage would be between 33.22 W and 33.45 W.

c)For this case $\epsilon = \pm 1$ since that's the tolerance 1C

d) $\delta_1 =|33.222-33.336|=0.116$

$\delta_2 =|33.449-33.336|=0.113$

So then we select the smalles value on this case $\delta =0.113$

e) For this case if we assume a tolerance of $\epsilon=\pm 1C$ for the temperature and a tolerance for the power input $\delta =0.113$ we see that:

$lim_{x \to a} f(x) =L$

Where a = 33.34 W, f(x) represent the temperature, x represent the input power and L = 203C

Step-by-step explanation:

For this case we have the following function

$T(w)= 0.1 w^2 +2.156 w +20$

Where T represent the temperature in Celsius and w the power input in watts.

Part a

For this case we need to find the value of w that makes the temperature 203C, so we can set the following equation:

$203= 0.1w^2 +2.156 w +20$

And we can rewrite the expression like this:

$0.1w^2 +2.156 w-183=$

And we can solve this using the quadratic formula given by:

$w =\frac{-b \pm \sqrt{b^2 -4ac}}{2a}$

Where a =0.1, b =2.156 and c=-183. If we replace we got:

$w_1 = \frac{-2.156 -\sqrt{2.156^2 -4(0.1)(-183)}}{2*0.1}=-54.896$

$w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-183)}}{2*0.1}=33.336$

And since the power can't be negative then the solution would be w = 33.34 watts.

Part b

For this case we can find the values of w for the temperatures 203-1= 202C and 203+1 = 204 C. And we got this:

$202= 0.1w^2 +2.156 w +20$

$0.1w^2 +2.156 w-182=$

$w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-182)}}{2*0.1}=33.222$

$204= 0.1w^2 +2.156 w +20$

$0.1w^2 +2.156 w-184=$

$w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-184)}}{2*0.1}=33.449$

So then the range of voltage would be between 33.22 W and 33.45 W.

Part c

For this case $\epsilon = \pm 1$ since that's the tolerance 1C

Part d

For this case we can do this:

$\delta_1 =|33.222-33.336|=0.116$

$\delta_2 =|33.449-33.336|=0.113$

So then we select the smallest value on this case $\delta =0.113$

Part e

For this case if we assume a tolerance of $\epsilon=\pm 1C$ for the temperature and a tolerance for the power input $\delta =0.113$ we see that:

$lim_{x \to a} f(x) =L$

Where a = 33.34 W, f(x) represent the temperature, x represent the input power and L = 203C

8 0

4 years ago

Helpppppppppppppppppppppp

Firlakuza [10]

I can’t help but you did just help me get an answer correct

5 0

4 years ago

Can someone please help me out

geniusboy [140]

On your paper, you wrote down a^2+b^2 = c^2 which is the pythagorean theorem

You are on the right track. You'll use this equation. In this case
a = 36
b = 18
c = 43
The value of a and b can be swapped. The value of c is always the longest side.

Let's see if a^2+b^2 = c^2 is true or not
a^2 + b^2 = c^2
36^2 + 18^2 = 43^2
1296 + 324 = 1849
1620 = 1849
The last equation is false so the first equation is false when a = 36, b = 18 and c = 43

We do NOT have a right triangle so we do NOT have a rectangle

Answer: No, the table is not "square" (ie, have right angles at the corners)

5 0

4 years ago

12) You can exchange $5 US dollars for L123 Honduran Lempiras. How many dollars can you exchange L2,300 for? ----​

12) You can exchange $5 US dollars for L123 Honduran Lempiras. How many dollars can you exchange L2,300 for? ----