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3 years ago

A rectangular prism has a length of 6cm and a width of 4cm. If its surface area is 208cm squared, what is its height?

Mathematics

1 answer:

bogdanovich [222]3 years ago

6 0

Answer:

Step-by-step explanation:

h=A

2﹣lw

l+w=208

2﹣6·4

6+4=8

it does show the division so i know u won't know what i did

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Simplify -3y+2x-5y-7x

Sphinxa [80]

<h3><em><u>Simplify -3y+2x-5y-7x is:</u></em></h3><h3>-8y - 5x</h3>

<em><u>Solution:</u></em>

<em><u>Given that, we have to simplify</u></em>

$-3y + 2x-5y-7x$

We can simplify the above expression by combining the like terms

Like terms are terms that has same variable ( with same exponent) with same or different coefficient

From given,

$-3y + 2x-5y-7x$

Here, like terms are -3y and -5y

And 2x and -7x

Group the like terms

$-3y-5y+2x-7x$

Combine the like terms

$-8y -5x$

Thus the given expression is simplified

6 0

2 years ago

Write all integers between:<br><br>a. -17 and -12<br><br>b. -1 and -6<br><br>

egoroff_w [7]

Answer:

look at explanation

Step-by-step explanation:

a) -13,-14,-15,-16

b) -2,-3,-4,-5

5 0

2 years ago

What is the lower quartile for 55,55,50,52,46,38,48,51,55,56

Feliz [49]

Answer:

Step-by-step explanation:

The first or lower quartile is the median of the lower half of the data: 48

4 0

2 years ago

Suppose after 2500 years an initial amount of 1000 grams of a radioactive substance has decayed to 75 grams. What is the half-li

krok68 [10]

Answer:

The correct answer is:

Between 600 and 700 years (B)

Step-by-step explanation:

At a constant decay rate, the half-life of a radioactive substance is the time taken for the substance to decay to half of its original mass. The formula for radioactive exponential decay is given by:

$A(t) = A_0 e^{(kt)}\\where:\\A(t) = Amount\ left\ at\ time\ (t) = 75\ grams\\A_0 = initial\ amount = 1000\ grams\\k = decay\ constant\\t = time\ of\ decay = 2500\ years$

First, let us calculate the decay constant (k)

$75 = 1000 e^{(k2500)}\\dividing\ both\ sides\ by\ 1000\\0.075 = e^{(2500k)}\\taking\ natural\ logarithm\ of\ both\ sides\\In 0.075 = In (e^{2500k})\\In 0.075 = 2500k\\k = \frac{In0.075}{2500}\\ k = \frac{-2.5903}{2500} \\k = - 0.001036$

Next, let us calculate the half-life as follows:

$\frac{1}{2} A_0 = A_0 e^{(-0.001036t)}\\Dividing\ both\ sides\ by\ A_0\\ \frac{1}{2} = e^{-0.001036t}\\taking\ natural\ logarithm\ of\ both\ sides\\In(0.5) = In (e^{-0.001036t})\\-0.6931 = -0.001036t\\t = \frac{-0.6931}{-0.001036} \\t = 669.02 years\\\therefore t\frac{1}{2} \approx 669\ years$

Therefore the half-life is between 600 and 700 years

5 0

2 years ago

Create a memory device to help you remember the quadratic formula. The device could be a song, drawing, rhyme, mnemonic device,

Lostsunrise [7]

Answer:

q-u-a-d-r-a-t-i-c f-o-r-m-u-l-a

Step-by-step explanation:

6 0

3 years ago