All categories

3 years ago

Determine how many shapes weigh 5 grams? [a]

Mathematics

1 answer:

Zielflug [23.3K]3 years ago

7 0

Answer:

Step-by-step explanation:

You might be interested in

Here are some letters and what they represent. All costs are in dollars. m represents the cost of a main dish. n represents the

k0ka [10]

The given question is incomplete

Here are some letters and what they represent. All costs are in dollars.

m represents the cost of a main dish.

n represents the number of side dishes.

s represents the cost of a side dish.

t represents the total cost of a meal.

Discuss with a partner: What does each equation mean in this situation? a. a. m = 7.50

b. m = s + 4.50

c . ns = 6

d. m + ns = t

2. Write a new equation that could be true in this situation.

Answer:

t = 3m -n(s+1)

Step-by-step explanation:

Using the information we can describe the value of each letters

using a and b

we have m = 7.50

and m = s + 4.50

7.50 = s + 4.50

s = 3

substitute s in c, we get

ns = 6
n = 6/3 = 2

using equation d we have

m + ns = t
7.50 + 6 = t
t = 13.50

so we an equation using the above data

t +ns = 3m - n

t = 3m -n(s+1)

4 0

3 years ago

X^3-8/x-2 simplify and I need an explanation please

Sergeu [11.5K]

Answer: $x^2+2x+4$

Step-by-step explanation:

The expression given in the exercise is:

$\frac{x^3-8}{x-2}$

If you descompose the number 8 into its prime factors, you get that:

$8=2*2*2=2^3$

Therefore, you can rewrite the numerator of the expression as following:

$=\frac{(x^3-2^3)}{(x-2)}$

For this exercise you need to remember that for a Difference of cubes:

$a^3-b^3=(a-b)(a^2+ab+b^2)$

Then, applying this, you get:

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

Now, it is necessary to remember the following:

$\frac{a}{a}=1$

Knowing the above, you can say that:

$\frac{(x-2)}{(x-2)}=1$

Therefore applying this, you get that the simplified expression is:

$=x^2+2x+4$

8 0

3 years ago

How many ounces of a silver alloy that costs $5.50 per ounce should be mixed with one that costs $7.00 per ounce to make a new 3

Pepsi [2]

Answer:

135.73 ounces of a silver alloy that costs $5.50 per ounce should be mixed.

Step-by-step explanation:

Given:

A silver alloy that costs $5.50 per ounce should be mixed with one that costs $7.00 per ounce to make a new 30 ounce alloy that costs $6.40 per ounce.

Now, to find the ounces of silver alloy.

Let the silver costs $5.50 per ounce be $x.$

And the silver costs $7.00 per ounce be $y.$

So, the total ounce make a new alloy:

$x+y=30\\y=30-x$ ....(1)

Now, the total costs of silver alloy:

$5.50x+7y=6.4$

Putting the value of $y$ from equation (1) in the place of $y$ :

$5.50x+7(30-x)=6.4$

$5.50x+210-7x=6.4$

$-1.5x+210=6.4$

Subtracting both sides by 210 we get:

$-1.5x=-203.6$

Dividing both sides by -1.5 we get:

$x=135.73$

Therefore, 135.73 ounces of a silver alloy that costs $5.50 per ounce should be mixed.

7 0

3 years ago

Determine the value of x in the figure.

asambeis [7]

You have to show the figure lol

4 0

3 years ago

Find the solution to the differential equation dB/dt+4B=20 with B(1)=30

natita [175]

Answer:

The solution of the differential equation is $B=5+25e^{-4t+4}$

Step-by-step explanation:

The differential equation $\frac{dB}{dt}+4B=20$ is a first order separable ordinary differential equation (ODE). We know this because a separable first-order ODE has the form:

$y'(t)=g(t)\cdot h(y)$

where g(t) and h(y) are given functions.

We can rewrite our differential equation in the form of a first-order separable ODE in this way:

$\frac{dB}{dt}+4B=20\\\frac{dB}{dt}=20-4B\\\frac{dB}{dt}=4(5-B)\\\frac{1}{5-B}\frac{dB}{dt}=4$

Integrating both sides

$\frac{1}{5-B}\frac{dB}{dt}=4\\\frac{1}{5-B}\cdot dB=4\cdot dt\\\\\int\limits {\frac{1}{5-B}} \, dB=\int\limits {4} \, dt$

The integral of left-side is:

$\int\limits {\frac{1}{5-B}} \, dB\\\mathrm{Apply\:u-substitution:}\:u=5-B\\\int\limits {\frac{1}{5-B}} \, dB=\int\limits {\frac{1}{u}} \, dB\\\mathrm{du=-dB}\\-\int\limits {\frac{1}{u}} \,du\\\mathrm{Use\:the\:common\:integral}:\quad \int \frac{1}{u}du=\ln \left(\left|u\right|\right)\\-\int\limits {\frac{1}{u}} \,du =-\ln \left|u\right|\\\mathrm{Substitute\:back}\:u=5-B\\-\ln \left|5-B\right|\\\mathrm{Add\:a\:constant\:to\:the\:solution}\\-\ln \left|5-B\right|+C$