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Alexxx [7]
3 years ago
6

There are 3 forks in the silverware drawer. There are 4 more knives than forks, and there are twice as many spoons as knives. Ho

w many pieces of silverware are there in all?​
Mathematics
2 answers:
frosja888 [35]3 years ago
5 0

Answer: There are 24 pieces of silverware in all.

Step-by-step explanation:

3+ 7 + 14 = 24

3 forks

There are 4 more knives than forks - 7 knives

There are twice as many spoons as knives - 7.2 = 14

Aleks04 [339]3 years ago
4 0

Answer:

24

Step-by-step explanation:

s= total silverware

s=3+3+4+2(7)

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What is the result of dividing x³-4 by x+2​
Dmitry [639]

Answer: x2 - 2x + 4 - 12/(x + 2)

Step-by-step explanation:

6 0
3 years ago
one third of the sum of two angles is 60 degree and one quarter of their difference is 28 degree. find the two angles
vlabodo [156]
1/3(a + b) = 60
a + b = 60 * 3 = 180

1/4(a - b) = 28
a - b = 28 * 4 = 112

a + b = 180
a - b = 112
---------------add
2a = 292
a = 292/2
a = 146

a + b = 180
146 + b = 180
b = 180 - 146
b = 34

so ur 2 angles are : 146 and 34
7 0
3 years ago
Solving separable differential equation DY over DX equals xy+3x-y-3/xy-2x+4y-8​
Ivanshal [37]

It looks like the differential equation is

\dfrac{dy}{dx} = \dfrac{xy + 3x - y - 3}{xy - 2x + 4y - 8}

Factorize the right side by grouping.

xy + 3x - y - 3 = x (y + 3) - (y + 3) = (x - 1) (y + 3)

xy - 2x + 4y - 8 = x (y - 2) + 4 (y - 2) = (x + 4) (y - 2)

Now we can separate variables as

\dfrac{dy}{dx} = \dfrac{(x-1)(y+3)}{(x+4)(y-2)} \implies \dfrac{y-2}{y+3} \, dy = \dfrac{x-1}{x+4} \, dx

Integrate both sides.

\displaystyle \int \frac{y-2}{y+3} \, dy = \int \frac{x-1}{x+4} \, dx

\displaystyle \int \left(1 - \frac5{y+3}\right) \, dy = \int \left(1 - \frac5{x + 4}\right) \, dx

\implies \boxed{y - 5 \ln|y + 3| = x - 5 \ln|x + 4| + C}

You could go on to solve for y explicitly as a function of x, but that involves a special function called the "product logarithm" or "Lambert W" function, which is probably beyond your scope.

8 0
2 years ago
In order to join a flying club,there is a $200 annual fee and $20 a month fee to maintain the flying club runaway.Write an equat
Fudgin [204]

Answer:

The equation modelling the situation is T=200+20m.

After <u>10 months</u> the cost would be $400.

Step-by-step explanation:

Given:

Annual Fee of flying club = $200

Monthly fee = $20

We need to write an equation for total cost.

Solution:

Let the number of months be denoted 'm'

Let Total cost be denoted by 'T'

Now we can say that:

Total cost is equal to sum of Annual Fee of flying club plus Monthly fee multiplied by number of months.

framing in equation form we get;

T=200+20m

Hence The equation modelling the situation is T=200+20m.

Now given:

We need to find the number of months when total cost will be $400

So we will substitute T= 400 in above equation.

200+20m=400

Subtracting both side by 200 we get;

200+20m-200=400-200\\\\20m=200

Dividing both side by 20 we get;

\frac{20m}{20}=\frac{200}{20}\\\\m=10 \ months

Hence After <u>10 months</u> the cost would be $400.

6 0
3 years ago
A rancher plans to make four identical and adjacent rectangular pens against a​ barn, each with an area of 400 msquared. What ar
Ilia_Sergeevich [38]

Answer:

base = 10\sqrt{5}m=22.36m and height = 8\sqrt{5}m = 17.89m

Step-by-step explanation:

In order to solve this problem, we must first understand how the pens are going to be placed. Refer to the attached diagram for an image of this.

From this diagram, we can start building the equations we are going to use to solve this. Since the problem doesn't specify the length of the barn, we will have no restriction on this. Taking this into account, we can start by building an equation for the area of each pen.

The area of a rectangle is given by the equation:

A=bh

where b is the base of the rectangle and h is the height of the rectangle.

with the given information we can now build our first equation:

400=bh

Since we have an equation with two variables, or two unknowns, we must have a second equation we can use to find the b and h values we need. In order to get this second equation, we must take into account that we need to minimize the amount of fencing to be used, so we can build an equation that will represent this fencing. When looking at the diagram, we can see that the fencing will be used to cover 4 bases and 5 heights, so our fencing equation should look like this:

F=4b+5h

Where F represents the amount of fencing to be used.

So now we can solve the first equation to substitute it into the second. Let's solve for the base b.

bh=400\\\\b=\frac{400}{h}

Which can now be substituted into the second equation so we get:

F=4(\frac{400}{h})+5h

which simplifies to

F=\frac{1600}{h} +5h

Now, this is the equation we need to minimize. In order to do so, we need to take the derivative of that and set it equal to zero, so we get:

F=\frac{1600}{h}+5h\\ \\F=1600h^{-1}+5h\\ \\F'=-1600h^{-2}+5\\ \\F'=-\frac{1600}{h^{2}}+5

And now we can set all this equal to zero, so we get:

-\frac{1600}{h^{2}}+5=0

Which can now be solved for h, so we get:

-\frac{1600}{h^{2}}+5=0\\ \\-\frac{1600}{h^{2} }=-5\\ \\1600=5h^{2}\\ \\h^{2}=\frac{1600}{5}\\  \\h^{2}=320\\ \\h=\sqrt{320}=8\sqrt{5}m=17.89m

That's where the first answer came from. We can now use this value to find the second answer together with the first equation:

b=\frac{400}{h}\\ \\b=\frac{400}{8\sqrt{5}}\\ \\b=\frac{50}{\sqrt{5}}

which simplifies to:

h=\frac{50}{\sqrt{5}}*\frac{\sqrt{5} }{\sqrt{5} }\\  \\h=\frac{50\sqrt{5} }{5}\\ \\h=10\sqrt{5}m = 22.36m

Aproximately and that's where te second answer came from.

So each  pen must measure:

22.36 m by 17.89 m

When multiplied you'll see that each pen will have an area of 400m^{2}

6 0
3 years ago
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