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3 years ago

Solve 5x(m+n) using the distributive property

Mathematics

1 answer:

Genrish500 [490]3 years ago

4 0

Answer: $5xm+5xn$

Step-by-step explanation:

For this exercise it is necessary to remember the following:

1) The Distributive Property states the following:

$a(b+c)=ab+ac\\\\a(b-c)=ab-ac$

2) The multiplication of signs:

$(+)(+)=+\\(-)(-)=+\\(-)(+)=-\\(+)(-)=-$

Knowing this, and having the following expression given in the exercise:

$5x(m+n)$

You can apply the Distributive property multiplying $m$ and $n$ , which are inside the parentheses, by $5x$ .

So, you get the following result:

$=(5x)(m)+(5x)(n)=5xm+5xn$

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The question is below

Ad libitum [116K]

Answer:

£59.25

Step-by-step explanation:

Hello!

To solve this problem, we must:

Solve for the length of the fence (aka height)
Find the area of the lawn (trapezoid)
Find the number of cans needed
Find the price of all the cans

Area of a trapezoid, and why the formula works:

A trapezoid is a quadrilateral with one set of parallel sides known as bases. The other two sides are known as the legs.

To find the area of a trapezoid, we use the formula:

$\frac{B_1 + B_2}{2}* h$

This works because if we used the formula, we would be duplicating the trapezoid to form a rectangle with a side length of B1 + B2, and a height of h. Since the trapezoid is half of that, we divide by 2.

Solve for height:

The height is unknown but can be found using the Pythagorean Theorem.

The difference between the bases is the length of the bottom leg of the right triangle, and 17 is the hypotenuse.

Difference = 20 - 12 = 8

Hypotenuse = 17

8² + fence² = 17²
64 + fence² = 289
225 = fence²
fence = 15

The height is 15

Solve for area:

Now we can solve for the area.

$A = \frac{B_1 + B_2}{2} * h$
$A = \frac{12 + 20}{2} * 15$
$A = \frac{32}{2} * 15$
$A = 16 * 15 = 240$

The area is 240

Cans:

The area of the lawn is 240 square meters. Each can cover 100 square meters.

240 ÷ 100 = 2.4

Since we can't use part of a can, we round up to three whole cans.

The price of 3 cans :

3 * 19.75
59.25

£59.25

The Pythagorean Theorem:

The Pythagorean theorem is a very common geometry formula used to find the length of the hypotenuse in a right triangle, given the lengths of the two other bases.

The formula is : $a^2 + b^2 = c^2$

a is a leg
b is a leg
c is the hypotenuse

Images attached for your reference

7 0

2 years ago

If a sphere has radius of 13 centimeters what is the volume

Travka [436]

The answer is 8788pi/3 Hope this helps! Mark brainly please!

7 0

2 years ago

Help Please need help

Olenka [21]

Table:

x y

-3    0.5

-2    1

-1     2.5

0      5

1      8.5

Now, y = ax^2 + bx + c

Start replacing x = 0 and y = 5

5 = a(0) + b(0) + c => c = 5.

Now, replace any other two pair of data:

(-2, 1) => -1 = a(-2)^2 + b(-2)

-1 = 4a - 2b

(1, 8.5) => 8.5 = a(1)^2 + b(1)

8.5 = a + b

Solve the system

4a - 2b = - 1
a + b = 8.5

=>

4a - 2b = - 1
2a + 2b = 17
-------------------------

6a = 16

=> a = 16 / 6 = 8/3

b = 8.5 - a = 17/2 - 8/3 = 35/6

=> y = (8/3)x^2 + (35/6)x + 5 <--------- answer

6 0

3 years ago

The logistic equation for the population (in thousands) of a certain species is given by:

Eva8 [605]

Answer:

b. 1.5

c. 1.5

d. No

Step-by-step explanation:

a. First, let's solve the differential equation:

$\frac{dp}{dt} =3p-2p^2$

Divide both sides by $3p-2p^2$ and multiply both sides by dt:

$\frac{dp}{3p-2p^2}=dt$

Integrate both sides:

$\int\ \frac{1}{3p-2p^2} dp =\int\ dt$

Evaluate the integrals and simplify:

$p(t)=\frac{3e^{3t} }{C_1+2e^{3t}}$

Where C1 is an arbitrary constant

I sketched the direction field using a computer software. You can see it in the picture that I attached you.

b. First let's find the constant C1 for the initial condition given:

$p(0)=3=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}$

Solving for C1:

$C_1=-1$

Now, let's evaluate the limit:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-1 } \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2-e^{-3x} }$

The expression $-e^{-3x}$ tends to zero as x approaches ∞ . Hence:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-1 } =\frac{3}{2} =1.5$

c. As we did before, let's find the constant C1 for the initial condition given:

$p(0)=0.8=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}$

Solving for C1:

$C_1=1.75$

Now, let's evaluate the limit:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}+1.75 } \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2+1.75e^{-3x} }$

The expression $-e^{-3x}$ tends to zero as x approaches ∞ . Hence:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}+1.75 } =\frac{3}{2} =1.5$

d. To figure out that, we need to do the same procedure as we did before. So, let's find the constant C1 for the initial condition given:

$p(0)=2=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}$

Solving for C1:

$C_1=-\frac{1}{2} =-0.5$

Can a population of 2000 ever decline to 800? well, let's find the limit of the function when it approaches to ∞:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-0.5 } \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2-0.5e^{-3x} }$