All categories

3 years ago

Solve for x. 2x = 15 + x a) x = -15 b) x = 15 c) x = -15/2 d) x = 15/2

Mathematics

2 answers:

Lorico [155]3 years ago

6 0

Answer:

x=15

Step-by-step explanation:

topjm [15]3 years ago

3 0

Answer:

x = 15

Step-by-step explanation:

You might be interested in

A triangle has 2 sides that equal 6 inches. The other side equals 4 inches. What kind of triangle is this

Tom [10]

Ισοσκελες τρίγωνο
....

7 0

3 years ago

Read 2 more answers

If a rectangle's length is 7 more than 4 times the width and the perimeter is 54 what are the dimensions of the rectangle?

Rus_ich [418]

Let w = width of the rectangle
4w+7 = length of the rectangle 

perimeter = 2 lengths + 2 widths 

2w + 2(4w+7) = 54 
2w + 8w + 14 = 54 
10w = 40 
w = 4 
4w+7 = 23 

dimensions of rectangle: 
width = 4 
length = 23 hope this helps

4 0

3 years ago

Use greatest common factor and the distributive property to write equivalent expressions in factored form for the following expr

Nitella [24]

Answer:

a. 4(d+3e)

b. 6(3x+5y)

c. 7(3a+4y)

d. 8(3f+7g)

Step-by-step explanation:

In each case we find the greatest common factor of the numbers. That is the greatest number that goes into both the numbers. Then we factor it out in front and inside parentheses we divide each original term by the greatest common factor:

a. 4d+12e, GCF: 4

$\displaystyle4\left(\frac{4d}{4}+\frac{12e}{4}\right)=4(d+3e)$

b. 18x+30y, GCF: 6

$\displaystyle6\left(\frac{18x}{6}+\frac{30y}{6}\right)=6(3x+5y)$

c. 21a+28y, GCF: 7

$\displaystyle7\left(\frac{21a}{7}+\frac{28y}{7}\right)=7(3a+4y)$

d. 24f+56g, GCF: 8

$\displaystyle8\left(\frac{24f}{8}+\frac{56g}{8}\right)=8(3f+7g)$

5 0

3 years ago

Of 8

Tomtit [17]

Answer:

98.5kg

Step-by-step explanation:

since we are rounding to the nearest whole number, our answer will use ±0.5

as we are calculating upper bound, we will add 0.5 to the original weight, so the answer is 98.5 kg

6 0

2 years ago

Verify that the given functions y1 and y2 satisfy the corresponding homogeneous equation; then find a particular solution of the

kkurt [141]

Answer:

Therefore the particular solution of the given differential equation is

$Y(x)=\frac{1}{6} x^2 ln x$

Step-by-step explanation:

The given ordinary differential equation is

$x^2y"-3xy' +4y=11x^2$

If y₁(x) =x² is a solution of ODE then it will be satisfy the ODE

y₁'(x)= 2x and y₁"(x)=2

Putting the value of y₁'(x) and y₁"(x) in x²y"-3xy'+4y=0 we get

x².2-3x.2x+4x²= 2x²-6x²+4x²=0

Therefore y₁(x) is a solution of the given differential equation.

Again,

y₂(x) =(x²ln x ) is a solution of ODE then it will be satisfy the ODE.

$y'_2=2x ln x+ x^2\times\frac{1}{x}$

$= 2x ln x+x$

$y"_2(x)=2x\times \frac{1}{x} +2lnx+1$ = 3+2 ln x

Putting the value of y₂'(x) and y₂"(x) in x²y"-3xy'+4y=0 we get

$x^2 (3+2 ln x)-3x( 2x ln \ x+x)+4x^2\ ln\ x$ =0

Therefore y₂(x) is a solution of the given differential equation.

The wronskian of y₁(x) and y₂(x) is

$W(y_1,y_2)(x)=\left|\begin{array}{cc}y_1&y_2\\y'_1&y'_2\end{array}\right|$

$=\left|\begin{array}{cc}x^2&x^2 ln x\\2x&2x ln x+x\end{array}\right|$

=x²(2x ln x+x)-x²ln x(2x)

=2x³ ln x +x³ - 2x³ln x

=x³≠0

Here $g(x)=\frac{y_2(x)}{y_2(x)} = ln x$

The particular solution is

$Y(x)=- y_1(x)\int\frac{y_2(x).g(x)}{W(y_1,y_2)(x)}dx + y_2(x)\int\frac{y_1(x).g(x)}{W(y_1,y_2)(x)}dx$

$=-x^2\int\frac{x^2 ln x . ln x}{x^3} dx+x^2ln x\int \frac{x^2ln x}{x^3} dx$

$=-x^2\int \frac{(lnx)^2}{x} dx+x^2 ln x\int \frac{ln x}{x} dx$

$=-x^2\int u^2 du+x^2ln x\int u dx$ Let ln x =u $\Rightarrow \frac{1}{x} dx=du$

$=-x^2 \frac{u^3}{3} +x^2 ln x \frac{u^2}{2}$

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

$=\frac{1}{6}x^2 (ln x)^3$

Therefore the particular solution of the given differential equation is

$Y(x)=\frac{1}{6} x^2 ln x$

8 0

3 years ago