Solve for x!!!!!!!!!

A small box of nails was 10 2⁄8 inches tall. If the large box of nails was 2 2⁄5 inches taller, how tall is the large box of nai

Romashka [77]

Answer:

253/20

Step-by-step explanation:

6 0

2 years ago

Solve the system algebraically. 3x - 2y - 1 = 0 y = 5x + 4 What is the solution? {(-9/7,-17/7)} {(9/7,17/7)} {(9/7,-17/7)}

lutik1710 [3]

3x - 2y - 1 = 0
y = 5x + 4

3x - 2(5x + 4) - 1 = 0
3x - 10x - 8 - 1 = 0
-7x - 9 = 0
-7x = 9
x = -9/7

y = 5x + 4
y = 5(-9/7) + 4
y = -45/7 + 4
y = -45/7 + 28/7
y = - 17/7

solution is (-9/7, -17/7)

3 0

2 years ago

Read 2 more answers

Write the point-slope form of the equation of the horizontal line that passes through the point (2, 1). Pease show how!

mote1985 [20]

Check the picture below.

using a second for it, since it's a horizontal line, then say hmmm we'll use the y-intercept, or (0, 1), so the equation of a line that passes through (2,1) and (0,1)

$\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ 2}}\quad ,&{{ 1}})\quad 
% (c,d)
&({{ 0}}\quad ,&{{ 1}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{1-1}{0-2}\implies \cfrac{0}{-2}
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-1=0(x-2)$

5 0

2 years ago

F(x) = x2 + 1 g(x) = 5-x (f+g)(x) = O x2 + x-4 x²+x+4 O x2-x+6 O x2 + x + 6

charle [14.2K]

Answer:

$\boxed{(f + g)(x) = {x}^{2} - x + 6}$

Given:

$f(x) = {x}^{2} + 1 \\ \\ g(x) = 5 - x$

To Find:

$(f + g)(x) = f(x) + g(x)$

Step-by-step explanation:

$= > f(x) + g(x) = ({x}^{2} + 1) + (5 - x) \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = {x}^{2} + 1 + 5 - x\\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = {x}^{2} + 6 - x\\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = {x}^{2} - x + 6$

5 0

3 years ago

Check whether the function yequalsStartFraction cosine 2 x Over x EndFraction is a solution of x y prime plus yequalsnegative 2

Jobisdone [24]

The question is:

Check whether the function:

y = [cos(2x)]/x

is a solution of

xy' + y = -2sin(2x)

with the initial condition y(π/4) = 0

Answer:

To check if the function y = [cos(2x)]/x is a solution of the differential equation xy' + y = -2sin(2x), we need to substitute the value of y and the value of the derivative of y on the left hand side of the differential equation and see if we obtain the right hand side of the equation.

Let us do that.

y = [cos(2x)]/x

y' = (-1/x²) [cos(2x)] - (2/x) [sin(2x)]

Now,

xy' + y = x{(-1/x²) [cos(2x)] - (2/x) [sin(2x)]} + ([cos(2x)]/x

= (-1/x)cos(2x) - 2sin(2x) + (1/x)cos(2x)

= -2sin(2x)

Which is the right hand side of the differential equation.

Hence, y is a solution to the differential equation.

6 0

3 years ago