4x+2y=10<br> Y=-2x <br> Find the slope (m) and y intercept

Four pounds of flour cost $2.60 how much flour do you get per dollar? round your answer to the nearest hundredth, if necessary.

gogolik [260]

Since 4 pounds of flour costs $2.60, we have that:

$\begin{gathered} 4\text{pounds}\to2.60 \\ x\to1 \end{gathered}$

Where x is the amount of flour you get for 1 dollar.

We can rewrite the expression above as follows:

$\begin{gathered} \frac{4\text{pounds}}{2.60}=\frac{x\text{ pounds}}{1} \\ \Rightarrow x\text{ pounds}=\frac{4pounds}{2.6}=1.53846\ldots \\ \Rightarrow x\approx1.54 \end{gathered}$

Then, we get 1.54 pounds of flour per dollar

7 0

1 year ago

If Line l^1 has equation y = 11 - 3x Find the Equation of the Line through (5,3) parallel to l^1 So y = ??????

gregori [183]

Answer:

Step-by-step explanation:

Given equation of line is y = 11 - 3x

The equation takes the form of slope intercept equation ; y = mx + b where, m = slope and b = intercept.

Comparing the equation;

Slope, m = - 3

We obtain the equation of line passing through the point (5, 3) and parallel to the given line ;

y - y1 = m(x - x1)

x1 = 5 ; y1 = 3

y - 3 = - 3(x - 5)

y - 3 = - 3x + 15

y = - 3x + 15 + 3

y = - 3x + 18

6 0

3 years ago

What is 3 3/8 ÷ 9 please help

denis23 [38]

Answer:

the exact form is 3/8

in a decimal it's 0.375

Step-by-step explanation:

3 0

4 years ago

A card is drawn from a standard deck of cards. Find the probability, given that the card is a non dash face card. P(nine)

sattari [20]

Answer:

Step-by-step explanation:Knowing that

There are 26 black cards and 26 red ones. Face card consists of jack (J), queen (Q), king (K). There are (26-6)=20 non face red cards. “Either ... or ...” here means bundle of both kinds of cards.

So, there are (26+20)=46 expected cards.

If you also ask for the probability, so the probability of the question asking above is equal to

23/26

6 0

3 years ago

Observe that x and e^x are solutions to the homogeneous equation associated with:

yanalaym [24]

To take advantage of the characteristic solutions $y_1(x)=x$ and $y_2(x)=e^x$ , you can try the method of variation of parameters, where we look for a solution of the form

$y=y_1u_1+y_2u_2$

with the condition that

${u_1}'y_1+{u_2}'y_2=0$

$\implies{u_1}'x+{u_2}'e^x=0$ $(\mathbf 1)$

Then

$y'={y_1}'u_1+y_1{u_1}'+{y_2}'u_2+y_2{u_2}'$

$\implies y'={y_1}'u_1+{y_2}'u_2$

$y''={y_1}''u_1+{y_1}'{u_1}'+{y_2}''u_2+{y_2}'{u_2}'$

Substituting into the ODE gives

$(1-x)({y_1}''u_1+{y_1}'{u_1}'+{y_2}''u_2+{y_2}'{u_2}')+x({y_1}'u_1+{y_2}'u_2)-y_1u_1+y_2u_2=2(x-1)^2e^{-x}$

Since

$y_1=x\implies{y_1}'=1\implies{y_1}''=0$

$y_2=e^x\implies{y_2}'=e^x\implies{y_2}''=e^x$

the above reduces to

$(1-x)({u_1}'+e^x{u_2}')=2(x-1)^2e^{-x}$

${u_1}'+e^x{u_2}'=2(1-x)e^{-x}$ $(\mathbf 2)$

$(\mathbf 1)$ and $(\mathbf 2)$ form a linear system that we can solve for ${u_1}',{u_2}'$ using Cramer's rule:

${u_1}'=\dfrac{W_1(x)}{W(x)},{u_2}'=\dfrac{W_2(x)}{W(x)}$

where $W(x)$ is the Wronskian determinant of the fundamental system and $W_i(x)$ is the same determinant, but with the $i$ -th column replaced with $(0,2(x-1)^2e^{-x})$ .