2x - y = 4<br> y = 2x + 4<br> Solution(s):<br><br> PLEASEEE

James estimated the total cost of groceries to be $50. The actual cost of the groceries is $65. What is the percent error? Round

Alex777 [14]

To get percent error you take actual-predicted and divide by the actual then multiply by 100 to get the percent.

65-50=15

15/60=0.25

0.25 times 100 =25%

7 0

3 years ago

Read 2 more answers

Find an equation equivalent to r = 10sin ø in rectangular coordinates.

kherson [118]

Answer: x^2 + y^2 -10y = 0

Step-by-step explanation:

Cartesian coordinates, also called the Rectangular coordinates, isdefined in terms of x and y. So, for the problem θ has to be eliminated or converted using basic foundations that are described by the unit circle and the right triangle trigonometry.

r= 10sin(θ)

Remember that:

x= r × cos(θ)

y= r × sin(θ)

r^2= x^2 + y^2

Multiply both sides of the equation by r. This will give:

r × r = 10r × sin(θ)

r^2 = 10r × sin(θ)

x^2 + y^2= 10r × sin(θ)

Because y= r × sin(θ), we can make a substitution. This will be:

x^2 + y^2= 10y

x^2 + y^2 -10y = 0

The above equation is the Rectangular coordinate equivalent to the given equation.

4 0

4 years ago

4xy-28x = 16 work and answer please

algol [13]

Lol do your own work. Pay attention in class, it'll pay off ;)

6 0

3 years ago

A post is supported by two wires (one on each side going in oppositedirections) creating an angle of 80° between the wires. The

Vladimir [108]

Using the Sine rule,

$\frac{\sin A}{A}=\frac{\sin B}{B}=\frac{\sin C}{C}$ $\begin{gathered} \text{Let A = 14m,} \\ Substituting the variables into the formula,Where the length of the wires are, AP = xm and BP = ym[tex]\begin{gathered} \frac{\sin80^0}{14}=\frac{\sin40^0}{x} \\ \text{Crossmultiply,} \\ x\times\sin 80^0=14\times\sin 40^0 \\ Divide\text{ both sides by }\sin 80^0 \\ x=\frac{14\sin40^0}{\sin80^0} \\ x=9.14m \end{gathered}$

Hence, the length of wire AP (x) is 9.14m.

For wire BP (y)m,

Sum of angles in a triangle is 180 degrees,

$A^0+P^0+B^0=180^0$ $\begin{gathered} \text{Where A}^0=\text{ unknown,} \\ P^0=80^0\text{ and,} \\ B^0=40^0 \\ A^0+80^0+40^0=180^0 \\ A^0+120^0=180^0 \\ A^0=180^0-120^0 \\ A^0=60^0 \end{gathered}$

Using the side rule to find the length of wire BP,

$\begin{gathered} \frac{\sin 60^0}{y}=\frac{\sin 80^0}{14} \\ \text{Crossmultiply,} \\ y\times\sin 80^0=14\times\sin 60^0 \\ \text{Didive both sides by }\sin 80^0 \\ y=\frac{14\times\sin 60^0}{\sin 80^0} \\ y=12.31m \end{gathered}$

Hence, the length of wire BP (y) is 12.31m

Therefore, the length of the wires are (9.14m and 12.31m).

4 0

1 year ago

X squared + ysquared-6y=-5

Oksanka [162]

Standard Form of the circle:

$x^{2} +(y-3) ^{2} =14$

For y:

y=3 (+/-) $\sqrt{-x} ^{2} +14$

For x:

x= $\sqrt{-y ^{2} } +6y+5

and

- \sqrt{-y} ^{2} +6y+5$

8 0

3 years ago

2x - y = 4 y = 2x + 4 Solution(s): PLEASEEE