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3 years ago

6

A pair of equations is shown below:

1 answer:

Naddika [18.5K]3 years ago

6 0

Happy Pride Month!

Part 1:

You can solve the pair of equation graphically by writing the y-intercepts which are (0,-1) and (0,-5) Then follow the slopes of each line till the intercept. The slopes are up 2 over 1 and up 4 over 1 respectivily

Part 2:

When you graph these you get the answer of (2,3)

x is 2 and y is 3

Have a good day!

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1 1. Find the x-intercepts for y= 2(x - 7)(x+2)?

givi [52]

Answer:

$b.) (7,0) and (-2,0)$

Step-by-step explanation:

To find the x-intercept, substitute 0 for y and solve for x:

$0=2(x-7)$

Divide both sides by 2:

$\frac{0}{2}=\frac{2(x-7)}{2}$

$0=x-7$

Add 7 to both sides to isolate x:

$0+7=x-7+7\\7=x$

The x-intercept for $y=2(x-7)$ is 7 at point $(7,0)$ .

$0=(x+2)$

Remove parentheses:

$0=x+2$

Subtract 2 from both sides:

$0-2=x+2-2$

$-2=x\\x=-2$

The x-intercept for $y=(x+2)$ is -2 at point $(-2,0)$ .

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3 years ago

What two numbers have a product of 234 and a sum of 31? You may have to use a guess and check method.

Dima020 [189]

Answer:

265

Step-by-step explanation:

234+31=265 I use additional

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3 years ago

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Steve likes to entertain friends at parties with "wire tricks." Suppose he takes a piece of wire 60 inches long and cuts it into

Alex_Xolod [135]

Answer:

a) the length of the wire for the circle = $(\frac{60\pi }{\pi+4}) in$

b)the length of the wire for the square = $(\frac{240}{\pi+4}) in$

c) the smallest possible area = 126.02 in² into two decimal places

Step-by-step explanation:

If one piece of wire for the square is y; and another piece of wire for circle is (60-y).

Then; we can say; let the side of the square be b

so 4(b)=y

b= $\frac{y}{4}$

Area of the square which is L² can now be said to be;

$A_S=(\frac{y}{4})^2 = \frac{y^2}{16}$

On the otherhand; let the radius (r) of the circle be;

2πr = 60-y

$r = \frac{60-y}{2\pi }$

Area of the circle which is πr² can now be;

$A_C= \pi (\frac{60-y}{2\pi } )^2$

$=( \frac{60-y}{4\pi } )^2$

Total Area (A);

A = $A_S+A_C$

= $\frac{y^2}{16} +(\frac{60-y}{4\pi } )^2$

For the smallest possible area; $\frac{dA}{dy}=0$

∴ $\frac{2y}{16}+\frac{2(60-y)(-1)}{4\pi}=0$

If we divide through with (2) and each entity move to the opposite side; we have:

$\frac{y}{18}=\frac{(60-y)}{2\pi}$

By cross multiplying; we have:

2πy = 480 - 8y

collect like terms

(2π + 8) y = 480

which can be reduced to (π + 4)y = 240 by dividing through with 2

$y= \frac{240}{\pi+4}$

∴ since $y= \frac{240}{\pi+4}$ , we can determine for the length of the circle ;

60-y can now be;

= $60-\frac{240}{\pi+4}$

= $\frac{(\pi+4)*60-240}{\pi+40}$

= $\frac{60\pi+240-240}{\pi+4}$

= $(\frac{60\pi}{\pi+4})in$

also, the length of wire for the square (y) ; $y= (\frac{240}{\pi+4})in$

The smallest possible area (A) = $\frac{1}{16} (\frac{240}{\pi+4})^2+(\frac{60\pi}{\pi+y})^2(\frac{1}{4\pi})$

= 126.0223095 in²

≅ 126.02 in² ( to two decimal places)

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A flower pot in the shape of a truncated cone with a height of 8 in. was formed by slicing off the tip of a cone along a plane p

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Hello :) the answer to your question is
B.60 in 3

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Please please please help

siniylev [52]

E quick has no idea about this one

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