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Rasek [7]
3 years ago
11

What is the equation of the line that passes through (0, 3) and (7,0)?

Mathematics
2 answers:
ehidna [41]3 years ago
6 0

The equation of the line that passes through (0,3) and (7,0) is y=−37x+3 y = − 3 7 x + 3.

Vadim26 [7]3 years ago
5 0

Answer:

Step-by-step explanation:

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12-2(n-1)=-18<br><br> aa help plzz
Luda [366]

Answer:

n=16

Step-by-step explanation:

Let's solve your equation step-by-step.

12−2(n−1)=−18

Step 1: Simplify both sides of the equation.

12−2(n−1)=−18

12+(−2)(n)+(−2)(−1)=−18(Distribute)

12+−2n+2=−18

(−2n)+(12+2)=−18(Combine Like Terms)

−2n+14=−18

−2n+14=−18

Step 2: Subtract 14 from both sides.

−2n+14−14=−18−14

−2n=−32

Step 3: Divide both sides by -2.

−2n /−2  =  −32 /−2

n=16

8 0
3 years ago
How do you solve for X in a fraction
Andrews [41]

By cross-multiplying, we get

15x + 30 = 24

15x = -6

x = -3 / 5



7 0
3 years ago
For every 5 hours that he works, Andrew is paid $60.20. How much is Andrew paid in dollars per hour?
masha68 [24]
Andrew is paid $12.04 per hour.
3 0
3 years ago
Read 2 more answers
Which expression is equivalent to RootIndex 3 StartRoot 32 x Superscript 8 Baseline y Superscript 10 Baseline EndRoot?
evablogger [386]

Answer:

The equivalent expression to the given expression is \sqrt[3]{32x^8y^{10}}=2x^2y^3\sqrt[3]{4x^2y}

Step-by-step explanation:

The given expression is \sqrt[3]{32x^8y^{10}}

To find the equivalent expression:

\sqrt[3]{32x^8y^{10}}=(32x^8y^{10})^{\frac{1}{3}}

We may write the above expression as below:

=(32^{\frac{1}{3}})((x^8)^{\frac{1}{3}})((y^{10})^{\frac{1}{3}})

=(2)((4)^{\frac{1}{3}})(x^6)\times (x^2)(x^{\frac{2}{3}})(y^3)(y^{\frac{1}{3}}) (using square root properties)

=(2\sqrt[3]{4})(x^2\sqrt[3]{x^2})(y^3\sqrt[3]{y}) (combining the like terms and doing multiplication )

=2x^2y^3\sqrt[3]{4x^2y}

Therefore \sqrt[3]{32x^8y^{10}}=2x^2y^3\sqrt[3]{4x^2y}

Therefore the equivalent expression to the given expression is \sqrt[3]{32x^8y^{10}}=2x^2y^3\sqrt[3]{4x^2y}

5 0
3 years ago
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