What are the x- and y-intercepts of −6x+4y=96 ?

1 answer:

3 0

* intercepts "x" --> y=0

-6x + 4(0) = 96
-6x = 96
x = $- \frac{96}{6}$

x = 16

(16,0)

*intercepts "y" --> x=0

-6(0) + 4y = 96
4y = 96
y = $\frac{96}{4}$

y = 24

(0,24)

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Step-by-step explanation:

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Given that 1 x2 dx 0 = 1 3 , use this fact and the properties of integrals to evaluate 1 (4 − 6x2) dx. 0

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So, the definite integral $\int\limits^1_0 {(4 - 6x^{2} )} \, dx= - 74$

Given that

$\int\limits^1_0 {x^{2} } \, dx = 13$

We find

$\int\limits^1_0 {(4 - 6x^{2} )} \, dx$

<h3>Definite integrals </h3>

Definite integrals are integral values that are obtained by integrating a function between two values.

So, $Integral \int\limits^1_0 {(4 - 6x^{2} )} \, dx$

So, $\int\limits^1_0 {(4 - 6x^{2} )} \, dx = \int\limits^1_0 {4} \, dx - \int\limits^1_0 {6x^{2} } \, dx \\= 4[x]^{1}_{0} - \int\limits^1_0 {6x^{2} } \, dx \\= 4[x]^{1}_{0} - 6\int\limits^1_0 {x^{2} } \, dx \\= 4[1 - 0] - 6\int\limits^1_0 {x^{2} } \, dx\\= 4[1] - 6\int\limits^1_0 {x^{2} } \, dx\\= 4 - 6\int\limits^1_0 {x^{2} } \, dx$