0.3y + y/z= y=10 and z=5

Mathematics

1 answer:

Xelga [282]3 years ago

3 0

Answer:

Step-by-step explanation:

write the equation

0.3y + y/z=

Then you fill in what you know

0.3(10) + 10/5=

then we multiply/divide

30 + 10/5

then we continue to multiply/divide untill there is nothing to multiply/divide anymore

30+2

then we add/subtract

Our answer is 32

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