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3 years ago

Which equations are in standard form? Check all that apply

Mathematics

1 answer:

professor190 [17]3 years ago

5 0

Answer:

2x + 3y = -6

-4x + 3y = 12

5x + 3y = }

x - y = 5

Step-by-step explanation:

Standard Form is $Ax+By=C$ .

2x + 3y = -6, -4x +3y = 12, 5x + 3y = '?', and x - y = 5 are the equations that are in standard form.

'2x+3 = 6' does not have the 'y' to make it standard form.

'y = 2x + 5' is slope-intercept form.

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Can someone help me with this problem? thank you!

abruzzese [7]

Answer:

When 6x - 4 = 8, x = 2.

Step-by-step explanation:

6x - 4 = 8

Add 4 to both sides.

6x = 12

Divide both sides by 6.

x = 2

Proof:

6x - 4 = 8

Substitute variable.

6(2) - 4 = 8

Multiply 6 and 2.

12 - 4 = 8

Subtract 4 from 12.

8 = 8

3 0

2 years ago

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HURRY PLEASE!! :(<br> Find the 15th term: a(1) = 7; d = 3

My name is Ann [436]

Yes that is correct ma dude

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3 years ago

Evaluate the following integral using trigonometric substitution.

wariber [46]

Answer:

Step-by-step explanation:

1. Given the integral function $\int\limits {\sqrt{a^{2} -x^{2} } } \, dx$ , using trigonometric substitution, the substitution that will be most helpful in this case is substituting x as $asin \theta$ i.e $x = a sin\theta$ .

All integrals in the form $\int\limits {\sqrt{a^{2} -x^{2} } } \, dx$ are always evaluated using the substitute given where 'a' is any constant.

From the given integral, $\int\limits {7\sqrt{49-x^{2} } } \, dx = \int\limits {7\sqrt{7^{2} -x^{2} } } \, dx$ where a = 7 in this case.

The substitute will therefore be $x = 7 sin\theta$

2.) Given $x = 7 sin\theta$

$\frac{dx}{d \theta} = 7cos \theta$

cross multiplying

$dx = 7cos\theta d\theta$

3.) Rewriting the given integral using the substiution will result into;

$\int\limits {7\sqrt{49-x^{2} } } \, dx \\= \int\limits {7\sqrt{7^{2} -x^{2} } } \, dx\\= \int\limits {7\sqrt{7^{2} -(7sin\theta)^{2} } } \, dx\\= \int\limits {7\sqrt{7^{2} -49sin^{2}\theta } } \, dx\\= \int\limits {7\sqrt{49(1-sin^{2}\theta)} } } \, dx\\= \int\limits {7\sqrt{49(cos^{2}\theta)} } } \, dx\\since\ dx = 7cos\theta d\theta\\= \int\limits {7\sqrt{49(cos^{2}\theta)} } } \, 7cos\theta d\theta\\= \int\limits {7\{7(cos\theta)} }}} \, 7cos\theta d\theta\\$