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3 years ago

I don’t understand this question?

Mathematics

1 answer:

Andrej [43]3 years ago

5 0

Answer:

no, because it has a constant rate of change

Step-by-step explanation:

this is because the x and y change at the same rate. non linear means that they dont change at the same rate.

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What is the area of the rhombus below?

Maurinko [17]

Answer:

130 Units 2

Step-by-step explanation:

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7 0

2 years ago

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What is the volume of a cube that measures 3.5 inches on each edge?

Bas_tet [7]

Answer:

42.9 in

Step-by-step explanation:

Volume of a cube = side cubed

V = (3.5)^3

V = 42.875

4 0

3 years ago

Help me please this is for the test tomorrow and need the answrt

Elza [17]

Answer:

224 cubes

Step-by-step explanation:

The volume of the cubes is 1/64. Because 1/4*1/4*1/4=1/64.

The volume of the prims is 14/4 because 7/4*2*1=14/4.

Take the 14/4 and divide it by 1/64.

Which equals 896/4 = 224

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3 years ago

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PLEASE HELP ME FIND THE ANSWER

n200080 [17]

B is true.
I think E is also true.

4 0

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;

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

$= \int\limits343 cos^{2} \theta \, d\theta$

8 0

3 years ago