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3 years ago

(i)Find the area of the parallelogram with base 14 cm and height 15 cm.

Mathematics

2 answers:

gulaghasi [49]3 years ago

7 0

Answer:

The area should be "210 cm to the second power"

viva [34]3 years ago

3 0

<h3>Given:</h3>

Base of the Parallelogram = 14 cm

Height of the Parallelogram = 15 cm

The Area of the Parallelogram.

<h3>Formula:</h3>

Area of a Parallelogram = base × height

<h3>Solution:</h3>

Area of the Parallelogram = 14 × 15 = 210 sq.cm

<h2>Answer:</h2>

Area of the Parallelogram is 210 sq.cm

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reply to this with answer Which best proves why the expressions 4 (x + 3) + 2 x and 6 (x + 2) must be equivalent expressions?

Ivan

Answer:

Given expressions are equivalent.

Step-by-step explanation:

Given expression:

4 (x + 3) + 2x and 6 (x + 2)

Find:

Given equivalent expressions or not

Computation:

By taking LHS

4 (x + 3) + 2x

4x + 12 + 2x

6x + 12

By taking RHS

6 (x + 2)

6x + 12

We say that

LHS = RHS

So,

Given expressions are equivalent.

3 0

2 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;

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