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3 years ago

Factorise

Mathematics

2 answers:

Rufina [12.5K]3 years ago

8 0

Answer:

6x - 9

3(2x−3)

2a(5b+3)

Step-by-step explanation:

Factor 6a+10ab

10ab+6a

Factor 6x−9

6x−9

=3(2x−3)

:))

s2008m [1.1K]3 years ago

7 0

Answer:

I think the answer is this3(2x-3)

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Differentiate f(x) = (2x + 5)^6 with respect to x

seraphim [82]

Step-by-step explanation:

d/dx [f(x)]

= d/dx [(2x + 5)⁶]

= 6(2x + 5)⁵ * d/dx (2x + 5)

= 12(2x + 5)⁵.

3 0

3 years ago

In evaluating a double integral over a region D, a sum of iterated integrals was obtained as follows:

BabaBlast [244]

Answer

$a=0$ , $b=2$

$g_1(x)=\frac{5x}{2}$ , $g_2(x)=7-x$

Step-by-step explanation:

Given that

$\int \int Df(x,y)dA=\int_0 ^5\int _0 ^ {\frac {2y}{5}} f(x,y)dxdy+\int_5^7\int_0^{7-y} f(x,y)dxdy\; \cdots (i)$

For the term $\int_0 ^5\int _0 ^ {\frac {2y}{5}} f(x,y)dxdy$ .

Limits for $x$ is from $x=0$ to $x=\frac {2y}{5}$ and for $y$ is from $y=0$ to $y=5$ and the region $D$ , for this double integration is the shaded region as shown in graph 1.

Now, reverse the order of integration, first integrate with respect to $y$ then with respect to $x$ . So, the limits of $y$ become from $y=\frac{5x}{2}$ to $y=5$ and limits of $x$ become from $x=0$ to $x=2$ as shown in graph 2.

So, on reversing the order of integration, this double integration can be written as

$\int_0 ^5\int _0 ^ {\frac {2y}{5}} f(x,y)dxdy=\int_0 ^2\int _ {\frac {5x}{2}}^5 f(x,y)dydx\; \cdots (ii)$

Similarly, for the other term $\int_5 ^7\int _0 ^ {7-y} f(x,y)dxdy$ .

Limits for $x$ is from $x=0$ to $x=7-y$ and limits for $y$ is from $y=5$ to $y=7$ and the region $D$ , for this double integration is the shaded region as shown in graph 3.

Now, reverse the order of integration, first integrate with respect to $y$ then with respect to $x$ . So, the limits of $y$ become from $y=5$ to $y=7-x$ and limits of $x$ become from $x=0$ to $x=2$ as shown in graph 4.

So, on reversing the order of integration, this double integration can be written as

$\int_5 ^7\int _0 ^ {7-y} f(x,y)dxdy=\int_0 ^2\int _5 ^ {7-x} f(x,y)dydx\;\cdots (iii)$

Hence, from equations $(i)$ , $(ii)$ and $(iii)$ , on reversing the order of integration, the required expression is

$\int \int Df(x,y)dA=\int_0 ^2\int _ {\frac {5x}{2}}^5 f(x,y)dydx+\int_0 ^2\int _5 ^ {7-x} f(x,y)dydx$

$\Rightarrow \int \int Df(x,y)dA=\int_0 ^2\left(\int _ {\frac {5x}{2}}^5 f(x,y)+\int _5 ^ {7-x} f(x,y)\right)dydx$

$\Rightarrow \int \int Df(x,y)dA=\int_0 ^2\int _ {\frac {5x}{2}}^{7-x} f(x,y)dydx\; \cdots (iv)$

Now, compare the RHS of the equation $(iv)$ with

$\int_a^b\int_{g_1(x)}^{g_2(x)} f(x,y)dydx$

We have,

$a=0, b=2, g_1(x)=\frac{5x}{2}$ and $g_2(x)=7-x$ .

3 0

3 years ago

Mean average of this question plz.

stira [4]

for this answer
I got 32

7 0

3 years ago

Multiply (x-3(4x+2) using the distributive property.

lilavasa [31]

D. (X-3)(4x)+(x-3)(2)

3 0

4 years ago

Read 2 more answers

Verify the basic identity. What is the domain of validity?cot theta = cos theta csc thetasee image

mr Goodwill [35]

Solution

Given the function below

$cot\theta=\cos\theta csc\theta$

Where

$csc\theta=\frac{1}{\sin\theta}$ $\begin{gathered} cot\theta=\cos\theta csc\theta \\ cot\theta=\cos\theta\times\frac{1}{\sin} \\ cot\theta=\frac{\cos\theta}{\sin\theta} \end{gathered}$

Also recall that the contangent formula is

$cot\theta=\frac{\cos\theta}{\sin\theta}$

Thus;

$\begin{gathered} cot\theta=\cos\theta csc\theta \\ \frac{\cos\theta}{\sin\theta}=\frac{\cos\theta}{\sin\theta} \end{gathered}$

From the above deduction,

sinθ is the denominator,

Hence, the domain of validity is defined for the set of real numbers except for sinθ = 0

6 0

2 years ago