23=30-g;6,7,8 what is the solution

(3m±4/5n)2<br>solve this question<br>

sukhopar [10]

Answer:

(3m-4/5)2

Final result :

(15m - 4)2

——————————

52

Step by step solution :

Step 1 :

4

Simplify —

5

Equation at the end of step 1 :

4

(3m - —)2

5

Step 2 :

Rewriting the whole as an Equivalent Fraction :

2.1 Subtracting a fraction from a whole

Rewrite the whole as a fraction using 5 as the denominator :

3m 3m • 5

3m = —— = ——————

1 5

Equivalent fraction : The fraction thus generated looks different but has the same value as the whole

Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator

Adding fractions that have a common denominator :

2.2 Adding up the two equivalent fractions

Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

3m • 5 - (4) 15m - 4

———————————— = ———————

5 5

Equation at the end of step 2 :

(15m - 4)

(—————————)2

5

Step 3 :

Final result :

(15m - 4)2

———

52

Step-by-step explanation:

5 0

2 years ago

The baseball field is 9/10 of a mile from Benson’s house. Benson runs 3/10 of a mile and walks 4/10 of a mile on his way to the

inysia [295]

Answer:

2/10 I believe

Step-by-step explanation:

3/10 + 4/10= 7/10

9/10 - 7/10 = 2/10

8 0

2 years ago

Find the value of X in the quadrilateral. (Picture)

Alex_Xolod [135]

Answer:

B

Hope this Helps! Have A Nice Day!!

5 0

3 years ago

Read 2 more answers

Use Simpson's Rule with n = 10 to approximate the area of the surface obtained by rotating the curve about the x-axis. Compare y

DiKsa [7]

The area of the surface is given exactly by the integral,

$\displaystyle\pi\int_0^5\sqrt{1+(y'(x))^2}\,\mathrm dx$

We have

$y(x)=\dfrac15x^5\implies y'(x)=x^4$

so the area is

$\displaystyle\pi\int_0^5\sqrt{1+x^8}\,\mathrm dx$

We split up the domain of integration into 10 subintervals,

[0, 1/2], [1/2, 1], [1, 3/2], ..., [4, 9/2], [9/2, 5]

where the left and right endpoints for the $i$ -th subinterval are, respectively,

$\ell_i=\dfrac{5-0}{10}(i-1)=\dfrac{i-1}2$

$r_i=\dfrac{5-0}{10}i=\dfrac i2$

with midpoint

$m_i=\dfrac{\ell_i+r_i}2=\dfrac{2i-1}4$

with $1\le i\le10$ .

Over each subinterval, we interpolate $f(x)=\sqrt{1+x^8}$ with the quadratic polynomial,

$p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m_i)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}$

Then

$\displaystyle\int_0^5f(x)\,\mathrm dx\approx\sum_{i=1}^{10}\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx$

It turns out that the latter integral reduces significantly to

$\displaystyle\int_0^5f(x)\,\mathrm dx\approx\frac56\left(f(0)+4f\left(\frac{0+5}2\right)+f(5)\right)=\frac56\left(1+\sqrt{390,626}+\dfrac{\sqrt{390,881}}4\right)$