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3 years ago

Help plz 15. and 16. mathexplain and answer is on side already

Mathematics

1 answer:

Darina [25.2K]3 years ago

5 0

15) x^2-2x-8=1/4x-1
x^2-2.25x-7=0
4x^2-9x-28=0
(x-4)(4x+7)=0
x-4= 0 or 4x+7=0
x=4 or x= -7/4=-1.75
16) answer hides in the f(x) equation
the trick is to think of the value that will make the (x-4) =0 and and the y value can be directly plugged off from the +c
therefore, ans is x=2 y= 4
(2,4)

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Evaluate the Riemann sum for f(x) = 3 - 1/2 times x between 2 and 14 where the endpoints are included with six subintervals taki

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Step-by-step explanation:

Given that :

we are to evaluate the Riemann sum for $f(x) = 3 - \dfrac{1}{2}x$ from 2 ≤ x ≤ 14

where the endpoints are included with six subintervals, taking the sample points to be the left endpoints.

The Riemann sum can be computed as follows:

$L_6 = \int ^{14}_{2}3- \dfrac{1}{2}x \dx = \lim_{n \to \infty} \sum \limits ^6 _{i=1} \ f (x_i -1) \Delta x$

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$\Delta x = \dfrac{b-a}{a}$

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$\Delta x = \dfrac{14-2}{6}$

$\Delta x = \dfrac{12}{6}$

$\Delta x =2$

Hence;

$x_0 = 2 \\ \\ x_1 = 2+2 =4\\ \\ x_2 = 2 + 2(2) \\ \\ x_i = 2 + 2i$

Here, we are using left end-points, then:

$x_i-1 = 2+ 2(i-1)$

Replacing it into Riemann equation;

$L_6 = \lim_{n \to \infty} \sum \imits ^{6}_{i=1} \begin {pmatrix}3 - \dfrac{1}{2} \begin {pmatrix} 2+2 (i-1) \end {pmatrix} \end {pmatrix}2$

$L_6 = \lim_{n \to \infty} \sum \imits ^{6}_{i=1} 6 - (2+2(i-1))$

$L_6 = \lim_{n \to \infty} \sum \imits ^{6}_{i=1} 6 - (2+2i-2)$

$L_6 = \lim_{n \to \infty} \sum \imits ^{6}_{i=1} 6 -2i$

$L_6 = \lim_{n \to \infty} \sum \imits ^{6}_{i=1} 6 - \lim_{n \to \infty} \sum \imits ^{6}_{i=1} 2i$

$L_6 = \lim_{n \to \infty} \sum \imits ^{6}_{i=1} 6 - 2 \lim_{n \to \infty} \sum \imits ^{6}_{i=1} i$

Estimating the integrals, we have :

$= 6n - 2 ( \dfrac{n(n-1)}{2})$

= 6n - n(n+1)

replacing thevalue of n = 6 (i.e the sub interval number), we have:

= 6(6) - 6(6+1)

= 36 - 36 -6

= -6

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