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2 years ago

Determine whether

Mathematics

1 answer:

lora16 [44]2 years ago

4 0

Answer:

(a) and (b) are not equivalent

Step-by-step explanation:

Given

$\frac{25^m}{5}$

Required

Determine an equivalent or nonequivalent expression

$(a)\ 25^{m-1$

We have:

$25^{m-1$

Apply law of indices

$25^{m-1} = \frac{25^m}{25}$

This is not equivalent to $\frac{25^m}{5}$

$(b)\ 25^{2m - 1}$

We have:

$25^{2m - 1}$

Apply law of indices

$25^{2m - 1} = \frac{25^{2m}}{25}$

This is not equivalent to $\frac{25^m}{5}$

$(c)\ 5^{2m-1}$

We have:

$5^{2m-1}$

Apply law of indices

$5^{2m-1} = \frac{5^{2m}}{5^1}$

$5^{2m-1} = \frac{5^{2m}}{5}$

Evaluate the numerator

$5^{2m-1} = \frac{25m}{5}$

This is an equivalent expression

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Find dy/dx by implicit differentiation. y cos x = 5x2 + 3y2

lbvjy [14]

Step 1:
Start by putting $\frac{d}{dx}$ in front of each term

$\frac{d}{dx}[y cos x]= \frac{d}{dx}[5x^2]+ \frac{d}{dx}[ 3y^2]$
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Step 2:

Deal with the terms in 'x' and the constant terms
$\frac{d}{dx}[ycosx]= 10x+ \frac{d}{dx} [3y^2]$
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Step 3:

Use the chain rule for the terms in 'y'
$\frac{d}{dx}[ycosx]=10x+6y \frac{dy}{dx}$
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Step 4:

Use the product rule on the term in 'x' and 'y'
$(y) \frac{d}{dx} cos x+(cos x) \frac{d}{dx}y =10x+6y \frac{dy}{dx}
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$y(-siny)+(cosx) \frac{dy}{dx} =10x+6y \frac{dy}{dx}$
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Step 5:

Rearrange to make $\frac{dy}{dx}$ the subject
$-y sin(y)+cos(x) \frac{dy}{dx} =10x+6y \frac{dy}{dx}$
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3 years ago

Which two ordered pairs would represent points on a graph of 5x + 6y = 12?

skelet666 [1.2K]

B. (0,2) and D. (6,-3)

4 0

3 years ago

Out of students appeared in an examination 75% passed in English, 55%. passed in Mathematics, 5% failed in both subjects and 21

rusak2 [61]

Based on the percentage that passed English and those who passed Mathematics and those who failed and passed both, the total number of students who appeared in the examination are 60 students.

The number of students who passed only in Math are 12 students.

<h3>What number of students sat in the exam?</h3>

This can be found as:

= Total who passed English only + Total who passed Math only + Total who failed both + Total who passed both

Assuming the total is n, the equation becomes:

n = 0.75n - 21 + 0.55n - 21 + 21 + 0.05n

n = 1.35n - 21

21 = 0.35n

n = 21 / 0.35

= 60 students

The number who passed mathematics only is:

= (60 x 55%) - students who passed both

= 33 - 21

= 12 students

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1 year ago

A rectangle is 4xcm wide and 2cm long what do you notice

deff fn [24]

The width of the rectangle is twice the length of the rectangle

<h3>How to compare the dimensions of the rectangle?</h3>

The given parameters are:

Width = 4 cm

Length = 2 cm

Express 4 cm as 2 * 2 cm

Width = 2 * 2 cm

Substitute Length = 2 cm in Width = 2 * 2 cm

Width = 2 * Length

This means that the width of the rectangle is twice the length of the rectangle

Hence, the true statement is that the width of the rectangle is twice the length of the rectangle