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3 years ago

Simplify the expression: a. b^2+49–(b–7)^2 b. 2a+6b)^2–24ab

1 answer:

5 0

Part a b2+49−(b−7)2

Distribute:

=b2+49+−b2+14b+−49

Combine Like Terms:

=b2+49+−b2+14b+−49

=(b2+−b2)+(14b)+(49+−49)

=14b

Answer:

=14b
part b

(2a+6b)2−24ab

Distribute:

=4a2+24ab+36b2+−24ab

Combine Like Terms:

=4a2+24ab+36b2+−24ab

=(4a2)+(24ab+−24ab)+(36b2)

=4a2+36b2

Answer:

=4a2+36b2

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dalvyx [7]

Answer:

13) $(5x)^{-\frac{5}{4}$ ⇒ $\frac{1}{\sqrt[4]{(5x)^5}}$

15) $(10n)^{\frac{3}{2}$ ⇒ $\sqrt{(10n)^3}$

Step-by-step explanation:

Given expression:

13) $(5x)^{-\frac{5}{4}$

15) $(10n)^{\frac{3}{2}$

Write the expressions in radical form.

Solution:

For an expression with exponents as fraction like

$(x)^{\frac{m}{n}$

the numerator $m$ represents the power it is raised to and the denominator $n$ represents the nth root of the expression.

For an expression with exponents as negative fraction like

$(x)^{-\frac{m}{n}$

We take the reciprocal of the term by rule for negative exponents.

So it is written as:

$\frac{1}{(x)^{\frac{m}{n}}}$

using the above properties we can write the given expressions in radical form.

13) $(5x)^{-\frac{5}{4}$

⇒ $\frac{1}{(5x)^{\frac{5}{4}}}$ [Using rule of negative exponents]

⇒ $\frac{1}{\sqrt[4]{(5x)^5}}$ [writing in radical form]

15) $(10n)^{\frac{3}{2}$

⇒ $\sqrt{(10n)^3}$ [Since 2nd root is given as $\sqrt{}$ in radical form]

3 0

3 years ago

Given the equation y − 4 = three fourths(x + 8) in point-slope form, identify the equation of the same line in standard form

Ahat [919]

Answer:

-3x + 4y = 40.

Step-by-step explanation:

Let's convert the equation to standard form:

y - 4 = 3/4(x + 8)

y - 4 = 3/4 x + 6

y = 3/4x + 10

Multiply through be 4:

4y = 3x + 40

-3x + 4y = 40 (answer).

4 0

3 years ago

Danny made a mistake in the following problem.

Olenka [21]

The mistake is made on line 3, after calculating 35/7, he should have started with the multiplication 6(2)

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4 years ago

Determine the open t-intervals on which the curve is concave downward or concave upward. (Enter your answer using interval notat

Rashid [163]

Solution :

We have been given a parametric curve :

x = sin t , y = cos t , 0 < t < π

In order to determine concavity of the given parametric curve, we need to evaluate its second derivative first.

Therefore,

$\frac{dx}{dt} = \cos t, \ \frac{dy}{dt}= - \sin t$

$\therefore \frac{dy}{dx}= \frac{- \sin t}{\cos t}$

$=- \tan t$

Taking double derivatives of the above equation:

$\frac{d^2y}{dx^2}= - \frac{d}{dx}(\tan t)$

$= - \sec^2 t \frac{dt}{dx}$

$= - \sec^2 t \left(\frac{1}{\cos t}\right)$

$= - \sec^3 t$

For the concave up, we have

$\frac{d^2y}{dx^2} > 0$

$\Rightarrow - \sec^3 t > 0$

∴ $t \ \epsilon \left( \frac{\pi }{2}, \pi \right)$

For the concave down, we have

$\frac{d^2y}{dx^2} < 0$

$\Rightarrow - \sec^3 t < 0$

$t \ \epsilon \left( 0,\frac{\pi }{2} \right)$

8 0

3 years ago

Find the equation of a line with a slope of -1 that passes through the point (2,3). Graph this line.

Len [333]

Answer:

y= -1x+5 or y= -x+5

Step-by-step explanation:

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3 years ago