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3 years ago

9

Which of the expressions below can be factored using the difference of squares method?

1 answer:

lara31 [8.8K]3 years ago

7 0

The expression that can be factored using the difference of squares method:
C) $100x^{2}-49y^{2}$
Factorization:
$100x^{2}-49y^{2} = (10x+7y)*(10x-7y)$

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Find the area of figure 1.<br><br> a -12 units2<br> b- 14 units2<br> c- 8 units2<br> d- 10 units2

Brilliant_brown [7]

Answer:

D

Step-by-step explanation:

if you look at the chart. it goes up five and has a width of 2. multiply 5 x 2 = you get 10

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3 years ago

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An angle is equal to one-third of its complement. Find the angle's measure.

artcher [175]

Let $x^\circ$ be the measure of the angle in question. Its complement has measure $(90-x)^\circ$ . We're told that this angle is congruent to 1/3 of its complement, so that

$x^\circ=\dfrac{(90-x)^\circ}3\implies3x^\circ=(90-x)^\circ\implies4x^\circ=90^\circ\implies x=22.5^\circ$

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3 years ago

Anna can buy 3 sweatshirts for a total of $45.H

Masteriza [31]

Answer: Anna could buy 5 sweatshirts for $60.

Step-by-step explanation:

lets figure out the price of one sweatshirt.

45/3 = 15

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5*15=60

so 5 sweatshirts cost $60

6 0

3 years ago

If the regular price of a new T-shirt is $13.

Verdich [7]

Answer: 30% of decrease

Step-by-step explanation:

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3 years ago

Find the solution of the given initial value problem. ty' + 2y = sin t, y π 2 = 9, t > 0 y(t) =

Helen [10]

For the ODE

$ty'+2y=\sin t$

multiply both sides by <em>t</em> so that the left side can be condensed into the derivative of a product:

$t^2y'+2ty=t\sin t$

$\implies(t^2y)'=t\sin t$

Integrate both sides with respect to <em>t</em> :

$t^2y=\displaystyle\int t\sin t\,\mathrm dt=\sin t-t\cos t+C$

Divide both sides by $t^2$ to solve for <em>y</em> :

$y(t)=\dfrac{\sin t}{t^2}-\dfrac{\cos t}t+\dfrac C{t^2}$

Now use the initial condition to solve for <em>C</em> :

$y\left(\dfrac\pi2\right)=9\implies9=\dfrac{\sin\frac\pi2}{\frac{\pi^2}4}-\dfrac{\cos\frac\pi2}{\frac\pi2}+\dfrac C{\frac{\pi^2}4}$

$\implies9=\dfrac4{\pi^2}(1+C)$

$\implies C=\dfrac{9\pi^2}4-1$

So the particular solution to the IVP is

$y(t)=\dfrac{\sin t}{t^2}-\dfrac{\cos t}t+\dfrac{\frac{9\pi^2}4-1}{t^2}$

or

$y(t)=\dfrac{4\sin t-4t\cos t+9\pi^2-4}{4t^2}$

6 0

3 years ago