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Tamiku [17]
3 years ago
6

Which statement is true about the product (9x2 – 4y2)(3x – 2y)?

Mathematics
1 answer:
nirvana33 [79]3 years ago
6 0

Answer:

It's the last choice.

Step-by-step explanation:

1.  (3x - 2y)(3x -2y)

= 9x^2 - 12xy + 4y^2

The product is  (9x^2 - 4y^2) (9x^2 - 12xy + 4y^2)

which is neither a  difference of 2 squares  or perfect square trinomial.

2.  (3x - 2y)(3x + 2y)

= 9x^2 - 6xy + 6xy - 4y^2

= 9x^2 - 4y^2

and (9x^2 - 4y^2(9x^2 - 4y^2) is a perfect square.

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4 0
4 years ago
The figure below shows a shaded rectangular region inside a large rectangle: (pic)
mariarad [96]

Answer:

D


Step-by-step explanation:

Probability of falling NOT in the shaded region is "area of white region" <em>divided by</em> "area of whole rectangle (big one)".


<u>Area of whole rectangle:</u>

Area of rectangle = length * width = 10 * 5 = 50


<u>Area of White Region:</u>

First, area of shaded region = length * width = 4 * 2 = 8

Now,

Area of white region = area of whole rectangle - area of shaded rectangle

Area of white region = 50 - 8 = 42


Hence, probability is \frac{42}{50}=0.84

0.84 = 84%

Answer choice D is right.





8 0
3 years ago
F(x)=2x^2-3x+7 evaluate f(-2)
Vilka [71]

Answer:

Step-by-step explanation:

Find two linear functions p(x) and q(x) such that (p (f(q(x)))) (x) = x^2 for any x is a member of R?

Let  p(x)=kpx+dp  and  q(x)=kqx+dq  than

f(q(x))=−2(kqx+dq)2+3(kqx+dq)−7=−2(kqx)2−4kqx−2d2q+3kqx+3dq−7=−2(kqx)2−kqx−2d2q+3dq−7  

p(f(q(x))=−2kp(kqx)2−kpkqx−2kpd2p+3kpdq−7  

(p(f(q(x)))(x)=−2kpk2qx3−kpkqx2−x(2kpd2p−3kpdq+7)  

So you want:

−2kpk2q=0  

and

kpkq=−1  

and

2kpd2p−3kpdq+7=0  

Now I amfraid this doesn’t work as  −2kpk2q=0  that either  kp  or  kq  is zero but than their product can’t be anything but  0  not  −1 .

Answer: there are no such linear functions.

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2 years ago
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