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Elodia [21]
3 years ago
14

Which equation is an identity?

Mathematics
2 answers:
OleMash [197]3 years ago
4 0
5w + 8 - w = 6w - 2(w - 4)

Hope this helps :)
sasho [114]3 years ago
3 0

Answer:

Option b - 5w + 8 - w = 6w - 2( w - 4)  

Step-by-step explanation:

To find : Which equation is an identity?    

Solution :

Option a - 11- (2v + 3) = -2v - 8          

If we open the parenthesis,

11- 2v -3 = -2v - 8    

-2v+8\neq -2v - 8    

No identity applied.

Option b - 5w + 8 - w = 6w - 2( w - 4)          

If we apply distributive property, a(b+c)=ab+ac

5w + 8 - w = 6w - 2w+8    

4w+8= 4w+8    

Distributive identity is applied.

Option c - 7m - 2 = 8m + 4 - m          

Solve by subtracting in RHS

7m - 2 \neq 7m + 4    

No identity applied.

Option d- 8y + 9 = 8y - 3          

This is not possible as 8y + 9 \neq 8y - 3

No identity applied.          

Therefore, Option b is correct.

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What are the steps for adding and subtracting mixed numbers with unlike denominators
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3 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.

6 0
2 years ago
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