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3 years ago

10

Which is the quotient of 1.54 + 0.44 ?

1 answer:

Anestetic [448]3 years ago

3 0

Answer

1.98

Step-by-step explanation:

1.54 + 0.44 = 1.98

hope it helps.

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140 + (-122) + (-10), I also need it with work.

Aleks04 [339]

Answer:

8

Step-by-step explanation:

140-122=18

18-10=8

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3 years ago

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2p^2+2p-4 solve using the AC method or the sum product method .. PLEASE HELP thank you :)))

Leviafan [203]

Answer: 2(p+2) * (p-1)

Step-by-step explanation:

2p^2 + 2p - 4

2 (p^2 + p - 2)

2 (p^2 + 2p - p - 2)

2(p x (p+2) - (p+2)

2 (p+2) * (p-1)

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3 years ago

What is the quotient if 4X x equals 7

eduard

The answer is 28 because the X is being multiplied by 4

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3 years ago

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PLSS HELP IM CONFUSED ON THIS QUESTION The figure below shows a transversal t which intersects the parallel lines PQ and RS:

n200080 [17]

1. 4 and 8 ==》 <em>corresponding angles</em>

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2. 5 and 6 ==》 <em>linear</em><em> </em><em>pair</em><em> </em><em>angles</em>

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3. 2 and 8 ==》 <em>alternate exterior angles </em>

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4. 3 and 6 ==》 <em>same side interior angle</em>s

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5. 2 and 4 ==》 <em>vertical</em><em> </em><em> </em><em>angles</em><em> </em>

8 0

3 years ago

Х- а<br>x-b<br>If f(x) = b.x-a÷b-a + a.x-b÷a - b<br>Prove that: f (a) + f(b) = f (a + b)

GenaCL600 [577]

Given:

Consider the given function:

$f(x)=\dfrac{b\cdot(x-a)}{b-a}+\dfrac{a\cdot(x-b)}{a-b}$

To prove:

$f(a)+f(b)=f(a+b)$

Solution:

We have,

$f(x)=\dfrac{b\cdot(x-a)}{b-a}+\dfrac{a\cdot (x-b)}{a-b}$

Substituting $x=a$ , we get

$f(a)=\dfrac{b\cdot(a-a)}{b-a}+\dfrac{a\cdot (a-b)}{a-b}$

$f(a)=\dfrac{b\cdot 0}{b-a}+\dfrac{a}{1}$

$f(a)=0+a$

$f(a)=a$

Substituting $x=b$ , we get

$f(b)=\dfrac{b\cdot(b-a)}{b-a}+\dfrac{a\cdot (b-b)}{a-b}$

$f(b)=\dfrac{b}{1}+\dfrac{a\cdot 0}{a-b}$

$f(b)=b+0$

$f(b)=b$

Substituting $x=a+b$ , we get

$f(a+b)=\dfrac{b\cdot(a+b-a)}{b-a}+\dfrac{a\cdot (a+b-b)}{a-b}$

$f(a+b)=\dfrac{b\cdot (b)}{b-a}+\dfrac{a\cdot (a)}{-(b-a)}$

$f(a+b)=\dfrac{b^2}{b-a}-\dfrac{a^2}{b-a}$

$f(a+b)=\dfrac{b^2-a^2}{b-a}$

Using the algebraic formula, we get

$f(a+b)=\dfrac{(b-a)(b+a)}{b-a}$ $[\because b^2-a^2=(b-a)(b+a)]$

$f(a+b)=b+a$

$f(a+b)=a+b$ [Commutative property of addition]

Now,

$LHS=f(a)+f(b)$

$LHS=a+b$

$LHS=f(a+b)$

$LHS=RHS$

Hence proved.

5 0

3 years ago