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2 years ago

9

Pls help on #10 will give brainliest

1 answer:

Free_Kalibri [48]2 years ago

4 0

The answer is A hope this helps

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Write 0.326 in two other ways.

Whitepunk [10]

0.326= 326/1000
0.326= 163/500
0.326= 32.6%

5 0

3 years ago

Read 2 more answers

Si 14 lápices cuestan 30

katrin2010 [14]

Then one pencil costs about 2.14 cents.
<span>Entonces un lápiz cuesta alrededor de 2.14 centavos.</span>

8 0

3 years ago

How many solutions does this system have?

Andrew [12]

That system has D. no solutions

8 0

2 years ago

Read 2 more answers

Knitting: You are knitting a blanket. You want the area of the blanket to be 24ft^2. You want the length of the blanket to be 2

andriy [413]

You can use factors to solve. Determine all the factor pairs of 24, find the two that are two numbers apart.
1, 24 X
2, 12 X
3, 8 X
4, 6 YES!

Algebraic way to solve using Quadratics:
l = 2 + w

A = lw

A = (2 + w)w Substitute (2 + w) for l

24 = (2 + w)w Substitute 24 in for the area

24 = 2w + w^2 Distribute

w^2 + 2w - 24 = 0 Set equal to 0 (put in standard form)

(w + 6) (w - 4) = 0 Factor

w + 6 = 0 and w - 4 = 0 Set each factor equal to 0.

So w= -6 or w = 4 ... -6 makes no sense for a length! So the width must be 4 and the length will be 4 + 2, which is 6.

5 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

2 years ago