Ask question

Ask question

All categories

allochka39001 [22]

3 years ago

7

I am making two kinds of cookies: chocolate chip and lemon cookies.

1 answer:

Shkiper50 [21]3 years ago

8 0

Answer:

Step-by-step explanation:

ljkml

You might be interested in

Helppppp!! im confused lol

Lady bird [3.3K]

Answer:

a

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Resolve into partial fractions<br><br>

lisabon 2012 [21]

Answer:

See below

Step-by-step explanation:

1)

$\frac{11 + x}{(2 - x)(x - 3)} = \frac{A}{2 - x} + \frac{B}{x - 3} \\ \\ \therefore \: \frac{11 + x}{(2 - x)(x - 3)} = \frac{A(x - 3) +B(2 - x) }{(2 - x)(x - 3)} \\ \\ \therefore \: 11 + x = Ax - 3A + 2B - Bx \\ \\ \therefore \: 11 + x = - 3A + 2B + Ax - Bx\\ \\ \therefore \: 11 + x = - 3A + 2B + (A - B)x \\ \\ equating \: the \: like \: terms \: on \: both \: sides \\ A - B = 1 \\ \therefore \: A = B + 1.....(1) \\ \\ - 3A + 2B = 11....(2) \\ from \: eq \: (1) \: and \: (2) \\ - 3(B + 1) + + 2B = 11 \\ \\ - 3B - 3 + 2B = 11 \\ \\ - B = 11 + 3 \\ \\ B = - 14 \\ A = - 14 + 1 = - 13 \\ \\ \frac{11 + x}{(2 - x)(x - 3)} = \frac{ - 13}{2 - x} + \frac{ - 14}{x - 3} \\ \\ \frac{11 + x}{(2 - x)(x - 3)} = - \frac{ 13}{2 - x} - \frac{ 14}{x - 3}$

2)

$\frac{12x + 11}{x^2 +x - 6}$

$=\frac{12x + 11}{x^2 +3x-2x - 6}$

$=\frac{12x + 11}{x(x +3) -2(x +3)}$

$\therefore \frac{12x + 11}{x^2 +x - 6}=\frac{12x + 11}{(x +3) (x - 2)}$

$\therefore \frac{12x + 11}{x^2 +x - 6}=\frac{A}{(x +3)}+\frac{B}{(x - 2)}$

$\therefore \frac{12x + 11}{x^2 +x - 6}=\frac{A(x-2)+B(x+3)}{(x-2)(x+3)}$

$\therefore 12x + 11 = A(x-2)+B(x+3)$

$\therefore 12x + 11 = Ax-2A+Bx+3B$

$\therefore 12x + 11 = Ax+Bx-2A+3B$

$\therefore 12x + 11 = (A+B) x-2A+3B$

Equating like terms on both sides:

$A + B = 12\implies A = 12-B... (1)$

$- 2A + 3B = 11... (2)$

Solving equations (1) & (2), we find:

A = 5, B = 7

$\therefore \frac{12x + 11}{x^2 +x - 6}=\frac{5}{(x +3)}+\frac{7}{(x - 2)}$

3 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

Decide if each equation is true or false.

julia-pushkina [17]

The third and fifth ones are false. Two negative numbers in the same equation make a positive not a negative

6 0

3 years ago

Read 2 more answers

You have $9.33. You buy 7 items that cost $0.87 each. How much money do you have left?

liberstina [14]

Answer:

$3.24

Step-by-step explanation:

0.87 x 7 = 6.09

9.33 - 6.09 = 3.24

3 0

3 years ago