All categories

3 years ago

Z=a/b+c/d solve for A

Mathematics

1 answer:

PilotLPTM [1.2K]3 years ago

6 0

Answer: $a=\frac{b(zd-c)}{d}$

Step-by-step explanation:

Having the following equation given in the exercise:

$z=\frac{a}{b}+\frac{c}{d}$

You can solve for "a" following this procedure:

1. You can apply the Subtraction property of equality and subtract $\frac{c}{d}$ from both sides of the equation:

$z-(\frac{c}{d})=\frac{a}{b}+\frac{c}{d}-(\frac{c}{d})\\\\z-\frac{c}{d}=\frac{a}{b}$

2. Now you must subtract the terms on the left side of the equation. Notice that the Least Common Denominator is "d". Then:

$\frac{zd-c}{d}=\frac{a}{b}$

3. Finally, you can apply the Multiplication property of equality and multiply both sides of the equation by "b". So, you get:

$(b)(\frac{zd-c}{d})=(\frac{a}{b})(b)\\\\a=\frac{b(zd-c)}{d}$

You might be interested in

What is the solution of

kobusy [5.1K]

Answer:

Third option: x=0 and x=16

Step-by-step explanation:

$\sqrt{2x+4}-\sqrt{x}=2$

Isolating √(2x+4): Addind √x both sides of the equation:

$\sqrt{2x+4}-\sqrt{x}+\sqrt{x}=2+\sqrt{x}\\ \sqrt{2x+4}=2+\sqrt{x}$

Squaring both sides of the equation:

$(\sqrt{2x+4})^{2}=(2+\sqrt{x})^{2}$

Simplifying on the left side, and applying on the right side the formula:

$(a+b)^{2}=a^{2}+2ab+b^{2}; a=2, b=\sqrt{x}$

$2x+4=(2)^{2}+2(2)(\sqrt{x})+(\sqrt{x})^{2}\\ 2x+4=4+4\sqrt{x}+x$

Isolating the term with √x on the right side of the equation: Subtracting 4 and x from both sides of the equation:

$2x+4-4-x=4+4\sqrt{x}+x-4-x\\ x=4\sqrt{x}$

Squaring both sides of the equation:

$(x)^{2}=(4\sqrt{x})^{2}\\ x^{2}=(4)^{2}(\sqrt{x})^{2}\\ x^{2}=16 x$

This is a quadratic equation. Equaling to zero: Subtract 16x from both sides of the equation:

$x^{2}-16x=16x-16x\\ x^{2}-16x=0$

Factoring: Common factor x:

x (x-16)=0

Two solutions:

1) x=0

2) x-16=0

Solving for x: Adding 16 both sides of the equation:

x-16+16=0+16

x=16

Let's prove the solutions in the orignal equation:

1) x=0:

$\sqrt{2x+4}-\sqrt{x}=2\\ \sqrt{2(0)+4}-\sqrt{0}=2\\ \sqrt{0+4}-0=2\\ \sqrt{4}=2\\ 2=2$

x=0 is a solution

2) x=16

$\sqrt{2x+4}-\sqrt{x}=2\\ \sqrt{2(16)+4}-\sqrt{16}=2\\ \sqrt{32+4}-4=2\\ \sqrt{36}-4=2\\ 6-4=2\\ 2=2$

x=16 is a solution

Then the solutions are x=0 and x=16

5 0

3 years ago

The midsegment of a trapezoid is always parallel to each base true or false

Deffense [45]

Answer:

True

Step-by-step explanation:

we know that

The Trapezoid Mid-segment Theorem states that : A line connecting the midpoints of the two legs of a trapezoid is parallel to the bases, and its length is equal to half the sum of lengths of the bases

see the attached figure to better understand the problem

EF is the mid-segment of trapezoid

EF is parallel to AB and is parallel to CD

EF=(AB+CD)/2

The mid-segment of a trapezoid is always parallel to each base

therefore

The statement is true

8 0

3 years ago

Type your answer in the box. Evaluate y3 – 2y (y – 5) + 13 when y = -3

kap26 [50]

Answer:

-62

Step-by-step explanation:

y³ – 2y (y – 5) + 13

when y = -3

y³ – 2y (y – 5) + 13

=(-3)³ – 2(-3) [ (-3) – 5] + 13 (by PEDMAS, evaluate parentheses first)

=(-27) + (6) (-8) + 13 (next do multiplication)

=(-27) + (-48) + 13

= -27 - 48 + 13

= -62

3 0

3 years ago

Pls help

bonufazy [111]

Answer:

$\underline{ \boxed{z\overline{z} = 34}}$

Step-by-step explanation:

$if \: z = -3 + 5i \\then \: \overline{z} = - (3 + 5i) \\ hence \: z\overline{z} = (-3 + 5i)( - 3 - 5i) \\ z\overline{z} = 9 + 15i - 15i - {(5i)}^{2} \\ z\overline{z} = 9 - (25 ){i}^{2} \\ z\overline{z} = 9 - (25)( - 1) \\ z\overline{z} = 9 + 25 \\ \underline{ \boxed{z\overline{z} = 34}}$