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2 years ago

Solve the equation for a? z=mab. a)a=z+mb b)a=z-mb c)a= mb/z d)a=z/mb

Mathematics

1 answer:

Bingel [31]2 years ago

4 0

Answer:

a = z/mb

Step-by-step explanation:

Given:

$\displaystyle \large{z=mab}$

To solve for a, divide both sides by the variables mb to isolate a-term.

$\displaystyle \large{\dfrac{z}{mb}=\dfrac{mab}{mb}}$

Simplify the expression and we finally have solved for a-term.

$\displaystyle \large{\dfrac{z}{mb}=a}\\\\\displaystyle \large{a=\dfrac{z}{mb}}$

Therefore, the answer to this question is a = z/mb.

Please let me know if you have any questions regarding my answer or explanation!

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Slove the following:

Aleksandr [31]

Answer:

let required no be x

$\frac{1}{7} = \frac{x}{42}$

by doing crisscrossed multiplication

$42 \times 1 = 7 \times x$

42=7x

dividing both side by 7

$\frac{42}{7} = 7x \div 7$

6=x

therefore x=6

therefore value of remaining no.is 6

7 0

3 years ago

Read 2 more answers

Please help! Thanks in advance.

alukav5142 [94]

In this problem, we need to plug in the given x values for $f(x)=0$ and find a and b.

When we plug in 1, we get: $2 \times {1}^{4} - 5 \times {1}^{3} - 14 \times {1}^{2} + a \times 1 + b = 0$

Simplify:
$2 - 5 - 14 + a + b = 0$
$- 17 + a + b = 0$
$a + b = 17$

We got our first statement about the values of the variables. If we find one more we can find those 2 variables.

We have another given root: 4.

Plug it in:
$2 \times {4}^{4} - 5 \times {4}^{3} - 14 \times {4}^{2} + a \times 4 + b = 0$
$512 - 320 - 224 + 4a + b = 0$
$- 32 + 4a + b = 0$
$4a + b = 32$

Now we have our second one. We can combine them:
$a + b = 17 \\ 4a + b = 32$

I use elimination method which is easier here.

Multiply the top equation by -1: $- a - b = - 17 \\ 4a + b = 32$

Add them up:
$3a = 15$

Simplify:
$a = 5$

Now we have a, we can plug in one of those equations to find b:
$5 + b = 17$
$b = 17 - 5$
$b = 12$

So, the answers are $a=5$ and $b=12$ .