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3 years ago

Solve w = t(r1 + r2) for t.

Mathematics

2 answers:

MatroZZZ [7]3 years ago

6 0

Answer:

The solution for t is $\frac{w}{(r_1+r_2)}$

Step-by-step explanation:

The given expression is $w = t(r_1+r_2)$

We need to solve for the 't' ,

To solve for t, divide $w = t(r_1+r_2)$ by $(r_1+r_2)$

$\frac{w}{(r_1+r_2)} = t$

Hence, The solution for t is $\frac{w}{(r_1+r_2)}$

kiruha [24]3 years ago

5 0

I think it would be:

w / (r1 + r2) = t

But I'm not sure

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Step-by-step explanation:

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4 years ago

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3 years ago

The cost for 7 dance lessons is 82$. The cost for 11 lessons is 122$. Write and solve an equation to find the cost of 4 lessons

Juliette [100K]

Answer:

$52

Step-by-step explanation:

Using the equation

y=mx+c

82=7x+c

122=11x+c

11x-7x=122-82

4x=40

x=10

To get c

82=7(10)+c

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6 0

3 years ago

An equation parallel and perpendicular to 4x+5y=19

UNO [17]

Answer:

Parallel line:

$y=-\frac{4}{5}x+\frac{9}{5}$

Perpendicular line:

$y=\frac{5}{4}x-\frac{1}{2}$

Step-by-step explanation:

we are given equation 4x+5y=19

Firstly, we will solve for y

$4x+5y=19$

we can change it into y=mx+b form

$5y=-4x+19$

$y=-\frac{4}{5}x+\frac{19}{5}$

so,

$m=-\frac{4}{5}$

Parallel line:

we know that slope of two parallel lines are always same

so,

$m'=-\frac{4}{5}$

Let's assume parallel line passes through (1,1)

now, we can find equation of line

$y-y_1=m'(x-x_1)$

we can plug values

$y-1=-\frac{4}{5}(x-1)$

now, we can solve for y

$y=-\frac{4}{5}x+\frac{9}{5}$

Perpendicular line:

we know that slope of perpendicular line is -1/m

so, we get slope as

$m'=\frac{5}{4}$

Let's assume perpendicular line passes through (2,2)

now, we can find equation of line

$y-y_1=m'(x-x_1)$

we can plug values

$y-2=\frac{5}{4}(x-2)$

now, we can solve for y

$y=\frac{5}{4}x-\frac{1}{2}$

4 0

3 years ago

I need help please and thank you

sergeinik [125]

Try googling it , sorry

8 0

3 years ago