Please help and explain

X^2+y^2=169, 3x+2y=39 please answer using substitution and each step

Oksanka [162]

Answer:

x=13,y=0

Step-by-step explanation:

We are given a system of equations

${x}^{2} + {y}^{2} = 169 - - eqn \: 1 \\ 3x + 2y = 39 - - eqn \: 2$

For equation 1,square all terms to reduce it

√x²+√y²=169

x+y=13

Make x the subject as required by the question to use substitution method

x=13-y

Plug x=13-y into eqn 2

3(13-y)+2y=39

39-3y+2y=39

39-y=39

y=39-39=0

Plug y=0 into equation 1

x+0=13

x=0

7 0

2 years ago

A factory makes 1200 shirts every 6 hours.the factory makes shirts for 9 hours a workday. what is the fewest number of workdays

Dominik [7]

1. Divide the target number of shirts, 12,600 by the company's rate of making the shirts. That is,
number of hours = 12,600 shirts/ (1200 shirts/6 hours)
= 63 hours
2. Determine the number of workdays for them to complete the shirts by dividing 63 hours by 9 hours/workday. This will give us an answer of 7 workdays.

8 0

3 years ago

Derivative of tan(2x+3) using first principle

kodGreya [7K]

$f(x)=\tan(2x+3)$

The derivative is given by the limit

$f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h$

You have

$\displaystyle\lim_{h\to0}\frac{\tan(2(x+h)+3)-\tan(2x+3)}h$
$\displaystyle\lim_{h\to0}\frac{\tan((2x+3)+2h)-\tan(2x+3)}h$

Use the angle sum identity for tangent. I don't remember it off the top of my head, but I do remember the ones for (co)sine.

$\tan(a+b)=\dfrac{\sin(a+b)}{\cos(a+b)}=\dfrac{\sin a\cos b+\cos a\sin b}{\cos a\cos b-\sin a\sin b}=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$

By this identity, you have

$\tan((2x+3)+2h)=\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}$

So in the limit you get

$\displaystyle\lim_{h\to0}\frac{\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}-\tan(2x+3)}h$
$\displaystyle\lim_{h\to0}\frac{\tan(2x+3)+\tan2h-\tan(2x+3)(1-\tan(2x+3)\tan2h)}{h(1-\tan(2x+3)\tan2h)}$
$\displaystyle\lim_{h\to0}\frac{\tan2h+\tan^2(2x+3)\tan2h}{h(1-\tan(2x+3)\tan2h)}$
$\displaystyle\lim_{h\to0}\frac{\tan2h}h\times\lim_{h\to0}\frac{1+\tan^2(2x+3)}{1-\tan(2x+3)\tan2h}$
$\displaystyle\frac12\lim_{h\to0}\frac1{\cos2h}\times\lim_{h\to0}\frac{\sin2h}{2h}\times\lim_{h\to0}\frac{\sec^2(2x+3)}{1-\tan(2x+3)\tan2h}$

The first two limits are both 1, and the single term in the last limit approaches 0 as $h\to0$ , so you're left with

$f'(x)=\dfrac12\sec^2(2x+3)$

which agrees with the result you get from applying the chain rule.

7 0

3 years ago

Ansley drove 8.7 miles to the park then she drove 12.43 miles to the museum how many miles did Ansley drive in all

Karo-lina-s [1.5K]

Answer: 21.13 miles

Explanation: The question asks “drive in all”. They gave you the miles driven, which is 8.7 and 12.43. If you add them together, you will get 21.13 miles.

5 0

2 years ago

Read 2 more answers

Suppose f is a one-to-one function with f(2)=8 and f’(2)=4. What is the value of (f^-1)’(8)?

yuradex [85]

Answer:

1/4

Step-by-step explanation:

The tangent line to f(x) at x=2 goes through the point (2, 8) and has slope 4.

The inverse function f⁻¹(x) is a reflection of the function f(x) across the line y=x. The corresponding tangent line will go through point (8, 2) and have slope 1/f'(2) = 1/4 at that point. That is, ...

(f⁻¹)'(8) = 1/f'(f⁻¹(8)) = 1/f'(2) = 1/4

5 0

3 years ago