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3 years ago

I already solved this problem. I just need help with the HINT part.

1 answer:

8 0

$x+y=25 \\ y=25-x~~Other~number~in~terms~of~x \\ \\ x^2+y^2=313 \\ x^2+2xy+y^2=313+2xy \\ (x+y)^2=313+2x(25-x) \\ \frac{25^2-313}{2} =25x-x^2 \\ x^2-25x+156 \\ \\ \Delta =(-25)^2-4\times 1\times 156 \\ \Delta =625-624 \\ \sqrt{\Delta}=1 \\ \\ x= \frac{25 \pm 1}{2} \\ x= \left \{ {{x_1=13} \atop {x_2=12}} \right.$

The number "x" is given by:
$y= \left \{ {{y_1=25-13 \rightarrow~y_1=12} \atop {y_1=25-12 \rightarrow ~y_1=13}} \right.$

Therefore, we have:

$\boxed {S=({12,13})} \rightarrow~The~curly~braces~does~not~want~appear~here.$

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1. Jillian Knits hats for her family and friends. In the week, she knits 4 hats, the next week, she has some help and together,

alexandr1967 [171]

Question 1: If shes making 4 hats alone and then Quadruples it shes making 16 hats per week if she quadruples it again she will be knitting 64 hats per week.

4x4=16

16x4=64hats

Question 2: There is an error because 5x5x5 = 5³

Question 3: 2⁴ - 8 x 2 / 4 = 12

2⁴ = 16 - 8 x 2 / 4

8x2 = 16

16/4 = 4

16-4 = 12

7 0

3 years ago

Let f(x) = 1/x^2 (a) Use the definition of the derivatve to find f'(x). (b) Find the equation of the tangent line at x=2

Verdich [7]

Answer:

(a) $f'(x)=-\frac{2}{x^3}$

(b) $y=-0.25x+0.75$

Step-by-step explanation:

The given function is

$f(x)=\frac{1}{x^2}$ .... (1)

According to the first principle of the derivative,

$f'(x)=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{x^2-(x+h)^2}{x^2(x+h)^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{x^2-x^2-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-h(2x+h)}{hx^2(x+h)^2}$

Cancel out common factors.

$f'(x)=lim_{h\rightarrow 0}\frac{-(2x+h)}{x^2(x+h)^2}$

By applying limit, we get

$f'(x)=\frac{-(2x+0)}{x^2(x+0)^2}$

$f'(x)=\frac{-2x)}{x^4}$

$f'(x)=\frac{-2)}{x^3}$ .... (2)

Therefore $f'(x)=-\frac{2}{x^3}$ .

(b)

Put x=2, to find the y-coordinate of point of tangency.

$f(x)=\frac{1}{2^2}=\frac{1}{4}=0.25$

The coordinates of point of tangency are (2,0.25).

The slope of tangent at x=2 is

$m=(\frac{dy}{dx})_{x=2}=f'(x)_{x=2}$

Substitute x=2 in equation 2.

$f'(2)=\frac{-2}{(2)^3}=\frac{-2}{8}=\frac{-1}{4}=-0.25$

The slope of the tangent line at x=2 is -0.25.

The slope of tangent is -0.25 and the tangent passes through the point (2,0.25).

Using point slope form the equation of tangent is

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

$y-0.25=-0.25(x-2)$

$y-0.25=-0.25x+0.5$

$y=-0.25x+0.5+0.25$

$y=-0.25x+0.75$

Therefore the equation of the tangent line at x=2 is y=-0.25x+0.75.

5 0

3 years ago

A jet travels at 483 miles per hour. Which equation represents

WINSTONCH [101]

Answer:

d=483t

Step-by-step explanation:

If i recall correctly

distance = velocity * time

This means that

d=483t

8 0

3 years ago

Sophia‘s favorite homemade cookie recipe requires one cup of chocolate chips for 10 servings if the number of cups required for

weqwewe [10]

Answer:

3 cups

Step-by-step explanation:

We can use a proportion to find how many cups of chocolate chips she needs for 30 servings. Assuming c = cups of chocolate chips and b = batches

$\frac{c}{b}$

$\frac{1}{10} = \frac{c}{30}$

We can now multiply the diagonal values that don't include the missing variable (30 and 1) and then divide it by the value that is diagonal to the variable (10)

$30 \cdot 1 = 30\\30 \div 10 = 3$

Therefore, she needs 3 cups of chocolate chips to make 30 servings.

8 0

3 years ago

Read 2 more answers

317 multiplied by d<br> 317*d

Goshia [24]

The answer would be 317d or you could
do 317(d)

3 0

3 years ago