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4 years ago

What was John’s age 4 years ago, if he will be z years old in 5 years?

Mathematics

1 answer:

enot [183]4 years ago

5 0

Answer:

z-9

Step-by-step explanation:

Right now, John is z-5. z-5 - 4 is equal to z -9.

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(4x-y=-15)<br>(3x-2y=-10)<br>

Andreyy89

Answer:

Step-by-step explanation:

$\left\{\begin{array}{ccc}4x-y=-15&\text{multiply both sides by (-2)}\\3x-2y=-10\end{array}\right\\\underline{+\left\{\begin{array}{ccc}-8x+2y=30\\3x-2y=-10\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad\qquad-5x=20\qquad\text{divide both sides by (-5)}\\.\qquad\qquad\boxed{x=-4}\\\\\\\text{Put the value of x to the first equation.}\\\\4(-4)-y=-15\\-16-y=-15\qquad\text{add 16 to both sides}\\-y=1\qquad\text{change the signs}\\\boxed{y=-1}$

5 0

3 years ago

Is the trend line linear? If so, write a linear equation that represents the trend line. Show your work. (3 points)

PolarNik [594]

I got 100% on my test today yay <3

5 0

3 years ago

Read 2 more answers

A survey of 500 high school students was taken to determine their favorite chocolate candy. Of the 500 students surveyed, 64 lik

Lyrx [107]

Answer:

466 student

Step-by-step explanation:

The number of students that like at most 2 kinds of these chocolate candies can be gotten from subtracting the number of students that like more than 2 candies from the total number of high school students.

Since 34 like all three kinds of chocolate candy, hence:

Number of students that like at most 2 kinds of these chocolate candies = 500 - 34 = 466

466 students like at most 2 kinds of these chocolate candies

6 0

3 years ago

Consider the two functions:

koban [17]

Answer:

a) The x value of the point where the two equations intersect in terms of a is $x=\frac{40}{4+5a}$

b) The value of the functions at the point where they intersect is $\frac{10 (28 + 15 a)}{4 + 5 a}$

c) The partial derivative of f with respect to $x$ is $\frac{\partial f}{\partial x} = -5a$ and the partial derivative of f with respect to $a$ is $\frac{\partial f}{\partial x} = -5x$

d) The value of $\frac{\partial f}{\partial x}(3,2) = -10$ and $\frac{\partial f}{\partial a}(3,2) = -15$

e) $\upsilon_1=-\frac{3}{4} = -0.75$ and $\upsilon_2=-\frac{3}{4} = -0.75$

f) equation $\upsilon_1 = \frac{-5a\cdot x}{70-5ax}=\frac{ax}{ax-14}$ and $\upsilon_2 = \frac{-5a\cdot a}{70-5ax}=\frac{a^2}{ax-14}$

Step-by-step explanation:

a) In order to find the $x$ we just need to equal the equations and solve for $x$ :

$f(x,a)=g(x)\\70-5xa = 30+4x\\70-30 = 4x+5xa\\40 = x(4+5a)\\\boxed {x = \frac{40}{4+5a}}$

b) Since we need to find the value of the function in the intersection point we just need to substitute the result from a) in one of the functions. As a sanity check , I will do it in both and the value (in terms of $a$ ) must be the same.

$f(x,a)=70-5ax\\f(\frac{40}{4+5a}, a) = 70-5\cdot a \cdot \frac{40}{4+5a}\\f(\frac{40}{4+5a}, a) = 70 - \frac{200a}{4+5a}\\f(\frac{40}{4+5a}, a) = \frac{70(4+5a) -200a}{4+5a}\\f(\frac{40}{4+5a}, a) =\frac{280+350a-200a}{4+5a}\\\boxed{ f(\frac{40}{4+5a}, a) =\frac{10(28+15a)}{4+5a}}$

and for $g(x)$ :

$g(x)=30+4x\\g(\frac{40}{4+5a})=30+4\cdot \frac{40}{4+5a}\\g(\frac{40}{4+5a})=\frac{30(4+5a)+80}{4+5a}\\g(\frac{40}{4+5a})=\frac{120+150a+80}{4+5a}\\\boxed {g(\frac{40}{4+5a})=\frac{10(28+15a)}{4+5a}}$

c) $\frac{\partial f}{\partial x} = (70-5xa)^{'}=70^{'} - \frac{\partial (5xa)}{\partial x}=0-5a\\\frac{\partial f}{\partial x} =-5a$

$\frac{\partial f}{\partial a} = (70-5xa)^{'}=70^{'} - \frac{\partial (5xa)}{\partial a}=0-5x\\\frac{\partial f}{\partial a} =-5x$

d) Then evaluating:

$\frac{\partial f}{\partial x} =-5a\\\frac{\partial f}{\partial x} =-5\cdot 2=-10$

$\frac{\partial f}{\partial a} =-5x\\\frac{\partial f}{\partial a} =-5\cdot 3=-15$

e) Substituting the corresponding values:

$\upsilon_1 = \frac{\partial f(3,2)}{\partial x}\cdot \frac{3}{f(3,2)} \\\upsilon_1 = -10 \cdot \frac{3}{40} = -\frac{3}{4} = -0.75$

$\upsilon_2 = \frac{\partial f(3,2)}{\partial a}\cdot \frac{3}{f(3,2)} \\\upsilon_2 = -15 \cdot \frac{2}{40} = -\frac{3}{4} = -0.75$

f) Writing the equations:

$\upsilon_1=\frac{\partial f (x,a)}{\partial x}\cdot \frac{x}{f(x,a)}\\\upsilon_1=-5a\cdot \frac{x}{70-5xa}\\\upsilon_1=\frac{-5ax}{70-5ax}=\frac{-5ax}{-5(ax-14)}\\\boxed{\upsilon_1=\frac{ax}{ax-14} }$

$\upsilon_2=\frac{\partial f (x,a)}{\partial x}\cdot \frac{a}{f(x,a)}\\\upsilon_2=-5a\cdot \frac{a}{70-5xa}\\\upsilon_2=\frac{-5a^2}{70-5ax}=\frac{-5a^2}{-5(ax-14)}\\\boxed{\upsilon_2=\frac{a^2}{ax-14} }$

8 0

4 years ago

PLS ANSWER TODAY TYSM :p<br><br> -5y+1-3y+2=-9

OLga [1]

We just combine the terms, like this:

-8y = -12

The answer is: y = 1.5 or 3/2.

Hope this helped! c:

7 0

4 years ago