Is this problem perpendicular parallel or neither<br> x+7x=-3 and y=7x+25

A test worth 125 points has 49 questions on it. Multiple Choice questions are worth 2 points each and short answer questions are

Charra [1.4K]

Answer:

There were 40 2-point questions and 9 3-point questions.

Step-by-step explanation:

To solve this situation, write two equations. One for the number of questions and one for the number of points.

Let x be the number of 2 point questions.

Let y be the number of 5 point questions.

x + y = 49

Now since x are each worth 2 points, the total points is 2x.

And since y are each worth 5 points, the total points is 5y.

So 2x + 5y = 125. Substitute one of the equations into the other to solve for the variables.

x + y = 49 becomes x = 49 - y. Substitute it.

2(49-y) + 5y = 125

98 - 2y + 5y = 125

98 + 3y = 125

3y = 27

y = 9

Substitute y = 9 back into the equation x = 49 - y to find x.

x = 49 - 9

x = 40

5 0

3 years ago

A positive real number is 2 less than another. When 4 times the larger is added to the square of the smaller, the result is 49.

Kobotan [32]

Answer:

The numbers are

$-2+3\sqrt{5}$ and $3\sqrt{5}$

Step-by-step explanation:

Let

x -----> the smaller positive real number

y -----> the larger positive real number

we know that

A positive real number is 2 less than another

so

$x=y-2$

$y=x+2$ ----> equation A

When 4 times the larger is added to the square of the smaller, the result is 49

so

$4y+x^2=49$ ----> equation B

substitute equation A in equation B

$4(x+2)+x^2=49$

solve for x

$x^2+4x-41=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$x^2+4x-41=0$

so

$a=1\\b=4\\c=-41$

substitute in the formula

$x=\frac{-4\pm\sqrt{4^{2}-4(1)(-41)}} {2(1)}$

$x=\frac{-4\pm\sqrt{180}} {2}$

$x=\frac{-4\pm6\sqrt{5}} {2}$

$x=-2\pm3\sqrt{5}$

so

The positive real number is

$x=-2+3\sqrt{5}$

Find the value of y

$y=x+2$

$y=-2+3\sqrt{5}+2$

$y=3\sqrt{5}$

6 0

3 years ago

What is the range of this<br> graph?

ozzi

Answer:

Step-by-step explanation: Another way to identify the domain and range of functions is by using graphs. Because the domain refers to the set of possible input values, the domain of a graph consists of all the input values shown on the x-axis.

5 0

3 years ago

Brainliest for right answer!!!<3

GenaCL600 [577]

The value of a should be 2000. Hope this helps.

6 0

3 years ago

Consider the function below. (If an answer does not exist, enter DNE.) f(x) = x3 − 27x + 3 (a) Find the interval of increase. (E

xxTIMURxx [149]

Answer:

(-∞,-3) and (3,∞)

Step-by-step explanation:

f(x) = x³ − 27x + 3

1. Find the critical points

(a) Calculate the first derivative of the function.

f'(x) = 3x² -27

(b) Factor the first derivative

f'(x)= 3(x² - 9) = 3(x + 3) (x - 3)

(c) Find the zeros

3(x + 3) (x - 3) = 0

x + 3 = 0 x - 3 = 0

x = -3 x = 3

The critical points are at <em>x = -3</em> and x = 3.

2. Find the local extrema

(a) x = -3

f(x) = x³ − 27x + 3 = (-3)³ - 27(-3) + 3 = -27 +81 + 3 = 57

(b) x = 3

f(x) = x³ − 27x + 3 = 3³ - 27(3) + 3 = 27 - 81 + 3 = -51

The local extrema are at (-3,57) and (3,-51).

3, Identify the local extrema as maxima or minima

Test the first derivative (the slope) over the intervals (-∞, -3), (-3,3), (3,∞)

f'(-4) = 3x² -27 = 3(4)² - 27 = 21

f'(0) = 3(0)² -27 = -27

f'(4) = 3(4)² - 27 = 51

The function is increasing on the intervals (-∞,-3) and (3,∞).

The graph below shows the critical points of your function.

6 0

3 years ago

Is this problem perpendicular parallel or neither x+7x=-3 and y=7x+25