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2 years ago

ANY MATH EXPERTS PLS HELP RN I GOT 30 MIN LEFT I PUT 100 POINTS

Mathematics

2 answers:

hjlf2 years ago

8 0

Answer:

Put the green dot that's on the left to the number six on the vertical line, or Y-axis, then the dot on the right goes 1 down, then 4 to the right

Step-by-step explanation:

Alika [10]2 years ago

8 0

Answer:

5 actually ...................

Step-by-step explanation:

You might be interested in

6x-5y=5 3x+5y=4 The x-coordinate of the solution to this system of equation is

Anna35 [415]

Answer:

x=1

Step-by-step explanation:

1) 6x-5y=5

2) 3x+5y=4

ets perform the following operation

1) +2), This leads to the following equation:

6x+3x-5y+5y=5+4

From where we obtain the solution for x

9x=9

x=1

3 0

3 years ago

∠1 and ∠2 are vertical angles. If m 1=3x-12 and m2=72°, find the value of x

inessss [21]

Answer:

x=28

Step-by-step explanation:

vertical angles are congruent

3x-12=72

3x=84

x=28

6 0

2 years ago

The school store opened on the first day of school with 45 notebooks and 15 pencils. Within two days it sold all of these items.

malfutka [58]

The number of pencils sold on the first day is 5 pencils.

The number of notebooks sold on the first day is 10 notebooks.

The number of pencils sold on the second day 10 pencils.

The number of notebooks sold on the second day 35 notebooks.

Step-by-step explanation:

Given data,

The total number of notebooks in the store = 45 notebooks.
The total number of Pencils in the store = 15 pencils.
The number of days that all items were sold = 2 days.

Now, you have to calculate the no.of notebooks and no. of pencils sold each day.

In the first day :

The number of sale of notebooks and pencils are given by,

Twice as many notebooks were sold as pencils.

Let us take, the number of pencils sold on the first day = x

And, the number of notebooks sold on the first day = 2x (Twice as pencils).

In the second day :

The number of sale of notebooks and pencils are given by,

For every 7 notebooks sold, 2 pencils were sold.

The number of pencils sold on the second day = 2y

The number of notebooks sold on the second day = 7y

The equation is framed for number of notebooks sold on each day :

The number of notebooks sold ⇒ 2x + 7y = 45 -------(1)

The equation is framed for number of pencils sold on each day :

The number of pencils sold: x + 2y = 15 ----------(2)

Solving the equations by multiplying eq(2) by 2 and subtract it from eq(1),

2x + 7y = 45

- (2x + 4y) = 30

3y = 15

⇒ y = 15/3

⇒ y = 5

The value of y is 5.

Substitute y=5 in eq (2),

⇒ x + 2(5) = 15

⇒ x + 10 = 15

⇒ x = 15 - 10

⇒ x = 5

The value of x is 5.

First day sale,

The number of pencils sold on the first day = x ⇒ 5 pencils

The number of notebooks sold on the first day = 2x ⇒ 10 notebooks

Second day sale,

The number of pencils sold on the second day = 2y ⇒ 10 pencils

The number of notebooks sold on the second day = 7y ⇒ 35 notebooks.

8 0

2 years ago

100 POINT QUESTION!!!!!!!!!!!!!!!!!!!!!!!!!!!!

babunello [35]

Answer:

Ok I don't know waht you mean but the answer is

Step-by-step explanation:

Answer:

x=3/2±(√11)2

x=1.5+1.65831i

x=1.5−1.65831i

Find the Solution for

2x^22−6x+10=0

using the Quadratic Formula where

a = 2, b = -6, and c = 10

x=(−b±√(b^2−4ac))/2a

x=(−(−6)±√((−6)2−4(2)(10)))2(2)

x=(6±√(36−80))/4

x=(6±√−44)/4

The discriminant b^2−4ac<0

so, there are two complex roots.

Simplify the Radical:

x=(6±2√11 i)/4

x=6/4±(2√11 i)4

Simplify fractions and/or signs:

x=3/2±(√11)2

which becomes

x=1.5+1.65831i

x=1.5−1.65831i

4 0

2 years ago

Determine the maximized area of a rectangle that has a perimeter equal to 56m by creating and solving a quadratic equation. What

sveticcg [70]

Answer:

Area of rectangle = $196\,m^2$

Length of rectangle = 14 m

Width of rectangle = 14 m

Step-by-step explanation:

Given:

Perimeter of rectangle is 56 m

To find: the maximized area of a rectangle and the length and width

Solution:

A function $y=f(x)$ has a point of maxima at $x=x_0$ if $f''(x_0)$

Let x, y denotes length and width of the rectangle.

Perimeter of rectangle = 2( length + width )

$=2(x+y)$

Also, perimeter of rectangle is equal to 56 m.

So,

$56=2(x+y)\\x+y=28\\y=28-x$

Let A denotes area of rectangle.

A = length × width

$A=xy\\=x(28-x)\\=28x-x^2$

Differentiate with respect to x

$\frac{dA}{dx}=28-2x$

Put $\frac{dA}{dx}=0$

$28-2x=0\\2x=28\\x=14$

Also,

$\frac{d^2A}{dx^2}=-2$

At x = 14, $\frac{d^2A}{dx^2}=-2$

So, x = 14 is a point of maxima

So,

$y=28-x=28-14=14$

Area of rectangle:

$A=xy=14(14)=196\,m^2$

Length of rectangle = 14 m

Width of rectangle = 14 m

5 0

2 years ago