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Vsevolod [243]
4 years ago
7

Trapezoid JKLM is shown on the coordinate plane below:

Mathematics
1 answer:
ololo11 [35]4 years ago
3 0

Answer:

( 1, -4)

Step-by-step explanation:

To find this answer you go over 2 to the right on the x-axis, then go down 6 on the y-axis. This gives ( 1, -4).

You might be interested in
A(t)=.892t^3-13.5t^2+22.3t+579 how to solve this
Minchanka [31]

Answer:

t = (5 ((446 sqrt(3188516012553) - 827891226)^(1/3) - 204292 (-1)^(2/3) (3/(413945613 - 223 sqrt(3188516012553)))^(1/3)))/(223 6^(2/3)) + 1125/223 or t = 1125/223 - (5 ((-2)^(1/3) (223 sqrt(3188516012553) - 413945613)^(1/3) - 204292 (-3/(413945613 - 223 sqrt(3188516012553)))^(1/3)))/(223 6^(2/3)) or t = 1125/223 - (5 ((827891226 - 446 sqrt(3188516012553))^(1/3) + 204292 (3/(413945613 - 223 sqrt(3188516012553)))^(1/3)))/(223 6^(2/3))

Step-by-step explanation:

Solve for t over the real numbers:

0.892 t^3 - 13.5 t^2 + 22.3 t + 579 = 0

0.892 t^3 - 13.5 t^2 + 22.3 t + 579 = (223 t^3)/250 - (27 t^2)/2 + (223 t)/10 + 579:

(223 t^3)/250 - (27 t^2)/2 + (223 t)/10 + 579 = 0

Bring (223 t^3)/250 - (27 t^2)/2 + (223 t)/10 + 579 together using the common denominator 250:

1/250 (223 t^3 - 3375 t^2 + 5575 t + 144750) = 0

Multiply both sides by 250:

223 t^3 - 3375 t^2 + 5575 t + 144750 = 0

Eliminate the quadratic term by substituting x = t - 1125/223:

144750 + 5575 (x + 1125/223) - 3375 (x + 1125/223)^2 + 223 (x + 1125/223)^3 = 0

Expand out terms of the left hand side:

223 x^3 - (2553650 x)/223 + 5749244625/49729 = 0

Divide both sides by 223:

x^3 - (2553650 x)/49729 + 5749244625/11089567 = 0

Change coordinates by substituting x = y + λ/y, where λ is a constant value that will be determined later:

5749244625/11089567 - (2553650 (y + λ/y))/49729 + (y + λ/y)^3 = 0

Multiply both sides by y^3 and collect in terms of y:

y^6 + y^4 (3 λ - 2553650/49729) + (5749244625 y^3)/11089567 + y^2 (3 λ^2 - (2553650 λ)/49729) + λ^3 = 0

Substitute λ = 2553650/149187 and then z = y^3, yielding a quadratic equation in the variable z:

z^2 + (5749244625 z)/11089567 + 16652679340752125000/3320419398682203 = 0

Find the positive solution to the quadratic equation:

z = (125 (223 sqrt(3188516012553) - 413945613))/199612206

Substitute back for z = y^3:

y^3 = (125 (223 sqrt(3188516012553) - 413945613))/199612206

Taking cube roots gives (5 (223 sqrt(3188516012553) - 413945613)^(1/3))/(223 2^(1/3) 3^(2/3)) times the third roots of unity:

y = (5 (223 sqrt(3188516012553) - 413945613)^(1/3))/(223 2^(1/3) 3^(2/3)) or y = -(5 (-1/2)^(1/3) (223 sqrt(3188516012553) - 413945613)^(1/3))/(223 3^(2/3)) or y = (5 (-1)^(2/3) (223 sqrt(3188516012553) - 413945613)^(1/3))/(223 2^(1/3) 3^(2/3))

Substitute each value of y into x = y + 2553650/(149187 y):

x = (5 ((223 sqrt(3188516012553) - 413945613)/2)^(1/3))/(223 3^(2/3)) - 510730/223 (-1)^(2/3) (2/(3 (413945613 - 223 sqrt(3188516012553))))^(1/3) or x = 510730/223 ((-2)/(3 (413945613 - 223 sqrt(3188516012553))))^(1/3) - (5 ((-1)/2)^(1/3) (223 sqrt(3188516012553) - 413945613)^(1/3))/(223 3^(2/3)) or x = (5 (-1)^(2/3) ((223 sqrt(3188516012553) - 413945613)/2)^(1/3))/(223 3^(2/3)) - 510730/223 (2/(3 (413945613 - 223 sqrt(3188516012553))))^(1/3)

Bring each solution to a common denominator and simplify:

x = (5 ((446 sqrt(3188516012553) - 827891226)^(1/3) - 204292 (-1)^(2/3) (3/(413945613 - 223 sqrt(3188516012553)))^(1/3)))/(223 6^(2/3)) or x = -(5 ((-2)^(1/3) (223 sqrt(3188516012553) - 413945613)^(1/3) - 204292 ((-3)/(413945613 - 223 sqrt(3188516012553)))^(1/3)))/(223 6^(2/3)) or x = -(5 ((827891226 - 446 sqrt(3188516012553))^(1/3) + 204292 (3/(413945613 - 223 sqrt(3188516012553)))^(1/3)))/(223 6^(2/3))

Substitute back for t = x + 1125/223:

Answer: t = (5 ((446 sqrt(3188516012553) - 827891226)^(1/3) - 204292 (-1)^(2/3) (3/(413945613 - 223 sqrt(3188516012553)))^(1/3)))/(223 6^(2/3)) + 1125/223 or t = 1125/223 - (5 ((-2)^(1/3) (223 sqrt(3188516012553) - 413945613)^(1/3) - 204292 (-3/(413945613 - 223 sqrt(3188516012553)))^(1/3)))/(223 6^(2/3)) or t = 1125/223 - (5 ((827891226 - 446 sqrt(3188516012553))^(1/3) + 204292 (3/(413945613 - 223 sqrt(3188516012553)))^(1/3)))/(223 6^(2/3))

6 0
3 years ago
Tania has 56 liters of milk. She uses the milk to make muffins and cakes. Each batch of muffin requires 7 liters of milk and eac
mihalych1998 [28]
Write the inequalities that are given by the :

<span>x: the number of batches of muffins

y: the number of batches of cakes

</span>Each batch of muffin requires 7 liters of milk and each batch of cakes require 4 liters of milk.

=> liters of milk use = 7x  + 4y

<span>Tania has 56 liters of milk.=> 7x + 4y ≤ 56

Which means that the amount of muffins and cakes made are limited by the availability of 56 liter of milk.

The inequality 7x + 4y ≤ 56 is graphed by drawing the line 7x + 4y = 56 and shading the region below that line.

The line 7x + 4y = 56 has these x and y intercepts:

y-intercept: x =0 => 4y = 56 => y = 56/4 => 14 => point (0,14)

x-intercept => y = 0 => 7x = 56 => x = 56/7= 8  => point (8,0)

So, the line passes through the poins (0,8) and (14,0) and the solution region is below that line.

Also, you know that x and y are restricted to be positive or zero =>

x ≥ 0

y ≥ 0.

So, the solution region is restricted to the first quadrant.

That implies that the answer is:
</span><span>
Line joining ordered pairs 0, 14 and 8, 0. Shade the portion of the graph below this line which lies within the first quadrant
</span>


4 0
4 years ago
HELP ME PLSSSS HELPPPPP
skad [1K]

Answer:

I think A

Step-by-step explanation:

Since the center of dilation is at A, only the A coordinate remains

8 0
3 years ago
Answer two questions about Systems AAA and BBB:
MA_775_DIABLO [31]

For the given systems A and B:

1) Replace the first equation of A by the sum between the two equations of A.

2) Yes, the systems are equivalent.

<h3>How to get system B from system A?</h3>

Here we have the two systems of equations:

A:

6x - 5y = 1

-2x + 2y = -1

B:

4x - 3y = 0

-2x + 2y = -1

The second equation is the same in both systems, so we only look at the first equations.

In A we have:

6x - 5y = 1

If we add the second equation of A, then we get:

(6x - 5y) + (-2x + 2y) = 1 + (-1)

4x - 3y = 0

This is the first equation of B.

Then we need to replace the first equation by the sum between the first and second equations.

2) Are the systems equivalent?

Yes, because we did not "modify" system A, we just rewrite it and we got system B, then both systems have the same solutions.

If you want to learn more about systems of equations:

brainly.com/question/847634

#SPJ1

4 0
2 years ago
HELPP WHATS THE ANSWERRR
sergejj [24]

Answer:

C

Step-by-step explanation:

X can’t have 2 y’s

6 0
3 years ago
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