All categories

3 years ago

I will give you the brainlessest if you awnser the following plz and thank you so much

Mathematics

1 answer:

kirill115 [55]3 years ago

8 0

2. x + y = 82
x - y = 24
---------------add
2x = 106
x = 53

x + y = 82
53 + y = 82
y = 82 - 53
y = 29

solution is : (53,29)
=======================
3. y = 2x
y = 4x + 6

2x = 4x + 6
2x - 4x = 6
-2x = 6
x = -3

y = 2x
y = 2(-3)
y = -6

solution is (-3,-6)
========================
4. 5x + 8y = -29
7x - 2y = -67...multiply by 4
----------------
5x + 8y = -29
28x - 8y = - 268 ..result of multiplying by 4
-----------------add
33x = - 297
x = - 9

5x + 8y = -29
5(-9) + 8y = -29
-45 + 8y = -29
8y = -29 + 45
8y = 16
y = 2

solution is : (-9,2)
=======================
5. y = -4x + 6
y = -5x - 4

-4x + 6 = -5x - 4
-4x + 5x = -4 - 6
x = -10

y = -4x + 6
y = -4(-10) + 6
y = 40 + 6
y = 46

solution is (-10,46)
=========================
6. H(m) = 2m + 12
H(m) = 3m + 10
=========================
7. -8x + 4y > -52
4y > 8x - 52
y > 2x - 13 <==
=========================
8. 3x - y = 28
3x + y = 14
---------------add
6x = 42
x = 7

3x - y = 28
3(7) - y = 28
21 - y = 28
-y = 28 - 21
-y = 7
y = -7

solution is (7,-7)
=====================
10. 5x - 5y > 70
-5y > -5x + 70
y < x - 14 <==
=====================
11. sorry...dont know
=====================
12. y = 4x + 4
y = -3x - 3

4x + 4 = -3x - 3
4x + 3x = -3 - 4
7x = -7
x = -1

y = 4x + 4
y = 4(-1) + 4
y = 0

solution is (-1,0)
=======================
13. -12x - 2y > - 42
-2y > 12x - 42
y < -6x + 21 <==
=======================
14. -5x + 2y = 9
3x + 5y = 7

solution is (-1,2)
========================
15. 3x + 6y = -2
15x + 30y = -10....divide by 5 to reduce = 3x + 6y = -2
is the same line....infinite solutions
========================
1. (the graph).....y < = 3x - 4....3rd one
========================
9. (2nd graph)....y < = -3x + 4....last one

You might be interested in

If the streets of a city are straight lines and the intersections are points, show how Principle 2 and Principle 3 might be illu

Naily [24]

Step-by-step explanation:

In the plane geometry Abeka, second edition, it is given :

Principle 2 states that between any two points, only one straight line can be drawn.

And according to principle 3 two straight lines interacts at one point only.

Thus this can be well illustrated by two straight lines which are represented by the streets of a city and these two streets intersects at a point.

6 0

3 years ago

Shane's family traveled 7/10 of the distance to his grandfather’s house on Saturday. They traveled 2/3 of the remaining distance

Naya [18.7K]

2/30 of the total distance was traveled on Sunday.

6 0

3 years ago

Peyton is 11 years old and knows 10-7=3 and 3+7=10 but finds it difficult to solve a math problem, such as 3x-5, because she can

Colt1911 [192]

Answer:

concrete operational stage

Step-by-step explanation:

Have a look at the attached photo below, it may be your full question:

The concrete operational stage is the third phase in Piaget's cognitive development theory. This period spans the period between childhood, it begins around 7 years of age and continues until about 11 years of age and is characterized by the development of logical thinking.

7 0

2 years ago

Alisha has sixty-seven dollars and thirty one cents in her checking account. She wrote a check for ninety-one dollars and fifty-

grin007 [14]

-24.22 will be overdrawn in Alisha checking account

4 0

3 years ago

Read 2 more answers

Use the given transformation x=4u, y=3v to evaluate the integral. ∬r4x2 da, where r is the region bounded by the ellipse x216 y2

exis [7]

The Jacobian for this transformation is

$J = \begin{bmatrix} x_u & x_v \\ y_u & y_v \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 3 \end{bmatrix}$

with determinant $|J| = 12$ , hence the area element becomes

$dA = dx\,dy = 12 \, du\,dv$

Then the integral becomes

$\displaystyle \iint_{R'} 4x^2 \, dA = 768 \iint_R u^2 \, du \, dv$

where $R'$ is the unit circle,

$\dfrac{x^2}{16} + \dfrac{y^2}9 = \dfrac{(4u^2)}{16} + \dfrac{(3v)^2}9 = u^2 + v^2 = 1$

so that

$\displaystyle 768 \iint_R u^2 \, du \, dv = 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2 \, du \, dv$

Now you could evaluate the integral as-is, but it's really much easier to do if we convert to polar coordinates.

$\begin{cases} u = r\cos(\theta) \\ v = r\sin(\theta) \\ u^2+v^2 = r^2\\ du\,dv = r\,dr\,d\theta\end{cases}$

Then

$\displaystyle 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2\,du\,dv = 768 \int_0^{2\pi} \int_0^1 (r\cos(\theta))^2 r\,dr\,d\theta \\\\ ~~~~~~~~~~~~ = 768 \left(\int_0^{2\pi} \cos^2(\theta)\,d\theta\right) \left(\int_0^1 r^3\,dr\right) = \boxed{192\pi}$

3 0

2 years ago