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2 years ago

Given:

Mathematics

2 answers:

Angelina_Jolie [31]2 years ago

4 0

Answer:

The correct match of statement with reasons is below.

Step-by-step explanation:

1. AB||DC and AB = CD Given

2. ∠1 = ∠4 If lines are parallel, then alternate interior

angles are equal.

3. AC=AC Reflexive

4.Triangle ABC is SAS

congruent to triangle CDA.

5. ∠2 = ∠3 CPCTE.

Hope I helped :3

maria [59]2 years ago

3 0

Answer:

The correct match of statement with reasons is below.

Step-by-step explanation:

1. AB||DC and AB = CD Given

2. ∠1 = ∠4 If lines are parallel, then alternate interior

angles are equal.

3. AC=AC Reflexive

4.Triangle ABC is SAS

congruent to triangle CDA.

5. ∠2 = ∠3 CPCTE.

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. Which ordered pairs in the form (x, y) are solutions to the equation 7x – 5y = 28?

Brut [27]

we know that

If a ordered pair (x,y) is a solution of the equation, then the ordered pair must satisfy the equation

we have the equation

$7x-5y=28$

Let's verify all the cases to determine the solution to the problem.

case A) point $(-6,-14)$

Substitute the values of x and y in the equation

$x=-6\\y=-14$

$7(-6)-5(-14)=28$

$-42+70=28$

$28=28$ -------> is true

therefore

The point $(-6,-14)$ is a solution of the equation

case B) point $(-1,-7)$

Substitute the values of x and y in the equation

$x=-1\\y=-7$

$7(-1)-5(-7)=28$

$-7+35=28$

$28=28$ -------> is true

therefore

The point $(-1,-7)$ is a solution of the equation

case C) point $(4,10)$

Substitute the values of x and y in the equation

$x=4\\y=10$

$7(4)-5(10)=28$

$28-50=28$

$-22=28$ -------> is not true

therefore

The point $(4,10)$ is not a solution of the equation

case D) point $(7,9)$

Substitute the values of x and y in the equation

$x=7\\y=9$

$7(7)-5(9)=28$

$49-45=28$

$4=28$ -------> is not true

therefore

The point $(7,9)$ is not a solution of the equation

therefore

the answer is

$(-6,-14)$

$(-1,-7)$

8 0

2 years ago

Read 2 more answers

I don't know how to solve it.

Nataly [62]

Step-by-step explanation:

Take the first derivative

$\frac{d}{dx} ( {x}^{3} - 3x)$

$3 {x}^{2} - 3$

Set the derivative equal to 0.

$3 {x}^{2} - 3 = 0$

$3 {x}^{2} = 3$

${x}^{2} = 1$

$x = 1$

$x = - 1$

For any number less than -1, the derivative function will have a Positve number thus a Positve slope for f(x).

For any number, between -1 and 1, the derivative slope will have a negative , thus a negative slope.

Since we are going to Positve to negative slope, we have a local max at x=-1

Plug in -1 for x into the original function

$( - 1) {}^{3} - 3( - 1) = 2$

So the local max is 2 and occurs at x=-1,

For any number greater than 1, we have a Positve number for the derivative function we have a Positve slope.

Since we are going to decreasing to increasing, we have minimum at x=1,

Plug in 1 for x into original function

${1}^{3} - 3(1)$

$1 - 3 = - 2$

So the local min occurs at -2, at x=1

8 0

1 year ago

Kieth bought 165 crayons that came in packs of 15. There were 9 colors of crayons how many packs of crayons did Kieth buy?

Andrej [43]

Answer:

11 packs

Step-by-step explanation:

165/15 = 11

6 0

2 years ago

Read 2 more answers

Which best describes how to solve the equation for x.

prohojiy [21]

Answer:

B. multiply both sides by c.

Step-by-step explanation:

To solve the equation for x, you have to isolate the x variable on one side of the equation. To do this, do the opposite of any operation applied to x to remove other variables from the x's side of the equation:

x/c = d

multiply both sides of the equation by c to isolate x:

(x/c) * c = d * c

x = dc

Now x can be solved for.

Hope this helps :)

7 0

1 year ago

In the PQRS triangle PQ=QR, QR side extended to S Show that PQ+RS=QS. -S Q R

Vitek1552 [10]

Answer:

Step-by-step explanation:

from the picture:

QP = QR

and

QR = RS

PQ + RS = QS

6 0

2 years ago