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3 years ago

HEEELp PLEASEEEEEEE tank uuuuuuuuuu

Mathematics

2 answers:

VARVARA [1.3K]3 years ago

8 0

9514 1404 393

Answer:

both figures are parallelograms

Step-by-step explanation:

The diagonals of a quadrilateral bisect each other if and only if that quadrilateral is a parallelogram. Hence, the left figure is a parallelogram.

The opposite angles of a quadrilateral are congruent if and only if that quadrilateral is a parallelogram. Hence the right figure is a parallelogram.

<em>Additional comment</em>

Proofs of these conditions can be offered, but these are the conclusions of the respective proofs. The bisected diagonals form pairs of triangles, provably congruent using SAS. (left figure) The angles total 360°, so any adjacent pair must total 180°, making opposite sides parallel. (right figure)

Kamila [148]3 years ago

7 0

Answer:

they both are because they are not symmetrical??

Step-by-step explanation:

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2y = 2x = 8 y = x + 4

padilas [110]

Answer:

I love algebra anyways

the ans is in the picture with the steps how i got it

(hope that helps can i plz have brainlist it will make my day :D hehe)

Step-by-step explanation:

4 0

3 years ago

3. Sophie and Jackie each have a collection of baseball cards, Jackie has 5 more cards than

IRISSAK [1]

Answer:

Sophie has 12.5 cards

Step-by-step explanation:

Let

x -----> number of cards Sophie has

y -----> number of cards Jackie has

we know that

x+y=30 -----> equation A

y=x+5 -----> equation B

Solve by substitution

Substitute equation B in equation A and solve for x

x+(x+5)=30

2x=30-5

x=12.5 cards

Note I assume the problem was invented without taking into account the result, because the amount of cards should be a whole number

5 0

3 years ago

Answer all these questions please

Zielflug [23.3K]

A = velocity / time
b) 40m/s
13m / 3s = 4.33 m/ss

4 0

2 years ago

Riley ordered a snow cone with 4 flavors.

frosja888 [35]

Riley has 4 flavors
caleb had 2 flavors
4-2 equals 2

6 0

3 years ago

Read 2 more answers

In a G.P the difference between the 1st and 5th term is 150, and the difference between the

liubo4ka [24]

Answer:

Either $\displaystyle \frac{-1522}{\sqrt{41}}$ (approximately $-238$ ) or $\displaystyle \frac{1522}{\sqrt{41}}$ (approximately $238$ .)

Step-by-step explanation:

Let $a$ denote the first term of this geometric series, and let $r$ denote the common ratio of this geometric series.

The first five terms of this series would be:

$a$ ,
$a\cdot r$ ,
$a \cdot r^2$ ,
$a \cdot r^3$ ,
$a \cdot r^4$ .

First equation:

$a\, r^4 - a = 150$ .

Second equation:

$a\, r^3 - a\, r = 48$ .

Rewrite and simplify the first equation.

$\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}$ .

Therefore, the first equation becomes:

$a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150$ ..

Similarly, rewrite and simplify the second equation:

$\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}$ .

Therefore, the second equation becomes:

$a\, r\, \left(r^2 - 1\right) = 48$ .

Take the quotient between these two equations:

$\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}$ .

Simplify and solve for $r$ :

$\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}$ .

$8\, r^2 - 25\, r + 8 = 0$ .

Either $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ or $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ .

Assume that $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}$ .

Similarly, assume that $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}$ .

4 0

3 years ago