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3 years ago

What is the name of this polygon?

Mathematics

2 answers:

Goryan [66]3 years ago

6 0

Answer:

That is a rhombus.

Brainliest is greatly appreciated

Answered by: Skylar

5/20/2020

1:37 PM (Eastern Time)

melisa1 [442]3 years ago

3 0

Answer:

This is a rhombus because it has 4 verticles, 4 edges and 2 lines of symmetry.

I know I'm late but hope this helps!!

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Question:

Marina CMI [18]

Since 2x+y>4 you get y>4-2x. Then you draw the line y=4-2x then you are interested in everything above the line not including it since it is y>4-2x so it is a dashed line. You need to shade the side that satifies the inequality and since it is every pair (x,y) where y>4-2x, you need to shade everything above the line which is the side that does not include the origin. So the answer is C.

7 0

3 years ago

The doubling period of a bacterial population is 20 20 minutes. At time t = 100 t=100 minutes, the bacterial population was 9000

Nuetrik [128]

Answer:

The initial population was 2810

The bacterial population after 5 hours will be 92335548

Step-by-step explanation:

The bacterial population growth formula is:

$P = P_0 \times e^{rt}$

where P is the population after time t, $P_0$ is the starting population, i.e. when t = 0, r is the rate of growth in % and t is time in hours

Data: The doubling period of a bacterial population is 20 minutes (1/3 hour). Replacing this information in the formula we get:

$2 P_0 = P_0 \times e^{r 1/3}$

$2 = e^{r \; 1/3}$

$ln 2 = r \; 1/3$

$ln 2 \times 3 = r$

$2.08 \% = r$

Data: At time t = 100 minutes (5/3 hours), the bacterial population was 90000. Replacing this information in the formula we get:

$90000 = P_0 \times e^{2.08 \; 5/3}$

$\frac{9000}{e^{2.08 \; 5/3}} = P_0$

$2810 = P_0$

Data: the initial population got above and t = 5 hours. Replacing this information in the formula we get:

$P = 2810 \times e^{2.08 \; 5}$

$P = 92335548$

3 0

3 years ago

There are 364 students who are enrolled in an introductory biology course. If there are five boys to every eight girls, how many

Butoxors [25]

Answer:

The number of boys in the course will be = 140

Step-by-step explanation:

Given:

Total number of students enrolled in an introductory biology course = 364

There are 5 boys to every 8 girls in the course.

To find how many boys are in the course.

Solution:

Since, there are 5 boys to every 8 girls in the course.

So, ratio of boys to girls enrolled in the course = 5 : 8

Let the number of boys in the course be = $5x$

Then number of girls in the course will be = $8x$

Total number of students would be given as:

⇒ <em>Number of boys + Number of girls</em>

⇒ $5x+8x$

⇒ $13x$

Total number of students given = 364.

Thus, we have:

$13x=364$

solving for $x$

Dividing both sides by 13.

$\frac{13x}{13}=\frac{364}{13}$

∴ $x=28$

So, number of boys in the course will be = $5\times 28$ = 140

5 0

3 years ago

Mr. miller has $25 in his savings and wants to put $10 every week into the account. write an equation showing how much money he

iris [78.8K]

The equation you write would be linear, and would be written in slope intercept form y=mx+b. Since Mr. Miller already has $25, we plug that in for "b" in the equation. We plug 10 in for "m", because "x" represents the number of weeks he has been saving. The equation would be y=10x+25. To find how much money Mr. Miller will have in 7 weeks, plug in 7 for x. y=10(7)+25 -> y=70+25 -> y=95 -> $95

7 0

3 years ago

Boa tarde gentiii Abaixo aparecem quatro sequências. Defina cada uma delas como RECURSIVA ou NÃO RECURSIVA. a) (5, 7, 9, 11, 13,

marusya05 [52]

Answer:

a) (5, 7, 9, 11, 13, 15, 17, 19, ...) Recursivo

b) (1, 4, 9, 16, 25, 36, 49, 64, ...) Recursivo

c) (1, 8, 27, 64, ...) Recursivo

d) (2, 5, 8, 11, 14, 17, ...) Recursivo

Step-by-step explanation:

Uma função recursiva é aquela em que os termos subsequentes da função são calculados com base nos termos anteriores

O comportamento recursivo é exibido pelos objetos quando consiste em seguir as partes;

1) Uma base que é predefinida

2) Um processo ou etapa recursiva que produz termos subsequentes pela aplicação de certos processos

Para a série;

a) (5, 7, 9, 11, 13, 15, 17, 19, ...)

Aqui 5 é a base e os termos subsequentes são encontrados adicionando 2 ao termo anterior, portanto, é uma função recursiva

aₙ = aₙ₋₁ + 2

b) (1, 4, 9, 16, 25, 36, 49, 64, ...)

Aqui 1 é a base e os termos subsequentes são encontrados ao quadrado da soma da raiz do termo anterior e 1, portanto, é uma função recursiva

aₙ = (√ (aₙ₋₁) + 1) ²

c) (1, 8, 27, 64, ...)

Aqui 1 é a base e os termos subsequentes são encontrados elevando à potência de três a soma da raiz cúbica do termo anterior e 1, portanto, é uma função recursiva

aₙ = (∛ (aₙ₋₁) + 1) ³

d) (2, 5, 8, 11, 14, 17, ...)

Aqui 2 é a base e os termos subsequentes são encontrados adicionando 3 ao termo anterior, portanto, é uma função recursiva.

7 0

3 years ago