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3 years ago

Trafik levhalasinin anlamı hangisidir?

Mathematics

1 answer:

noname [10]3 years ago

8 0

<h3>Selammm ♡</h3>

Bence cevap C diye düşünüyorum çünkü trafik levhası kaza yapmayalım diye bizi uyarır, yaya ve araçların olduğu yollarda, trafik düzeninin sağlayan önemli gereçlerdir. Bu levhalar sayesinde kargaşalar engellenmekte, can ve mal güvenliği de korunmaktadır.

<h3>Başarılar</h3>

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What is another way of representing (6 x 1/10)

Leona [35]

Answer:

Step-by-step explanation:

3/5x is a simplified version if that helps.

7 0

3 years ago

prove that if f is integrable on [a,b] and c is an element of [a,b], then changing the value of f at c does not change the fact

Neko [114]

Answer with Step-by-step explanation:

We are given that if f is integrable on [a,b].

c is an element which lie in the interval [a,b]

We have to prove that when we change the value of f at c then the value of f does not change on interval [a,b].

We know that limit property of an integral

$\int_{a}^{b}f dt=\int_{a}^{c}fdt+\int_{c}^{b} fdt$

$\int_{a}^{b} fdt=f(b)-f(a)$ ....(Equation I)

Using above property of integral then we get

$\int_{a}^{b}fdt=\int_{a}^{c}fdt+\int_{c}^{b} fdt$ ......(Equation II)

Substitute equation I and equation II are equal

Then we get

$\int_{a}^{b}fdt= f(c)-f(a)+{f(b)-f(c)}$

$\int_{a}^{b}fdt=f(c)-f(a)+f(b)-f(c)=f(b)-f(a)$

$\int_{a}^{c}fdt+\int_{c}^{b}fdt=f(b)-f(a)$

Therefore, $\int_{a}^{b}fdt=\int_{a}^{c}fdt+\int_{c}^{b}fdt$ .

Hence, the value of function does not change after changing the value of function at c.

6 0

3 years ago

Miguel draws a square on a coordinate plane. One vertex is located at (5,4). The length of each side is 3 units. Circle the lett

Ulleksa [173]

Answer:

A, B, and E.

Step-by-step explanation:

We know that one vertex is at (5, 4), and each side of our square is 3 units long.

Then the distance between the known vertex and another vertex is 3 units (if those vertexes are connected by a side of the square) or (√2)*3 units (if those vertexes are connected by the diagonal of the square).

Also remember that the distance between two points (a, b) and (b, c) is:

distance = √( (a - c)^2 + (b - d)^2)

So we need to find the distance between our point and all the ones given in the options:

A) the distance between (5, 4) and (5, 1) is:

distance = √( (5 - 5)^2 + (4 - 1)^2) = 3

Then point (5, 1) can be a vertex.

B) The distance between (5, 4) and (5, 7) is:

distance = √( (5 - 5)^2 + (4 - 7)^2) = 3

Then (5, 7) can be a vertex.

C) The distance between (5, 4) and (7, 8) is:

distance = √( (5 - 7)^2 + (4 - 8)^2) = √( 2^2 + 4^2) = √20

Point (7, 8) can not be a vertex.

D) The distance between (5, 4) and (2, 6) is:

distance = √( (5 - 2)^2 + (4 - 6)^2) = √( 3^2 + 2^2) = √13

Point (2, 6) can not be a vertex.

E) The distance between (5, 4) and (2, 1) is:

distance = √( (5 - 2)^2 + (4 - 1)^2) = √( 3^2 + 3^2) = √18 = √(2*9) = √2*√9 = √2*3

Then point (2, 1) can be a vertex.

8 0

3 years ago

A photo processing place can develop 416 photos in 8 hours. How many photos can it develop every hour?

Rina8888 [55]

Answer:

52 photos per hour

Step-by-step explanation:

8 goes into 400 50 times, and goes into the remaining 16 twice. Therefore, 416 photos/8 hour=52

5 0

3 years ago

Geometry math question

Brut [27]

Look at the picture.

Let |AE| = |AD| = b

We have a proportion:

$\dfrac{x}{13}{y}=\dfrac{19}{y+a}=\dfrac{x}{y+2a}$

Solve for y from first proportion

$\dfrac{13}{y}=\dfrac{19}{y+a}\ \ \ |\text{cross multiply}\\\\19y=13(y+a)\\\\19y=13y+13a\ \ \ \ |-13y\\\\6y=13a\ \ \ |:6\\\\y=\dfrac{13}{6}a$

Substitute to the second proportion

$\dfrac{19}{y+a}=\dfrac{x}{y+2a}\\\downarrow\\\dfrac{19}{\frac{13}{6}a+a}=\dfrac{x}{\frac{13}{6}a+2a}\\\\\dfrac{19}{\frac{13}{6}a+\frac{6}{6}a}=\dfrac{x}{\frac{13}{6}a+\frac{12}{6}a}\\\\\dfrac{19}{\frac{19}{6}a}=\dfrac{x}{\frac{25}{6}a}\ \ \ \ |\dot a\neq0\\\\\dfrac{19}{\frac{19}{6}}=\dfrac{x}{\frac{25}{6}}\\\\19\cdot\dfrac{6}{19}=x\cdot\dfrac{6}{25}\\\\6=\dfrac{6x}{25}\ \ \ |\cdot25\\\\6x=6\cdot25\ \ \ |:6\\\\x=25$

Answer: B. 25

7 0

4 years ago