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3 years ago

Does Anyone have idea how to solve this problem?

Mathematics

2 answers:

garik1379 [7]3 years ago

6 0

A.

5 * 2 = 10

8 * 2 = 16

10 cm by 16 cm is the new dimensions. Area = 160 sq cm. Perimeter = 52 cm.

B.

5 * 3 = 15

8 * 3 = 24

15 cm by 24 cm is the new dimensions. Area = 360 sq cm. Perimeter = 78 cm.

C.

The areas for the different sized photos both still maintain a ratio of 5:8 cm.

yarga [219]3 years ago

6 0

It is a because the length and width corresponds with rectangles

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A basketball player’s free throw percentage is 70%. She has just been awarded 3 free throws after getting fouled on a 3-point sh

Vlada [557]

P(3 Free Throws) = P(One Free Throw & One Free Throw & One Free Throw)

P(3 Free Throws) = P(One Free Throw) x P(One Free Throw) x P(One Free Throw)

P(3 Free Throws) = (7/10) x (7/10) x (7/10)

P(3 Free Throws) = (7x7x7)/(10x10x10)

P(3 Free Throws) = 343/1000

P(3 Free Throws) = 0.343

The probability is 0.343 to make all 3 free throws.

6 0

3 years ago

Read 2 more answers

Question 3 (10 points)

tekilochka [14]

Step-by-step explanation:

Circumference formula:

C = 2πr = πd

Find π:

25.12 = 8π ⇒ π = 25.12/8 ⇒ π = 3.14
9.42 = 3π ⇒ π = 9.42/3 ⇒ π = 3.14

Area formula:

A = πr² = πd²/4

Find π:

50.24 = π*8²/4 ⇒ 50.24 = 16π ⇒ π = 50.24/16 ⇒ π = 3.14
7.065 = π*3²/4 ⇒ 7.065 = 9π/4 ⇒ π = 4*7.065/9 ⇒ π = 3.14

The value of π is same - 3.14 from each case we have

5 0

3 years ago

Pls help this is pretty urgent

laiz [17]

Answer:

(a) See below.

(b) x = 0 or x = 1

Step-by-step explanation:

Given rational function:

$f(x)=\dfrac{\ln |x-1|}{x}$

Part (a)

Substitute x = 2 into the given rational function:

$\begin{aligned}\implies f(2) & =\dfrac{\ln |2-1|}{2}\\\\ & =\dfrac{\ln 1}{2}\\\\ & =\dfrac{0}{2}\\\\ & =0\end{aligned}$

Therefore, as the function is defined at x = 2, the function is continuous at x = 2.

Part (b)

Given interval: [-2, 2]

Logs of negative numbers or zero are undefined. As the numerator is the natural log of an absolute value, the numerator is undefined when:

|x - 1| = 0 ⇒ x = 1.

A rational function is undefined when the denominator is equal to zero, so the function f(x) is undefined when x = 0.

So the function is discontinuous at x = 0 or x = 1 on the interval [-2, 2].

Part (c)

x = 1 is a vertical asymptote. As the function exists on both sides of this vertical asymptote, it is an infinite discontinuity. Since the function doesn't approach a particular finite value, the limit does not exist. Therefore, x = 1 is a non-removable discontinuity.

A hole exists on the graph of a rational function at any input value that causes both the numerator and denominator of the function to be equal.

$\implies f(0)=\dfrac{\ln |0-1|}{0}=\dfrac{\ln 1}{0}=\dfrac{0}{0}$

Therefore, there is a hole at x = 0.

The removable discontinuity of a function occurs at a point where the graph of a function has a hole in it. Therefore, x = 0 is a removable discontinuity.