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3 years ago

Mathematics

1 answer:

zhuklara [117]3 years ago

7 0

Step-by-step answer:

Given:

A triangle

Perimeter = 60 cm

longest side = 4* shortest side (x)

Solution:

longest side = 4x

shortest side = x

third (intermediate side = 60 -x -4x = 60-5x

The triangle inequality specifies that the sum of the two shorter sides must be greater than the longest side to form a triangle. Hence

x + y > 4x

x + 60-5x > 4x

60 - 4x > 4x

8x < 60

x < 60/8 = 7.5, or

x < 7.5

Therefore to form a triangle, x (shortest side) must be less than 7.5 cm.

Examine the options: both 7 and 5 are both less than 7.5 cm.

40, 30 and 25 all have a problem because the longest side (4 times longer) will exceed the perimeter of 60.

Now also examine cases where 4x is NOT the longest side, in which case we need

4x>=y

4x >= 60-5x

9x >=60

x >= 6.67

so x=5 will not qualify, because 4x will no longer be the longest side.

The only valid option is x=7 cm

The side lengths for x=7 and x=5 are, respectively,

(7, 25, 28)

5, 20, 35 (in which case, the longest side is no longer 4x=20, so eliminated)

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When would the product of the denominators and the least common denominator of the denominators be the same?

Bas_tet [7]

Answer:

Example 1:

Find the common denominator of the fractions.

16 and 38

We need to find the least common multiple of 6 and 8 . One way to do this is to list the multiples:

6,12,18,24−−,30,36,42,48,...8,16,24−−,32,40,48,...

The first number that occurs in both lists is 24 , so 24 is the LCM. So we use this as our common denominator.

Listing multiples is impractical for large numbers. Another way to find the LCM of two numbers is to divide their product by their greatest common factor ( GCF ).

Example 2:

Find the common denominator of the fractions.

512 and 215

The greatest common factor of 12 and 15 is 3 .

So, to find the least common multiple, divide the product by 3 .

12⋅153=3 ⋅ 4 ⋅ 153=60

If you can find a least common denominator, then you can rewrite the problem using equivalent fractions that have like denominators, so they are easy to add or subtract.

Example 3:

Add.

512+215

In the previous example, we found that the least common denominator was 60 .

Write each fraction as an equivalent fraction with the denominator 60 . To do this, we multiply both the numerator and denominator of the first fraction by 5 , and the numerator and denominator of the second fraction by 4 . (This is the same as multiplying by 1=55=44 , so it doesn't change the value.)

512=512⋅55=2560215=215⋅44=860

512+215=2560+860 =3360

Note that this method may not always give the result in lowest terms. In this case, we have to simplify.

=1130

The same idea can be used when there are variables in the fractions—that is, to add or subtract rational expressions .

Example 4:

Subtract.

12a−13b

The two expressions 2a and 3b have no common factors, so their least common multiple is simply their product: 2a⋅3b=6ab .

Rewrite the two fractions with 6ab in the denominator.

12a⋅3b3b=3b6ab13b⋅2a2a=2a6ab

Subtract.

12a−13b=3b6ab−2a6ab =3b − 2a6ab

Example 5:

Subtract.

x16−38x

16 and 8x have a common factor of 8 . So, to find the least common multiple, divide the product by 8 .

16⋅8x8=16x

The LCM is 16x . So, multiply the first expression by 1 in the form xx , and multiply the second expression by 1 in the form 22 .

x16⋅xx=x216x38x⋅22=616x

Subtract.

x16−38x=x216x−616x =x2 − 616x\

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2 years ago

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sp2606 [1]

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3 years ago

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Norma-Jean [14]

Answer:

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$A=20000(1.08)^{t}$

$A=40(3)^{t}$

$A=1700(1.07)^{t}$

Exponential decay functions are:

$A=80(\frac{1}{2})^{t}=80(0.5)^{t}$

$A=1600(0.8)^{t}$

$A=1700(0.93)^{t}$

Step-by-step explanation:

Given:

An exponential function is of the form $y=ab^{x}$ , where, $a\ne 0$ .

Now, if a > 0 and b > 1, then the exponential function represent exponential growth.

If a > 0 and 0 < b < 1, then the exponential function represent exponential decay.

Let us check each function now.

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Here, $a = 20000, b = 1.08$

As 1.08 > 1, the function is exponential growth.

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Here, $a = 80, b = 0.50$

As 0.5 < 1, the function is exponential decay.

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Here, $a = 1600, b = 0.8$

As 0.8 < 1, the function is exponential decay.

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Here, $a = 40, b = 3$

As 3 > 1, the function is exponential growth.

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Here, $a = 1700, b = 1.07$

As 1.07 > 1, the function is exponential growth.

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As 0.93 < 1, the function is exponential decay.

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3 years ago

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Hatshy [7]

Answer:

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Step-by-step explanation:

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2 years ago

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melisa1 [442]

Answer:

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Step-by-step explanation:

In the given-question, the information is missing. first, we declare the missing information, after that we define its solution:

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please find the attachment.

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