The perimeter of this isosceles triangle is 22 cm. If one side is 6 cm, what are the possible lengths of the other two sides? Part B Explain how you know. Provide at least one reason for your answer.

Answers

Answer 1

Answer: we know that

the isosceles triangle has two equal sides

let's analyze two cases

first case
perimeter=22 cm
let's suppose that the known side of 6 cm is one of the two equal sides
perimeter=6+6+x
22=6+6+x-----> x=22-12----> x=10 cm

answer first case
the possible lengths of the other two sides are
6 cm
10 cm

second case
let's suppose that the known side of 6 cm is the side that is not equal
perimeter=22 cm
perimeter=6+x+x
22=6+x+x----> 2x=22-6----> 2x=16-----> x=8 cm

answer second case
the possible lengths of the other two sides are
8 cm
8 cm

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D+16/3=17 the steps??!!

Answers

Answer:

Step-by-step explanation:

D+16/3=17 the steps?

1. D+16/3=17

Divide 16/3.

D+6 = 17

2. subtract 6 from both side.

D = 11.

Answer is D=35

STEP1:

d + 16
Simplify ——————
3
Equation at the end of step
1
:

(d + 16)
———————— - 17 = 0
3
STEP
2
:
Rewriting the whole as an Equivalent Fraction

2.1 Subtracting a whole from a fraction

Rewrite the whole as a fraction using 3 as the denominator :

17 17 • 3
17 = —— = ——————
1 3
Equivalent fraction : The fraction thus generated looks different but has the same value as the whole

Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator

Adding fractions that have a common denominator :

2.2 Adding up the two equivalent fractions
Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

(d+16) - (17 • 3) d - 35
————————————————— = ——————
3 3
Equation at the end of step
2
:

d - 35
—————— = 0
3
STEP
3
:

When a fraction equals zero :

3.1 When a fraction equals zero ...
Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.

Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator.

Here's how:

d-35
———— • 3 = 0 • 3
3
Now, on the left hand side, the 3 cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
d-35 = 0

Solving a Single Variable Equation:

3.2 Solve : d-35 = 0

Add 35 to both sides of the equation :
d = 35

Find an integer that leaves a remainder of 2 when divided by either 3 or 5, but that is divisible by 4.

Answers

We want an integer $x$ such that

$x\equiv\begin{cases}2&\pmod3\n0&\pmod4\n2&\pmod5\end{cases}$

Note that the moduli are all relatively prime, so we can use the Chinese remainder theorem right away. As a first step, let's suppose

$x=4\cdot5\cdot2+3\cdot5\cdot0+3\cdot4\cdot2$

Taken modulo 3, the last two terms immediately vanish, and $4\cdot5\cdot2=40\equiv1\pmod3$ . We want a remainder of 2, so we just multiply this term by 2.

$x=4\cdot5\cdot2^2+3\cdot5\cdot0+3\cdot4\cdot2$

Next, taken modulo 4, all terms vanish, so we're good here.
Then, taken modulo 5, the first two terms vanish and we're left with $3\cdot4\cdot2\equiv24\equiv4\pmod5$ . We want a remainder of 2. To rectify this, we can first multiply this term by the inverse of 4 modulo 5, then multiply again by 2. This guarantees that

$3\cdot(4\cdot4^(-1))\cdot2^2\equiv2\pmod5$

The inverse of 4 modulo 5 is 4, since $4^2\equiv16\equiv1\pmod5$ , so we end up with

$x=4\cdot5\cdot2^2+3\cdot5\cdot0+3\cdot4^2\cdot2^2=272$

You can confirm for yourself that 272 satisfies the desired conditions. The CRT says that any integer of the form

$272\pmod{3\cdot4\cdot5}\equiv32\pmod60$

will work, i.e. $32+60n$ where $n\in\mathbb Z$ , and in particular 32 is the smallest positive solution.

Approximate the logarithm using the properties of logarithms, given logb 2 ≈ 0.3562,

logb 3 ≈ 0.5646,
and
logb 5 ≈ 0.8271.
(Round your answer to four decimal places.)
㏒b(2b^3)

Answers

Final answer:

By using the properties of logarithms, namely ㏒b(a * c) = ㏒b(a) + ㏒b(c) and ㏒b(a^p) = p * ㏒b(a), we find the approximation of the logarithm ㏒b(2b^3) to be 3.3562.

Explanation:

To approximate the value of ㏒b(2b^3), we can use the properties of logarithms. The first one is that ㏒b(a * c) = ㏒b(a) + ㏒b(c), where a and c are the numbers we wish to find the logarithm of. We can apply this property to your given expression as follows:

㏒b(2b^3) = ㏒b(2) + ㏒b(b^3).

The second logarithmic property we need is that ㏒b(a^p) = p * ㏒b(a), where a is the base, and p is the exponent. We can apply this property to the second term on the right side:

㏒b(b^3) = 3 * ㏒b(b).
Since ㏒b(b) is equal to 1 (because b raised to the power of 1 is b), then our equation becomes:
㏒b(2b^3) = ㏒b(2) + 3.

Now, we can substitute the values given for ㏒b(2), which is 0.3562:

㏒b(2b^3) = 0.3562 + 3 = 3.3562.

So, approximating ㏒b(2b^3) to four decimal places, we get 3.3562.

Learn more about Logarithms here:

brainly.com/question/37245832

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