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3 years ago

Find the value of x. Show your work.

Mathematics

1 answer:

r-ruslan [8.4K]3 years ago

5 0

Exterior angle sum of an polygon (shape) is 360 degrees (this is a property)

x + 98 + 41 + 76 + 86 = 360

Therefore x = 59

Please double check the equation I have written. Your picture is a bit blurry therefore I can really tell what number is 6 and 8. Sorry

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Solve for x: 3/4x+5/4=4x?

butalik [34]

Answer:

x = 5/13 = 0.385

Step-by-step explanation:

Step 1 :

Simplify —

Equation at the end of step 1 :

3 5

((— • x) + —) - 4x = 0

4 4

Step 2 :

Simplify —

Equation at the end of step 2 :

3 5

((— • x) + —) - 4x = 0

4 4

Step 3 :

Adding fractions which have a common denominator :

3.1 Adding fractions which have a common denominator

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

3x + 5 3x + 5

—————— = ——————

4 4

Equation at the end of step 3 :

(3x + 5)

———————— - 4x = 0

Step 4 :

Rewriting the whole as an Equivalent Fraction :

4.1 Subtracting a whole from a fraction

Rewrite the whole as a fraction using 4 as the denominator :

4x 4x • 4

4x = —— = ——————

1 4

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 :

4.2 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:

(3x+5) - (4x • 4) 5 - 13x

————————————————— = ———————

4 4

Equation at the end of step 4 :

5 - 13x

——————— = 0

Step 5 :

When a fraction equals zero :

5.1 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:

5-13x

————— • 4 = 0 • 4

Now, on the left hand side, the 4 cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :

5-13x = 0

Solving a Single Variable Equation :

5.2 Solve : -13x+5 = 0

Subtract 5 from both sides of the equation :

-13x = -5

Multiply both sides of the equation by (-1) : 13x = 5

Divide both sides of the equation by 13:

x = 5/13 = 0.385

One solution was found :

x = 5/13 = 0.385

Processing ends successfully

plz mark me as brainliest :)

4 0

3 years ago

Read 2 more answers

1/4x + 2/3y = 5 1/2x + 3/2y = 11 which of the following expressions represents the multiplicative property of equality

Mandarinka [93]

Because

(1/4x+2/3y=5)12=3x+8y=60

And

(1/2x+3/2y=11)3=x+3y=22

If you need/want more detailed explanations please ask.

5 0

3 years ago

6n + n^7 is divisible by 7 and prove it in mathematical induction<br>

kompoz [17]

Answer:

Apply induction on $n$ (for integers $n \ge 1$ ) after showing that $\genfrac{(}{)}{0}{}{7}{j} = (7!) / (j! \, (7 - j)!)$ is divisible by $7$ for $j \in \lbrace 1,\, \dots,\, 6 \rbrace$ .

Step-by-step explanation:

Lemma: $\genfrac{(}{)}{0}{}{7}{j} = (7!) / (j! \, (7 - j)!)$ is divisible by $7$ for $j \in \lbrace 1,\, 2,\, \dots,\, 6\rbrace$ .

Proof: assume that for some $j \in \lbrace 1,\, 2,\, \dots,\, 6\rbrace$ , $\genfrac{(}{)}{0}{}{7}{j}$ is not divisible by $7$ .

The combination $\genfrac{(}{)}{0}{}{7}{j} = (7!) / (j! \, (7 - j)!)$ is known to be an integer. Rewrite the factorial $7!$ to obtain:

$\displaystyle \begin{pmatrix}7 \\ j\end{pmatrix} = \frac{7!}{j! \, (7 - j)!} = \frac{7 \times 6!}{j!\, (7 - j)!}$ .

Note that $7$ (a prime number) is in the numerator of this expression for $\genfrac{(}{)}{0}{}{7}{j}\!$ . Since all terms in this fraction are integers, the only way for $\genfrac{(}{)}{0}{}{7}{j}$ to be non-divisible by $7\!$ is for the denominator $j! \, (7 - j)!$ of this expression to be an integer multiple of $7\!\!$ .

However, since $1 \le j \le 6$ , the prime number $\!7$ would not a factor of $j!$ . Similarly, since $1 \le 7 - j \le 6$ , the prime number $7\!$ would not be a factor of $(7 - j)!$ , either. Thus, $j! \, (7 - j)!$ would not be an integer multiple of the prime number $7$ . Contradiction.

Proof of the original statement:

Base case: $n = 1$ . Indeed $6 \times 1 + 1^{7} = 7$ is divisible by $7$ .

Induction step: assume that for some integer $n \ge 1$ , $(6\, n + n^{7})$ is divisible by $7$ . Need to show that $(6\, (n + 1) + (n + 1)^{7})$ is also divisible by $7\!$ .

Fact (derived from the binomial theorem $(\ast)$ ):

$\begin{aligned} & (n + 1)^{7} \\ &= \sum\limits_{j = 0}^{7} \left[\genfrac{(}{)}{0}{}{7}{j}\, n^{j}\right] && (\ast)\\ &= \genfrac{(}{)}{0}{}{7}{0} \, n^{0} + \genfrac{(}{)}{0}{}{7}{7} \, n^{7} + \sum\limits_{j = 1}^{6} \left[\genfrac{(}{)}{0}{}{7}{j}\, n^{j}\right] \\ &= 1 + n^{7} + \sum\limits_{j = 1}^{6} \left[\genfrac{(}{)}{0}{}{7}{j}\, n^{j}\right]\end{aligned}$ .

Rewrite $(6\, (n + 1) + (n + 1)^{7})$ using this fact:

$\begin{aligned} & 6\, (n + 1) + (n + 1)^{7} \\ =\; & 6\, (n + 1) + \left(1 + n^{7} + \sum\limits_{j = 1}^{6} \left[\genfrac{(}{)}{0}{}{7}{j}\, n^{j}\right]\right) \\ =\; & 6\, n + n^{7} + 7 + \sum\limits_{j = 1}^{6} \left[\genfrac{(}{)}{0}{}{7}{j}\, n^{j}\right]\right) \end{aligned}$ .

For this particular $n$ , $(6\, n + n^{7})$ is divisible by $7$ by the induction hypothesis.

$\sum\limits_{j = 1}^{6} \left[\genfrac{(}{)}{0}{}{7}{j}\, n^{j}\right]$ is also divisible by $7$ since $n$ is an integer and (by lemma) each of the coefficients $\genfrac{(}{)}{0}{}{7}{j} = (7!) / (j! \, (7 - j)!)$ is divisible by $7\!$ .

Therefore, $6\, (n + 1) + (n + 1)^{7}$ , which is equal to $6\, n + n^{7} + 7 + \sum\limits_{j = 1}^{6} \left[\genfrac{(}{)}{0}{}{7}{j}\, n^{j}\right]\right)$ , is divisible by $7$ .

In other words, for any integer $n \ge 1$ , if $(6\, n + n^{7})$ is divisible by $7$ , then $6\, (n + 1) + (n + 1)^{7}$ would also be divisible by $7\!$ .

Therefore, $(6\, n + n^{7})$ is divisible by $7$ for all integers $n \ge 1$ .

6 0

2 years ago

2/5 tons =____ pounds

katrin [286]

The correct answer is 800 pounds

7 0

3 years ago

Read 2 more answers

Beck used a compass and straightedge to accurately construct line segment OS, as shown in the figure below:

Setler79 [48]

Answer is: <span>m<POS=60 m<POQ=120</span>

4 0

2 years ago