Which expression is equivalent to the given expression?

Line Q is parallel to the line represented by the equation 5x – 4y = 10. Line Q passes through the point (12, –7). Which linear

marysya [2.9K]

A, plug in 12 for x and -7 for y to get you’re answer

3 0

3 years ago

Find the common ratio and an explicit form in each of the following geometric sequences.

Margaret [11]

Answer:

The explicit form is $a_{n}=162(2/3)^{n-1}$

Step-by-step explanation:

The explicit form of a geometric sequence is given by:

$a_{n}=ar^{n-1}$

where an is the nth term, a is the first term of the sequence and r is the common ratio.

In this case:

a=162

The value of the common ratio is obtained by dividing one term by the previous term.

For the first and second terms:

108/162=2/3

For the second and third terms (In order to prove that 2/3 is the common ratio)

72/108=2/3

Therefore:

r=2/3

Replacing a and r in the formula:

$a_{n}=162(2/3)^{n-1}$

7 0

3 years ago

(a) If G is a finite group of even order, show that there must be an element a = e, such that a−1 = a (b) Give an example to sho

Dahasolnce [82]

Answer:

See proof below

Step-by-step explanation:

First, notice that if a≠e and a^-1=a, then a²=e (this is an equivalent way of formulating the problem).

a) Since G has even order, |G|=2n for some positive number n. Let e be the identity element of G. Then A=G\{e} is a set with 2n-1 elements.

Now reason inductively with A by "pairing elements with its inverses":

List A as A={a1,a2,a3,...,a_(2n-1)}. If a1²=e, then we have proved the theorem.

If not, then a1^(-1)≠a1, hence a1^(-1)=aj for some j>1 (it is impossible that a^(-1)=e, since e is the only element in G such that e^(-1)=e). Reorder the elements of A in such a way that a2=a^(-1), therefore a2^(-1)=a1.

Now consider the set A\{a1,a2}={a3,a4,...,a_(2n-1)}. If a3²=e, then we have proved the theorem.

If not, then a3^(-1)≠a1, hence we can reorder this set to get a3^(-1)=a4 (it is impossible that a^(-1)∈{e,a1,a2} because inverses are unique and e^(-1)=e, a1^(-1)=a2, a2^(-1)=a1 and a3∉{e,a1,a2}.

Again, consider A\{a1,a2,a3,a4}={a5,a6,...,a_(2n-1)} and repeat this reasoning. In the k-th step, either we proved the theorem, or obtained that a_(2k-1)^(-1)=a_(2k)

After n-1 steps, if the theorem has not been proven, we end up with the set A\{a1,a2,a3,a4,...,a_(2n-3), a_(2n-2)}={a_(2n-1)}. By process of elimination, we must have that a_(2n-1)^(-1)=a_(2n-1), since this last element was not chosen from any of the previous inverses. Additionally, a_(2n1)≠e by construction. Hence, in any case, the statement holds true.

b) Consider the group (Z3,+), the integers modulo 3 with addition modulo 3. (Z3={0,1,2}). Z3 has odd order, namely |Z3|=3.

Here, e=0. Note that 1²=1+1=2≠e, and 2²=2+2=4mod3=1≠e. Therefore the conclusion of part a) does not hold

7 0

3 years ago

How do I solve these equations

Sever21 [200]

You need to know the properties of each function.
Tan is opposite (y) over adjacent (x). Sin>0 means that sin is positive, therefore, it is located on either quadrant 1 or 2. Tan=4/3 so it is located in quadrant 1.
The side of the triangle must be Adjacent=3 opposite=4 and hypotenuse=5
Now, you are asked to find the half angle of Cos which is
$\sqrt{\frac{1+cos}{2} }$
By following the formula, cos=3/5 then:
$\sqrt{ \frac{1- \frac{3}{5} }{2} }$ Multiply everything (inside the square $\sqrt{ \frac{5+3}{10} }$ root) by 5
$\frac{2 \sqrt{2} }{ \sqrt{10} }$
$\frac{2 \sqrt{20} }{10}$ Then just simplify
$\frac{ \sqrt{20} }{5}$ The answer is Square root of 20 over 5

3 0

3 years ago

Show that the line 4y = 5x-10 is perpendicular to the line 5y + 4x = 35

Shkiper50 [21]

Step-by-step explanation:

<h2><em><u>concept :</u></em></h2><h2 /><h2><em><u>concept :When two lines are perpendicular, then the product of their slopes is equivalent to -1.</u></em></h2><h2 /><h2><em><u>concept :When two lines are perpendicular, then the product of their slopes is equivalent to -1.Equation of line in the form y = mx + c have m as slope of line and c as y-intercept.</u></em></h2><h2 /><h2><em><u>concept :When two lines are perpendicular, then the product of their slopes is equivalent to -1.Equation of line in the form y = mx + c have m as slope of line and c as y-intercept.Solution:</u></em></h2><h2 /><h2><em><u>concept :When two lines are perpendicular, then the product of their slopes is equivalent to -1.Equation of line in the form y = mx + c have m as slope of line and c as y-intercept.Solution:Given equations of lines are</u></em></h2><h2 /><h2><em><u>concept :When two lines are perpendicular, then the product of their slopes is equivalent to -1.Equation of line in the form y = mx + c have m as slope of line and c as y-intercept.Solution:Given equations of lines are4y = 5x-10</u></em></h2><h2 /><h2><em><u>concept :When two lines are perpendicular, then the product of their slopes is equivalent to -1.Equation of line in the form y = mx + c have m as slope of line and c as y-intercept.Solution:Given equations of lines are4y = 5x-10or, y = (5/4)x(5/2).</u></em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>(</em><em>1</em><em>)</em></h2><h2 /><h2><em><u>5y + 4x = 35</u></em></h2><h2 /><h2><em><u>5y + 4x = 35ory = (-4/5)x + 7.</u></em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>(</em><em>2</em><em>)</em></h2><h2 /><h2><em><u>Let m and n be the slope of equations i and ii, respectively.</u></em></h2><h2 /><h2><em><u>Let m and n be the slope of equations i and ii, respectively.Here, m = 5/4</u></em></h2><h2 /><h2><em><u>Let m and n be the slope of equations i and ii, respectively.Here, m = 5/4n= -4/5</u></em></h2><h2 /><h2><em><u>Let m and n be the slope of equations i and ii, respectively.Here, m = 5/4n= -4/5therefore, mx n = -1</u></em></h2><h2 /><h2><em><u>Let m and n be the slope of equations i and ii, respectively.Here, m = 5/4n= -4/5therefore, mx n = -1Hence, the lines are perpendicular.</u></em></h2>

8 0

3 years ago