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3 years ago

Please answer attached Parallel lines with Multiple transversals

Mathematics

1 answer:

love history [14]3 years ago

3 0

Answer:

Step-by-step explanation:

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A. 90 B. 105 C. 225 D. 315

Simora [160]

Answer:

225 is the correct one

Step-by-step explanation:

u multiply 15m*9 months u get 135 and then u add the existing 90m and then it becomes 225mp3

15m*9+90

7 0

3 years ago

This is the homework question (Please see picture). If my X and Y's are correct how do I graph it?

Nataliya [291]

Answer:

to graph it. make a horizontal line for the x axis and a vertical line for the y axis and place your coordinate points were the are supposed to be and label them.

Step-by-step explanation:

Make the y and x axis.
label the coordinate points.
Example:

The coordinate points(3,7) means that 3 would be on the horizontal line and 7 would be on the vertical line meaning the coordinate point(3,7) would be located were the two numbers meet.

5 0

3 years ago

Question -

Sauron [17]

$\rule{200}4$

Answer : The required ratio is (14m-6):(8m+23) .

$\rule{200}4$

Here we are given that the ratio of sum of first n terms of two AP's is (7n + 1):(4n + 27) .

That is.

$\small\sf\longrightarrow \dfrac{S_1}{S_2}=\dfrac{7n +1}{4n +27} \\$

As , we know that the sum of n terms of an AP is given by ,

$\small\sf\longrightarrow \pink{ S_n =\dfrac{n}{2}[2a +(n-1)d]} \\$

Assume that ,

First term of 1st AP = a
First term of 2nd AP = a'
Common difference of 1st AP = d
Common difference of 2nd AP = d'

Using this we have ,

$\small\sf\longrightarrow \dfrac{S_1}{S_2}=\dfrac{\dfrac{n}{2}[2a + (n-1)d]}{\dfrac{n}{2}[2a' +(n-1)d'] } \\$

$\small\sf\longrightarrow \dfrac{7n+1}{4n+27}=\dfrac{2a + (n -1)d}{2a' + (n -1)d' } . . . . . (i) \\$

Now also we know that the nth term of an AP is given by ,

$\longrightarrow\sf\small \pink{ T_n = a + (n-1)d}\\$

Therefore,

$\longrightarrow\sf\small \dfrac{T_{m_1}}{T_{m_2}}= \dfrac{ a + (n-1)d }{a'+(n-1)d'}. . . . . (ii)\\$

$\longrightarrow\sf\small \dfrac{T_1}{T_2}=\dfrac{2a + (2n-2)d}{2a'+(2n-2)d'} . . . . . (iii)\\$

From equation (i) and (iii) ,

$\longrightarrow\sf\small n-1 = 2m-2\\$

$\longrightarrow\sf\small n = 2m -2+1 \\$

$\longrightarrow\sf\small n = 2m -1 \\$

Substitute this value in equation (i) ,

$\longrightarrow \sf\small \dfrac{2a+ (2m-1-1)d}{2a' +(2m-1-1)d'}=\dfrac{7(2m-1)+1}{4(2m-1) +27}\\$

Simplify,

$\longrightarrow\sf\small \dfrac{ 2a + (2m-2)d}{2a' +(2m-2)d'}=\dfrac{14m-7+1}{8m-4+27}\\$

$\longrightarrow\sf\small \dfrac{2[a + (m-1)d]}{2[a' + (m-1)d']}=\dfrac{ 14m-6}{8m+23}\\$

$\longrightarrow\sf\small \dfrac{[a + (m-1)d]}{[a' + (m-1)d']}=\dfrac{ 14m-6}{8m+23}\\$

From equation (ii) ,

$\longrightarrow\sf\small \underline{\underline{\blue{ \dfrac{T_{m_1}}{T_{m_2}}=\dfrac{ 14m-6}{8m+23}}}}\\$

$\rule{200}4$

8 0

3 years ago

In the polynomial below, what number should replace the question mark to produce a difference of squares?

Stella [2.4K]

A polynomial is an algebraic equation that consists of more than one term. Otherwise, it is called a monomial. The technique in producing a difference of the squares is completing of the squares method. For a quadratic formula with a general form of ax2 + bx + c = 0, you can determine c such that the formula could be factored into two monomials. The equation would be (b/2)^2.

For example, if the equation is x^2 + 2x+ __ = 0, c is solved as follows:

c = (2/2)^2 = 1

The equation would be x^2 + 2x +1 = 0. When you simplify and factor it out, the equation would become (x+1)^2 = 0.

3 0

3 years ago

The vertices of the trapezoid are J(4m, 4n), K(4q, 4n), M(4p, 0), and L(0, 0). Find the midpoint of the midsegment of the trapez

Lilit [14]

To get the midsegment, namely HN, well, we need H and N

hmm so.... notice the picture you have there, is just an "isosceles trapezoid", namely, it has two equal sides, the left and right one, namely JL and KM

the midpoint of JL is H and the midpoint of KM is N

thus

$\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ \square}}\quad ,&{{ \square}})\quad 
% (c,d)
&({{ \square}}\quad ,&{{ \square}})
\end{array}\qquad
% coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)\\\\
-----------------------------\\\\$

$\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
J&({{ 4m}}\quad ,&{{ 4n}})\quad 
% (c,d)
L&({{ 0}}\quad ,&{{ 0}})
\end{array}\qquad
% coordinates of midpoint 
\left(\cfrac{0+4m}{2}\quad ,\quad \cfrac{0+4n}{2} \right)
\\\\\\
\left( \cfrac{4m}{2},\cfrac{4n}{2} \right)\implies \boxed{(2m,2n)\impliedby H}\\\\
-----------------------------\\\\$

$\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
K&({{ 4q}}\quad ,&{{ 4n}})\quad 
% (c,d)
M&({{ 4p}}\quad ,&{{ 0}})
\end{array}\qquad
% coordinates of midpoint 
\left(\cfrac{4p+4q}{2}\quad ,\quad \cfrac{0+4n}{2} \right)
\\\\\\
\left( \cfrac{2(2p+2q)}{2},\cfrac{4n}{2} \right)\implies \boxed{[(2p+2q), 2n]\impliedby N}\\\\
-----------------------------\\\\$

$\bf \textit{so, the midpoint of HN is }
\\\\\\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
H&({{ 2m}}\quad ,&{{ 2n}})\quad 
% (c,d)
N&({{ 2p+2q}}\quad ,&{{ 2n}})
\end{array}\\\\\\
% coordinates of midpoint 
\left(\cfrac{(2p+2q)+2m}{2}\quad ,\quad \cfrac{2n+2n}{2} \right)
\\\\\\
\left( \cfrac{2(p+q+m)}{2},\cfrac{4n}{2} \right)\implies (p+q+m)\quad ,\quad 2n$

5 0

3 years ago

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