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3 years ago

PLEASE HELP ME ASAP!!!!

Mathematics

1 answer:

marishachu [46]3 years ago

5 0

Answer:

$\displaystyle y = -4x$

Step-by-step explanation:

0 = 0 ± b

$\displaystyle 0 = b \\ \\ y = -4x$

* Parallel Lines have SIMILAR <em>RATE OF CHANGES</em> [<em>SLOPES</em>], so −4 remains the way it is.

I am joyous to assist you anytime.

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Answer I’ll give u brain thingy

Zanzabum

Answer:

It means that for every 1 unit of time, the equation shows an increase of 8.38 units of distance.

Step-by-step explanation:

Slope is $\frac{rise}{run}$ . Any number that does not have a denominator technically has a denominator of 1. Therefore, your slope is actually $\frac{8.38}{1}$ . That means you have a rise, or the vertical progression, of 8.38 while your run, or your horizontal progression, is 1. For every unit your line travels horizontally, it travels 8.38 units vertically. I hope this makes sense.

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3 years ago

What transformations to the linear parent function, f(x) = x, give the function

77julia77 [94]

Answer:

A. Shift down 2 units.

B. Vertically stretch by a factor of 4.

Step-by-step explanation:

Given the function

f(x)=x

If we stretch y vertically by a factor of m, we have: y=m·f (x)

Therefore:

Vertically stretching f(x) by a factor of 4, we have: 4x.

Next, if we take down f(x) by k units we have: y= f(x)-k

Therefore: Taking down 4x by 2 units, we obtain:

g(x)=4x-2

Therefore, Options A and B applies.

7 0

3 years ago

Number 1d please help me analytical geometry

lesantik [10]

For a) is just the distance formula

$\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
A&({{ x}}\quad ,&{{ 1}})\quad 
% (c,d)
B&({{ -4}}\quad ,&{{ 1}})
\end{array}\qquad 
% distance value
\begin{array}{llll}

d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
\sqrt{8} = \sqrt{({{ -4}}-{{ x}})^2 + (1-1)^2}
\end{array}$
-----------------------------------------------------------------------------------------
for b) is also the distance formula, just different coordinates and distance

$\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
A&({{ -7}}\quad ,&{{ y}})\quad 
% (c,d)
B&({{ -3}}\quad ,&{{ 4}})
\end{array}\ \ 
\begin{array}{llll}

d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
4\sqrt{2} = \sqrt{(-3-(-7))^2+(4-y)^2}
\end{array}$
--------------------------------------------------------------------------
for c) well... we know AB = BC.... we do have the coordinates for A and B
so... find the distance for AB, that is $\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
A&({{ -3}}\quad ,&{{ 0}})\quad 
% (c,d)
B&({{ 5}}\quad ,&{{ -2}})
\end{array}\qquad 
% distance value
\begin{array}{llll}

d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}\\\\
d=\boxed{?}

\end{array}$

now.. whatever that is, is = BC, so the distance for BC is

$\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
B&({{ 5}}\quad ,&{{ -2}})\quad 
% (c,d)
C&({{ -13}}\quad ,&{{ y}})
\end{array}\qquad 
% distance value
\begin{array}{llll}

d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}\\\\
d=BC\\\\
BC=\boxed{?}

\end{array}$

so... whatever distance you get for AB, set it equals to BC, BC will be in "y-terms" since the C point has a variable in its ordered points

so.. .solve AB = BC for "y"
------------------------------------------------------------------------------------

now d) we know M and N are equidistant to P, that simply means that P is the midpoint of the segment MN

so use the midpoint formula

$\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
M&({{-2}}\quad ,&{{ 1}})\quad 
% (c,d)
N&({{ x}}\quad ,&{{ 1}})
\end{array}\qquad
% coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)=P
\\\\\\
$

$\bf \left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)=(1,4)\implies 
\begin{cases}
\cfrac{{{ x_2}} + {{ x_1}}}{2}=1\leftarrow \textit{solve for "x"}\\\\
\cfrac{{{ y_2}} + {{ y_1}}}{2}=4
\end{cases}$

now, for d), you can also just use the distance formula, find the distance for MP, then since MP = PN, find the distance for PN in x-terms and then set it to equal to MP and solve for "x"

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3 years ago

Pam draws three scalene triangles. In each figure, she measures each angle, as shown.

Ganezh [65]

Answer:

the answer is c.

Step-by-step explanation:

there's no explanation

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3 years ago

The price of gourmet coffee beans at the corner market has increased 10% per year over the past several years if the current cos

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3 years ago

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