All categories

3 years ago

Inuicure ine answer choice that best completes the state

Mathematics

1 answer:

stira [4]3 years ago

6 0

Answer:

Step-by-step explanation:

If the numbers are unknown but there is a scale factor of 3, you'll have to multiply x and y by 3.

You might be interested in

Find the slope of the line through the given points. d (3, 7) and (3, −8)

Romashka [77]

Answer:

Undefined

Step-by-step explanation:

Use rise over run (y2 - y1) / (x2 - x1), where (x1, y1) and (y1, y2) are points on the line

Plug in the given points:

(y2 - y1) / (x2 - x1)

(-8 - 7) / (3 - 3)

-15/0

= undefined

The slope is undefined, because you cannot divide by 0.

6 0

3 years ago

Find the greatest common factor for the group of numbers. 12 , 28 , 24

irakobra [83]

The greatest common factor, or GCF, is quite self-explanatory. Between two numbers, you have to find the greatest number that can be divided by both numbers without any remainder. This is useful in solving addition and subtraction operations involving fractions. The technique to do this is to place the numbers to the right. Then, place the factors to the left. I'm gonna show the solution so that you can understand better.

4 | 12 28 24
----------------------
| 3 7 6
----------------------

Since all numbers can be divided by 4, you place it on the left side. Then, place the quotients on the next row below. Since, there is no more common factor, it ends with the 2nd row. Then, multiply all the left numbers with the numbers in the very last column.

GCF = 4×3×7×6
GCF = 504

8 0

4 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"

7 0

3 years ago

Que es el teorema del factor

Bingel [31]

Answer:

En álgebra, el teorema del factor es un teorema que vincula factores y ceros de un polinomio. Es un caso especial del teorema del resto polinómico.

Step-by-step explanation:

6 0

3 years ago

Can u tell me the the answer to 44×4 multiply in columns

m_a_m_a [10]

Answer:

176

hope it helps :)

mark brainliest!!

5 0

2 years ago