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Lena [83]
3 years ago
7

josh is 49 inches tall. his father is 70 inches tall. what percent of his father's height is josh's height

Mathematics
2 answers:
ANTONII [103]3 years ago
7 0

Answer:

70%

Step-by-step explanation:


DedPeter [7]3 years ago
6 0

Answer:

Josh is 70% of his fathers height

Step-by-step explanation:

49/70=0.70

0.70 into a percent is 70%

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Will give brainliest if correct.
Sonja [21]
The answer is AB should be 2 feet not 3. :) Hopes this helps, have a good day.
8 0
2 years ago
Read 2 more answers
Can you please show me your math so I know how you got it?
yan [13]
So the answer is 14 ounces
Explanation
Since 16.8 is the original plus 20% we could say 16.8=120%

And since X is the original we could say X=100
So using that info we could put it into a proportion
120/100=16.8/x or
120. 16.8
——=. ———-
100. X
Then we cross multiply and divide
So,
120x=16.8(100)
120x=1680 then divide both sides by 120
X=14


7 0
2 years ago
What is the factor of the expression -30-21x
timama [110]

Answer:

GCF = -3

-3(7x+10)

Step-by-step explanation:

-21x - 30 =

-3(7x+10)

5 0
3 years ago
Read 2 more answers
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
for the following questions, determine how many solutions each equation has. if one solution, state the value of x. x+6+8=2x-x+1
vichka [17]

Answer:

infinite solutions

Step-by-step explanation:

x+6+8=2x-x+14

x+6+8=x+14

x+14=x+14

14=14

or

x=x

plug in any number

2+6+8=2(2)-2+14

16=16

another example

8+6+8=2(8)-8=14

22=22

5 0
3 years ago
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