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3 years ago

Will someone please explain this to me?

Mathematics

2 answers:

svlad2 [7]3 years ago

6 0

Answer:

hi there!

the correct answer is: 29 degrees

Step-by-step explanation:

if you refer to the picture I have drawn below angle b is supplementary to the the 119 angle (i know I accidentally wrote 118) that means in order to find angle b you need to do 180-119 which equals 61

then since all angles in a triangle add up to 180, you need to perform (180-90-61) which equals 29

Elza [17]3 years ago

4 0

Step-by-step explanation:

m<DAC is the exterior angle of the right angle triangle, where m<ACB and m<ABC are the interior angles of the triangle opposite to the exterior angle

Sum of the opposite interior angles is equal to the exterior angle

m<ACB+ m<ABC=m<DAC

a°+90°= 119°

a°=119-90°

a°=29°

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You can replace any letter in function notation with any number.

Vikentia [17]

A. That would be True

5 0

3 years ago

The screen at Dallas' AT&T Stadium is 175 feet wide. A movie theatre screen with

Mazyrski [523]

Answer:

17150 ft²

Step-by-step explanation:

Given the ratio of sides = a : b , then

ratio of areas = a² : b²

Here ratio of sides = 175 : 40 = 35 : 8 , then

ratio of areas = 35² : 8² = 1225 : 64

let x be the area of AT & T Stadium then by proportion

$\frac{x}{1225}$ = $\frac{896}{64}$ ( cross- multiply )

64x = 896 × 1225 = 1097600 ( divide both sides by 64 )

x = 17150

The AT & T screen has an area of 17150 ft²

4 0

2 years ago

Find the area of a right triangle with a base of 38 cm and height of 26 cm.

Marianna [84]

Answer:

494cm2

Step-by-step explanation:

Área = (base × height)/2

Área=(38×26)/2=988/2=494cm²

7 0

2 years ago

Read 2 more answers

Find the Midpoint of a segment with the following endpoints. Type your

exis [7]

Answer:

3/2;5/2

Step-by-step explanation:

-1+4/2 ; 4+1/2

3/2 ; 5/2

6 0

2 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

Will someone please explain this to me?​

Will someone please explain this to me?