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hichkok12 [17]
2 years ago
11

Help me with hanger math...

Mathematics
1 answer:
andrew11 [14]2 years ago
6 0
The answer is 3 First you subtract 2 from 17 and then you divide 15 by 5 to get the answer of 3
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Find the equation of the line that is perpendicular to 3y-x=7 and passes through the point (2,-1
boyakko [2]
Hello
<span>an equation for the line in​ point-slope form and general form is :
y = ax+b   a : </span>slope  ; the <span>Passing through (x' ; y')
</span>y' = ax'+b 
y-y' =a(x-x')      and : x' =2   y' = - 1
calculate a :
let :  y = ax+b .....(D)
   ....<span>3y-x=7</span>....(D') or  y = (1/3)x+7/3
.(D) perpendicular to(D') : slope (D) × slope (D') = -1 

 slope (D') = 1/3
slop(D)×(1/3) = -1
slope (D) = -3
equation for the line    :   y-y' =a(x-x') 
y+1 =(-3) (x-2)
7 0
3 years ago
How could you use subtraction to find the area of a figure
wel

Answer:

brainly.com/question/7699717

Step-by-step explanation:

8 0
2 years ago
Solve the matrix equation for a, b, c, and d. [1 2] [a b] [6 5][3 4] [c d]= [19 8]
Anit [1.1K]

Answer:

The answer is "\bold{\left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}7&-2\\ -\frac{1}{2}&\frac{7}{2}\end{array}\right]}".

Step-by-step explanation:

\bold{\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]}

Solve the L.H.S part:

\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right]\\\\\\\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]

After calculating the L.H.S part compare the value with R.H.S:

\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]= \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]} \\\\

\to a+2c =6....(i)\\\\\to b+2d =5....(ii)\\\\\to 3a+4c =19....(iii)\\\\\to 3b+4d = 8 ....(iv)\\\\

In equation (i) multiply by 3 and subtract by equation (iii):

\to 3a+6c=18\\\to 3a+4c=19\\\\\text{subtract}... \\\\\to 2c = -1\\\\\to  c= - \frac{1}{2}

put the value of c in equation (i):

\to a+ 2 (- \frac{1}{2})=6\\\\\to a- 2 \times \frac{1}{2}=6\\\\\to a- 1=6\\\\\to a =6 +1\\\\\to a = 7\\

In equation (ii) multiply by 3 then subtract by equation (iv):

\to 3b+6d=15\\\to 3b+4d=8\\\\\text{subtract...}\\\\\to 2d = 7\\\\\to d= \frac{7}{2}\\

put the value of d in equation (iv):

\to 3b+4 (\frac{7}{2})=8\\\\\to 3b+4 \times \frac{7}{2}=8\\\\\to 3b+14=8\\\\\to 3b =8-14\\\\\to 3b = -6\\\\\to b= \frac{-6}{3}\\\\\to b= -2

The final answer is "\bold{\left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}7&-2\\ -\frac{1}{2}&\frac{7}{2}\end{array}\right]}".

4 0
3 years ago
Can you draw an isosceles triangle with only one 80 degree angle?
aev [14]

Answer:

The triangle is 50,50 80

Step-by-step explanation:

The three angles of a triangle equal 180 degrees

We only want one 80 degree angle.  That means the other two angles have to be the same, since it is an isosceles triangle.  Call the other angle x

x+x+80 = 180

2x+80 =180

Subtract 80 from each side

2x+80-80=180-80

2x =100

Divide by 2

2x/2=100/2

x = 50

Each of the other angles in the triangle would have to equal 50 degrees

The triangle is 50,50 80

6 0
3 years ago
Is 3.67676767 a irrational number
Nikitich [7]
I’m pretty sure it’s irrational
3 0
3 years ago
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