Evaluate the expression using substitution ( m=3, y =-5) 5m + 3y/5 + 4( m-y)

Mathematics

1 answer:

sdas [7]2 years ago

6 0

Answer:

Step-by-step explanation:

5(3) +3(-5)/5 +4(3--5)

15 + -15/5 +4(8)

15 -3 +32

12+32

You might be interested in

is it possible to draw a tringle with 90 Angle and one leg that is 8 centimeters long and one leg that is 6 centimeters long? is

Delvig [45]

Yes it is The hypotenuse is 10 cms long.
Because its a right angled triangle ( 90 angle) the hypotenuse will be 10 cms long - it will obey the Pythagoras theorem:-

10^2 = 8^2 + 6^2

There is only one right angled triangle with these measurements but there are many of them which have sides in same ratio

for example

3, 4 and 5

9,12 and 15

12, 16 and 20 and so on..

5 0

3 years ago

What is a equivalent ratio to 8:9

LenKa [72]

Answer:16:18

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

If Jose divides his cards into 8 equal stacks, how many cards will there be in each stack? Will there be any cards left over?

gavmur [86]

Idk how many start cards there are o this problem cant be solved...

8 0

3 years ago

What is the difference in simplest form? <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bn%5E2%2B3n%2B2%7D%7Bn%5E2%2B6n%2B8%7D-%5

Paha777 [63]

$\bf \cfrac{n^2+3n+2}{n^2+6n+8}-\cfrac{2n}{n+4}\implies \cfrac{n^2+3n+21}{(n+4)(n+2)}-\cfrac{2n}{n+4}
\\\\\\
\textit{so our LCD will just be (n+4)(n+2)}
\\\\\\
\cfrac{n^2+3n+2~~~~-~~~~(n+2)(2n)}{(n+4)(n+2)}
\\\\\\
\cfrac{n^2+3n+2~~~~-~~~~(2n^2+4n)}{(n+4)(n+2)}
\\\\\\
\cfrac{n^2+3n+2~~~~-~~~~2n^2-4n}{(n+4)(n+2)}
\implies 
\cfrac{-n^2-n+2}{(n+4)(n+2)}
\\\\\\
\cfrac{-(n^2+n-2)}{(n+4)(n+2)}\implies \cfrac{-\underline{(n+2)}(n-1)}{(n+4)\underline{(n+2)}}\implies \cfrac{-(n-1)}{n+4}\implies \cfrac{1+n}{n+4}$

3 0

3 years ago

Two more questions , 15 points.

Mumz [18]

The formula that is useful for solving both of these problems is ...

$(v_2)^2-(v_1)^2=2ad\qquad\text{a=acceleration, d=distance}\\\\ \text{where $v_1$ and $v_2$ are initial and final velocities}$

9. Given v₁=15, a=9.8, d=10, find v₂.

... (v₂)² = 15² + 2·9.8·10

... v₂ = √421 ≈ 20.5 . . . . m/s

10. Given d=12 m when a=-9.8 m/s² and v₂=0, find d when a=0.17·(-9.8 m/s²).

The formula tells us that d=(v₁)²/(2a), which is to say that the distance is inversely proportional to the acceleration. If acceleration is 0.17 times that on earth, distance will be 1/0.17 ≈ 5.88 times that on earth.

(12 m)/0.17 ≈ 70.6 m

7 0

3 years ago