All categories

3 years ago

Answer

Mathematics

1 answer:

denis-greek [22]3 years ago

5 0

Answer:

x = 1, y = 3

Step-by-step explanation:

Use the y and substitute it into the 2nd equation. Solve for x. Then use that answer to solve for y.

You might be interested in

What should i know about algebra 1 before i start

Contact [7]

Answer:nothin much

Step-by-step explanation:

8 0

2 years ago

Consider the expression 25 – 10 ÷ 2 + 3.

Tems11 [23]

Given:

The expression is:

$25-10\div 2+3$

To find:

Part A: The expression using parentheses so that the expression equals 23.

Part B: The expression using parentheses so that the expression equals 3.

Solution:

Part A:

In option A,

$(25-10)\div 2+3=15\div 2+3$

$(25-10)\div 2+3=7.5+3$ [Using BODMAS]

$(25-10)\div 2+3=10.5$

In option B,

$25-10\div (2+3)=25-10\div 5$

$25-10\div (2+3)=25-2$ [Using BODMAS]

$25-10\div (2+3)=23$

In option C,

$(25-10)\div (2+3)=15\div 5$

$(25-10)\div (2+3)=3$

In option D,

$25-(10\div 2)+3=25-5+3$

$25-(10\div 2)+3=28-5$ [Using BODMAS]

$25-(10\div 2)+3=23$

After the calculation, we have $25-10\div (2+3)=23$ and $25-(10\div 2)+3=23$ .

Therefore, the correct options are B and D.

Part B: From part A, it is clear that

$(25-10)\div (2+3)=3$

Therefore, the correct option is C.

8 0

2 years ago

A bus travels at a constant speed. The following table shows the distance the bus travels (in kilometers) over a period of time

Licemer1 [7]

The equation is two different types and a number and in the question you have to multiple

6 0

2 years ago

Find the number of two-letter permutations of the letters.<br><br> Q, I, E, R, T, Y, U

fenix001 [56]

A sword is Saskatchewan. Mdkdodn idioms has nsito day you can come to sleep early tomorrow and you yo me

5 0

2 years ago

A car has mass 1500 kg and is traveling at a speed of 35 miles/hour. what is its kinetic energy in joules? (Be sure to convert m

photoshop1234 [79]

Answer:

Factor by which kinetic energy increase = 4 times

Step-by-step explanation:

Given,

Mass of the car, v1 = 1500 kg

initial speed of car = 35 miles/h

$=\dfrac{35\times 1609.34}{3600}\ m/s$

= 15.64 m/s

Initial kinetic energy of the car is given by,

$k_1\ =\ \dfrac{1}{2}.m.v_1^2$

$=\ \dfrac{1}{2}\times 1500\times (15.64)^2\ joule$

= 183606.46 J

Final velocity of car v2 = 70 miles/hour

$=\dfrac{70\times 1609.34}{3600}$

= 31.29 m/s

So, final kinetic energy of car is given by

$k_2\ =\ \dfrac{1}{2}.m.v_2^2$

$=\ \dfrac{1}{2}\times 1500\times (31.29)^2$

= 734425.84 J

Now, the ratio of final to initial kinetic energy can be given by,

$\dfrac{k_2}{k_1}=\ \dfrac{734425.84}{183606.46}$

$=>\ k_2\ =\ 4k_1$

Hence, the kinetic energy will increase by 4 times.

7 0

2 years ago