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Elina [12.6K]
3 years ago
7

Which pair of lines have the same slopes and y-intercepts?

Mathematics
1 answer:
kirza4 [7]3 years ago
4 0

Answer:

parallel lines have the same slopes

You might be interested in
A linear equation in two variables has infinitely many solution the set of all solutions to a linear equation in two variables f
Greeley [361]
<span>1, During a 1-hr television program, there were 22 commercials, Some commercials were 15 sec and some were 30 sec long. Find the number of 15- sec commercials and the number of 30 sec commercial if the total playing time for commercial was 9.5 minutes.

</span>a) <span>show how to formulate your system of equations for your problem,

Answer:
- state the variables: x number of 15 sec commercials, y number of 30 sec commercials
- translate the word statement into algebraic language

b) state the two equations:

- translate the word statement into algebraic equations:

</span><span>* there were 22 commercials => x + y = 22
* the total playing time for commercial was 9.5 minutes => 15x + 30y = 9.5*60 </span><span>

Answer:
Equation (1) x + y = 22
Equation (2) 15x + 30y =  570

c) state which method you will use to solve either addition or substitution

Answer:
adition: multiply the first equation times - 15 and add the two equations.

d) solve your system showing and explaining each step of the process

Answer:
- muliply eq (1) times - 15

-15x - 15y = - 330

- add that to the eq (2):

-15x + 15x - 15y + 30y = -330 + 570

- add like terms: 15y = 240

- divide both members by 15: 15y/15 = 240 / 15 => y = 16

- replace y = 16 in eq I(1) => x + 16 = 22 => x = 22 - 16 = 6

Solution: x = 6, y = 16

e) Be sure to check your solution!

x + y = 22
6 + 16 = 22
22 = 22 --> check

15x + 30y = 9.5*60
15(6) + 30(16) = 570
90 + 480 = 570
570 = 570 ---> check

f) Once a solution is found explain what this solution means in the context of the problem, write the answer to the word problem in words</span><span>

The solution means that d</span><span>uring a 1-hr television program, there were</span><span> 6 commercials of 15 sec and 16 commercials of 30 sec.

2, How many quarts of water should be mixed with a 30% vinegar solution to obtain 12 qt of a 25% vinegar solution. (Hint: water is 0% vinegar).</span>

a) variables:

- state the variables
x: number of quarters of water
y: number of quarters of vinegar

- translate the words into mathematical language

12 qt 25% vinegar solution

=> total number of quarts = 12 = x + y

balance in vinegar:

vinegar from water + vinegar from 30% vinegar solution = vinegar in the 25% vinegar solution

0 + 0.3y = 0.25 (x + y)

c) Equations

Eq (1) x + y = 12

Eq (2): 0.3y = 0.25x + 0.25y

=> 0.25x - 0.05y = 0

d) solve

- multiply eq (1) times 0.05 and add to eq (2)

0.05x + 0.05y = 0.6
0.25x - 0.05y = 0
------------------------
0.30x = 0.6

x = 0.6 / 0.3 = 2

x + y = 12 => y = 12 - y = 12 - 2 = 10

Solution: x = 2, y = 10

e) check:

x + y = 12
2 qt + 10qt = 12 qt
12 qt = 12 qt ---> check

vinegar:

10*0.3 = 12*0.25
3 = 3 -> check

f) Explanation of the solution

The solution means that you have to mix 2 qts of water with 10 qts of 30% vinegar solution to obtain 12 qt of 25% vinegar solution.

5 0
3 years ago
PLEASE ANSWER CORRECTLY, I WILL GIVE BRAINLIEST &lt;3 THANKS IN ADVANCE!
guajiro [1.7K]

Answer:

1 : 15

2: 12

3: 52

4: 27.3

Step-by-step explanation:

For #1:

if line m and n are perpendicular then they will create right angles

Right angles have a measure of 90 degrees

That being said we can find the measure of the missing angle by subtracting the measure of the known angle (75) from 90

so k = 90 - 75

90 - 75 = 15

Hence k = 15.

For #3 ( very similar to #1, only difference is the values of the angles )

so R = 90 - 38

90 - 38 = 52

Hence, R = 52

For #2 and #4

Complementary angles have a sum of 90°

So like the previous questions we can find the measure of the missing angle by subtract the measure of the given angle from 90

∠V = 90 - 78

90 - 78 = 12

Hence ∠V = 12

∠Y = 90 - 62.7

90 - 62.7 = 27.3

Hence ∠Y = 27.3

6 0
2 years ago
Use a system of equations to solve this problem.Bronze is a mix, or alloy, of tin and copper. A metal worker needs 100 g of bron
d1i1m1o1n [39]

Answer:

x = 50 grs

y = 50 grs

Step-by-step explanation:

" x "   is quantity of bronze of 5% tin

" y "   is quantity of bronze of 45% tin

The worker needs 100 grs of bronze and this bronze must be 25% tin

The mathematical expression for the 100 grs condition is:

x + y = 100

And the The mathematical expression for the % condition is:

0,05*x  + 0.45*y  = 0,25 * ( x + y )

The system is:

x + y = 100

0,05*x + 0,45*y = 0,25* ( x + y )

Solving this system we get values of x and y

y = 100 - x

0,05*x  + 0,45* ( 100 - x ) = 0,25 * ( x + 100 - x )

0,05*x  + 45 - 0,45*x = 0,25*( 100 )

0,05*x  + 45 - 0,45*x  = 25

- 0,4*x  = -45 + 25

x = 20/0,4            x = 50  grs

and y = 50 grs

8 0
3 years ago
My brother is currently three times as old as me, but in five years, my brother
wolverine [178]

Answer: Your brother is 15 years old right now.

Step-by-step explanation:

Let be "m" your currently age and "b" your brother's currently age.

Knowing that right now he is three times (this indicates multiplication) as old as you, you can write the following equation:

b=3m

In five years your age will be:

 m+5

And your brother's age will be:

b+5

Then, if in five years, your  brother will be twice as old as you, you can set up the second equation:

b+5=2(m+5)

Then, the System of equations is:

\left \{ {{b=3m} \atop {b+5=2(m+5)}} \right.

 Apply the Substitution method and substitute the first equation into the second one and solve for "m":

3m+5=2(m+5)\\\\3m+5=2m+10\\\\m=10-5\\\\m=5

Substitute the value of "m" into the first equation and evaluate in order to find "b":

b=3(5)\\\\b=15

4 0
3 years ago
Why the derivative of (x^2/a^2) = (2x/a^2)? ​
Neko [114]

I assume you're referring to a function,

f(x) = \dfrac{x^2}{a^2}

where <em>a</em> is some unknown constant. By definition of the derivative,

\displaystyle f'(x) = \lim_{h\to0}{f(x+h)-f(x)}h

Then

\displaystyle f'(x) = \lim_{h\to0}{\frac{(x+h)^2}{a^2}-\frac{x^2}{a^2}}h \\\\ f'(x) = \frac1{a^2} \lim_{h\to0}{(x+h)^2-x^2}h \\\\ f'(x) = \frac1{a^2} \lim_{h\to0}{(x^2+2xh+h^2)-x^2}h \\\\ f'(x) = \frac1{a^2} \lim_{h\to0}{2xh+h^2}h \\\\ f'(x) = \frac1{a^2} \lim_{h\to0}(2x+h) = \boxed{\frac{2x}{a^2}}

8 0
2 years ago
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