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raketka [301]
3 years ago
14

What is the slope of the line segment? 1/4 -1/4 4 -4

Mathematics
2 answers:
oksian1 [2.3K]3 years ago
6 0

Answer:

its 4

Step-by-step explanation:

I just did the test and got it correct

Vinvika [58]3 years ago
3 0
Slope = 16/4  = 4 Answer
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Need help on this ASAP!!
dlinn [17]
D=3G divided by h hope this helped!
5 0
3 years ago
What two numbers add to 92 and multiply to 160?
KatRina [158]
None But, 80 x 2 = 160 and 80 + 2 = 82, are you sure you mean 92?

3 0
3 years ago
A swimming pool whose volume is 10 comma 000 gal contains water that is 0.03​% chlorine. Starting at tequals​0, city water conta
Katarina [22]

Answer:

C(60) = 2.7*10⁻⁴

t = 1870.72 s

Step-by-step explanation:

Let x(t) be the amount of chlorine in the pool at time t. Then the concentration of chlorine is  

C(t) = 3*10⁻⁴*x(t).

The input rate is 6*(0.001/100) = 6*10⁻⁵.

The output rate is 6*C(t) = 6*(3*10⁻⁴*x(t)) = 18*10⁻⁴*x(t)

The initial condition is x(0) = C(0)*10⁴/3 = (0.03/100)*10⁴/3 = 1.

The problem is to find C(60) in percents and to find t such that 3*10⁻⁴*x(t) = 0.002/100.  

Remember, 1 h = 60 minutes. The initial value problem is  

dx/dt= 6*10⁻⁵ - 18*10⁻⁴x =  - 6* 10⁻⁴*(3x - 10⁻¹)               x(0) = 1.

The equation is separable. It can be rewritten as dx/(3x - 10⁻¹) = -6*10⁻⁴dt.

The integration of both sides gives us  

Ln |3x - 0.1| / 3 = -6*10⁻⁴*t + C    or    |3x - 0.1| = e∧(3C)*e∧(-18*10⁻⁴t).  

Therefore, 3x - 0.1 = C₁*e∧(-18*10⁻⁴t).

Plug in the initial condition t = 0, x = 1 to obtain C₁ = 2.9.

Thus the solution to the IVP is

x(t) = (1/3)(2.9*e∧(-18*10⁻⁴t)+0.1)

then  

C(t) = 3*10⁻⁴*(1/3)(2.9*e∧(-18*10⁻⁴t)+0.1) = 10⁻⁴*(2.9*e∧(-18*10⁻⁴t)+0.1)

If  t = 60

We have

C(60) = 10⁻⁴*(2.9*e∧(-18*10⁻⁴*60)+0.1) = 2.7*10⁻⁴

Now, we obtain t such that 3*10⁻⁴*x(t) = 2*10⁻⁵

3*10⁻⁴*(1/3)(2.9*e∧(-18*10⁻⁴t)+0.1) = 2*10⁻⁵

t = 1870.72 s

7 0
3 years ago
Han's house is 450 meters from school. Lin's house is 135 meters closer to school. how far is Lin's house from school
Alenkinab [10]

Answer:

Lins house is 315 metres from school

Step-by-step explanation:

It's simple, just do 450-135.

3 0
4 years ago
Read 2 more answers
Find the angle that makes them supplementary.
8090 [49]

Answer:

Mathematics NCERT Grade 7, Chapter 5: Lines and Angles is all about different lines, line segments, and angles.

A line segment has two end points.

A ray has only one end point (its vertex).

A line has no end points on either side.  

This chapter throws light on topics such as related angles, pair of lines. The chapter deals with different types of angles and lines. The types of angles discussed in this chapter are as follows:

Pairs of Angles          Condition

Two complementary angles Measures add up to 90°

Two supplementary angles Measures add up to 180°

Two adjacent angles Have a common vertex and a common arm but no common interior

Linear pair Adjacent and supplementary

A linear pair is a pair of adjacent angles whose non-common sides are opposite rays.

All the types are discussed in detail and are supplemented with examples and short questions.  

Under the topic pair of lines, various sub-sections are discussed namely:

Intersecting lines

Transversal  

Angles made by a transversal

When two lines intersect (looking like the letter X) we have two pairs of opposite angles. They are called vertically opposite angles. They are equal in measure.

A transversal is a line that intersects two or more lines at distinct points.

Six angles discussed in this section:

1. Interior angles

2. Exterior angles

3. Pairs of Corresponding angles

4. Pairs of Alternate interior angles

5. Pairs of Alternate exterior angles

6. Pairs of interior angles on the same side of the transversal

Transversal of parallel lines

5 0
3 years ago
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