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2 years ago

PLEASE HELP SOON!!!!!!!!!!!!!!!

Mathematics

2 answers:

qwelly [4]2 years ago

4 0

Answer:

Step-by-step explanation:

10 - (-20) turns into 10+20 and 8+(-3) is just 8-3

Luden [163]2 years ago

4 0

Answer:

Step-by-step explanation:

remove brackets

8-3+10+20=35

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Two trains started moving at the same time, train A from Boston to New York, and train B from New York to Boston. The distance b

Ede4ka [16]

Answer:

After 1.7 hours

Explanation:

1. Calculate the average speed of train B

60mph = 3/4 (s) ⇒ s = 60 (4/3) mph = 80 mph

2. Build a table

When the two trains meet:

Train Average speed Distance time (distance/average speed)

mph

A 60 x x/60

B 80 240 - x (240 - x) / 80

3. Write the equation

The time, when the two trains meet, is the same for both trains:

$\dfrac{x}{60}=\dfrac{240-x}{80}\\ \\ \\ 80x=14,400-60x\\ \\ 80x+60x=14,400\\ \\ 140x=14,400\\ \\ x=14,400/140\\ \\ x=102.85miles$

4. Calculate the time:

x/60 = 102.86 / 60 = 1.7 hours

The two trains will meet after 1.7 hours

7 0

2 years ago

If a person rides their bike 40 miles in 5 hours, then they rode their bike at a speed of ___ Mph?

monitta

Answer:

8 mph

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Determine whether the sets have a subset relationship. R = {right triangles} O = {obtuse triangles} a) R and O are disjoint. b)

DerKrebs [107]

Answer:

Step-by-step explanation:

A right angled triangle has one angle equal to 90°

an obtuse angle triangle has one angle greater tan 90°

a) An o and right-angled triangle btuse angled triangle can have one angle equal.

b) R and O are not equivalent by definition

c) subset of R and Oare not all triangles as R and O are categories of all triangles. R and O are subset of all triangles

d) Acute triangles have all acute angles. So subset of R and O can't be all acute traingles

e) All triangles with two acute angles are may or may not have third angle as obtuse angle or 90° angle

f) none of the above

5 0

3 years ago

Given f(x) =

sergejj [24]

Answer:

Step-by-step explanation:

We are given the function:

$\displaystyle f(x) = \left\{ \begin{array}{ll} 2\cos(\pi x) \text{ for } x \leq -1 \\ \\ \displaystyle \frac{2}{\cos(\pi x)}\text{ for } x > -1 \end{array} \right.$

And we want to find:

$\displaystyle \lim_{x\to -1}f(x)$

So, we need to determine whether or not the limit exists. In other words, we will find the two one-sided limits.

Left-Hand Limit:

$\displaystyle \lim_{x\to-1^-}f(x)$

Since we are approaching from the left, we will use the first equation:

$\displaystyle =\lim_{x\to -1^-}2\cos(\pi x)$

By direct substitution:

$=2\cos(\pi (-1))=2\cos(-\pi)=2(-1)=-2$

Right-Hand Limit:

$\displaystyle \lim_{x\to -1^+}f(x)$

Since we are approaching from the right, we will use the second equation:

$=\displaystyle \lim_{x\to -1^+}\frac{2}{\cos(\pi x)}$

Direct substitution:

$\displaystyle =\frac{2}{\cos(\pi (-1))}=\frac{2}{\cos(-\pi)}=\frac{2}{(-1)}=-2$

So, we can see that:

$\displaystyle \displaystyle \lim_{x\to-1^-}f(x)=\displaystyle \lim_{x\to -1^+}f(x) =-2$

Since both the left- and right-hand limits exist and equal the same thing, we can conclude that:

$\displaystyle \lim_{x \to -1}f(x)=-2$

Our answer is A.

8 0

2 years ago

Integers between-4 and 2,inclusive

rjkz [21]

The integers between -4 and 2, inclusive would be -4, -3, -2, -1, 0, 1, and 2

3 0

3 years ago