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3 years ago

Evaluate the expression for a = 14, b = 29, c = 1, d = 1: a + b - C-d

Mathematics

2 answers:

grandymaker [24]3 years ago

7 0

Answer:

Solution,

a=14

b=29

c=1

D=1

Now,

 $a + b - c - d \\ = 14 + 29 - 1 - 1 \\ = 43 - 1 - 1 \\ =42 - 1 \\ = 41$ 

hope this helps...

Good luck on your assignment...

Elina [12.6K]3 years ago

4 0

The correct answer is 41

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PLEASE HELP PLEASE!!!

defon

Answer:

We conclude that at x = 0 and x = -1, the value of f(x) = 2ˣ - 1 and g(x) = 1/2x is the same.

Therefore, the solution to f(x) = g(x) is:

x = -1

x - 0

Step-by-step explanation:

Given the table

x f(x) = 2ˣ - 1 g(x) = 1/2x

-2 -3/4 -1

-1 -1/2 -1/2

0 0 0

1 1 1/2

2 3 1

If we carefully observe, we can determine that

at x = 0, the value of f(x) = 2ˣ - 1 and g(x) = 1/2x is the same.

In other words,

at x = 0

f(0) = 0

g(0) = 0

Thus,

at x = 0

f(x) = g(x)

Also at x = -1, the value of f(x) = 2ˣ - 1 and g(x) = 1/2x is the same.

In other words,

at x = -1

f(-1) = -1/2

g(-1) = -1/2

Thus,

at x = -1

f(x) = g(x)

Summary:

Thus, we conclude that at x = 0 and x = -1, the value of f(x) = 2ˣ - 1 and g(x) = 1/2x is the same.

Therefore, the solution to f(x) = g(x) is:

x = -1

x - 0

6 0

2 years ago

Let f(x)=−3x. The graph of f(x) is transformed into the graph of g(x) by a vertical stretch of 4 units and a translation of 4 u

Serjik [45]

G(x) = -3x - 4

g(x) = |3x| + 4

I hope I answered the questions right

6 0

3 years ago

Read 2 more answers

One morning, John drive 5 hours before stopping to eat. After lunch, he. increased his speed by 10 mph. If he completed a 500-mi

Marrrta [24]

So he drove 5hrs, before lunch, now, he was going at speed, say "r", after he ate, he sped up by 10mph, so, his rate is whatever "r" is, plus 10, or " r + 10 ", after lunch

we know the whole trip took 10hrs, so, is 5hrs before lunch and 5hrs after lunch

we also know the whole trip was 500miles, so if he drove, say "d" miles before lunch, after lunch he drove the slack after lunch, or " 200 - d "

recall, your d = rt, distance = rate * time

$\bf \begin{array}{lccclll}
&distance&rate&time\\
&-----&-----&-----\\
\textit{before lunch}&d&r&5\\
\textit{after lunch}&500-d&r+10&5
\end{array}
\\\\\\

\begin{cases}
\boxed{d}=5r\\
500-d=(r+10)5\\
----------\\
500-\boxed{5r}=(r+10)5
\end{cases}
\\\\\\
\cfrac{500-5r}{5}=r+10\implies 100-r=r+10$

and I'm sure you know what that is

6 0

2 years ago

A high school track is shaped as a rectangle with a half circle on either side.

Ilia_Sergeevich [38]

Answer:

Option (3)

Step-by-step explanation:

Track is shaped as a rectangle with a semicircle on either side.

Therefore, length of the track = 2(Length of rectangle) + perimeter of a complete circle formed by joining the semicircles on each side.

= 2(96) + 2πr

= $192+2\pi (\frac{35}{2})$