Ask question

Ask question

All categories

3 years ago

11

Solve this inequality. Check your solution. d + 5 < 7

2 answers:

Ostrovityanka [42]3 years ago

4 0

D+5<7
d<7-5
d<2
This is your answer.

luda_lava [24]3 years ago

3 0

D + 5 < 7
d < 7 - 5
d < 2 <=== solution ......do d (is less then) 2

check...
lets let d = 1 lets let d = 0
d + 5 < 7 d + 5 < 7
1 + 5 < 7 0 + 5 < 7
6 < 7......true 5 < 7....true

You might be interested in

Consider the NASCAR race, the Indianapolis 500. Its name refers to the total race distance of 500 miles. There are 44 cars are r

lukranit [14]

Answer: $5289.35

Step-by-step explanation:

Since the cars have an average mileage of 6.78 miles per gallon and the race is 500 miles, the number of gallons each car will take will be:

= 500/6.78

= 73.75 gallons

Since they're 44 cars, the total gallons of fuel will be:

= 44 × 73.75

= 3245 gallons

Since the fuel cost is $1.63 per gallon, the total fuel cost, will be:

= $1.63 × 3245

= $5289.35

8 0

3 years ago

Sophie Ruth is eating a 50-gram chocolate bar which contains 30%, percentcocoa.

Scorpion4ik [409]

X/50=30/100
15 grams of cocoa

7 0

3 years ago

Randy ordered 3 T-shirts for $25.67 each and one skirt for $23.55 and an 8% tax. What is the total cost of the order?

Setler [38]

The answer is $108.60 because $25.67 x 3= 77.01+ $23.55= 100.56 x 8% = $8.0448 + $100.56= $108.60

8 0

3 years ago

What is the steps to solve 25 milligrams per day for one week

topjm [15]

I don't really know what the question is that needs to be solved, but
here's a suggestion that might help:

Since there are seven days in a week, you could multiply 25 milligrams
by 7, and that will tell you the number of milligrams in a whole week.

8 0

3 years ago

Suppose that P(n) is a propositional function. Determine for which nonnegative integers n the statement P(n) must be true if a)

wel

Solution :

a). $P(0)$ is true

Then , $P(0+2)=P(2)$ is true.

$P(2+2)=P(4)$ is true

$P(4+2)=P(6)$ is true.

Therefore, we see that $P(n)$ is true for all the even integers : $\{0, 2,4,6,...\}$

b). $P(0)$ is true

Then , $P(0+3)=P(3)$ is true.

$P(3+3)=P(6)$ is true

$P(6+3)=P(9)$ is true.

Therefore, we see that $P(n)$ is true for all the multiples of 3 : $\{0, 3,6,9,12,...\}$

c). $P(0)$ and $P(1)$ is true, then $P(0+2)=P(2)$ is true

$P(1)$ and $P(2)$ is true, then $P(1+2)=P(3)$ is true.

$P(2)$ and $P(3)$ is true, then $P(2+2)=P(4)$ is true.

So, we observe that $P(n)$ is true for all the non- negative integers : $\{0, 1,2,3,4,5,6,...\}$ .

d). $P(0)$ is true,

So, $P(0+2)$ and $P(0+3)$ is true or $P(2)$ and $P(3)$ is true.

Now, $P(2)$ is true.

Again, $P(2+2)$ and $P(2+3)$ is true or $P(4)$ and $P(5)$ is true.

Now, $P(3)$ is true.

Again, $P(3+2)$ and $P(3+3)$ is true or $P(5)$ and $P(6)$ is true.

Thus,

$P(n)$ is true for all the non- negative integers except 1 : $\{0, 2,3,4,5,6,...\}$ .

3 0

3 years ago