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4 years ago

Write an expression for an increased by 3

Mathematics

2 answers:

choli [55]4 years ago

8 0

Please explain. What do you mean by increased by 3? Do you mean something like "5 + 3 = 8"?

Airida [17]4 years ago

5 0

5x + 6
6 more than 5 times a number
Hope this is correct!

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PLEASE ANSWER DUE IN 20 MIN

professor190 [17]

Answer:

Read

Step-by-step explanation:

Mean: add all the values and divide by the number of values

Median: when in order from least to greatest, the middle value

Mode: The value that ocoours most often

5 0

3 years ago

If a=(1 2 3 4) find A×A and the relation determined by (I) y=2x (II) x+y

kiruha [24]

Answer:

HOPE IT HELPS PLZ MARK ME BRAINLIEST

Step-by-step explanation:

A={1,2,3,4,5,6}

R={(x,y):y is divisible by x}

We know that any number (x) is divisible by itself.

(x,x)∈R

∴R is reflexive.

Now,(2,4)∈R [as 4 is divisible by 2]

But,(4,2)∈ /

R. [as 2 is not divisible by 4]

∴R is not symmetric.

Let (x,y),(y,z)∈R. Then, y is divisible by x and z is divisible by y.

∴z is divisible by x.

⇒(x,z)∈R

∴R is transitive.

Hence, R is reflexive and transitive but not symmetric.

4 0

3 years ago

Bt= 8500 *(8/27)^t/3After a special medicine is introduced into a Petri dish full of bacteria, the number of bacteria remaining

ozzi

Answer:

Every 1.71 seconds, the bacteria loses $\frac{1}{2}$

Step-by-step explanation:

Given

$B(t) = 8500 * (\frac{8}{27})^\frac{t}{3}\\$

Required [Missing from the question]

Every __ seconds, the bacteria loses $\frac{1}{2}$

First, we model the function from t/3 to t.

$B(t) = 8500 * (\frac{8}{27})^\frac{t}{3}\\$

Apply law of indices

$B(t) = 8500 * (\frac{8^\frac{1}{3}}{27^\frac{1}{3}})^t$

Evaluate each exponent

$B(t) = 8500 * (\frac{2}{3})^t$ --- This gives the number of bacteria at time t

At time 0, we have:

$B(0) = 8500 * (\frac{2}{3})^0$

$B(0) = 8500 * 1$

$B(0) = 8500$

Let r be the time 1/2 disappears.

When 1/2 disappears, we have:

$B(r) = \frac{B(0)}{2}$

$B(r) = \frac{8500}{2}$

$B(r) = 4250$

So, we have:

$B(t) = 8500 * (\frac{2}{3})^t$

Substitute r for t

$B(r) = 8500 * (\frac{2}{3})^r$

Substitute $B(r) = 4250$

$4250 = 8500 * (\frac{2}{3})^r$

Divide both sides by 8500

$\frac{4250}{8500} = (\frac{2}{3})^r$

$\frac{1}{2} = (\frac{2}{3})^r$

Take log of both sides

$log(\frac{1}{2}) = log (\frac{2}{3})^r$

Apply law of logarithm

$log(\frac{1}{2}) = r\ log (\frac{2}{3})$

Make r the subject

$r = log(\frac{1}{2}) / log (\frac{2}{3})$

$r = \frac{-0.3010}{-0.1761}$

$r = 1.71$

Hence, it reduces by 1/2 after every 1.71 seconds

6 0

3 years ago

Can anyone tell me how to do this? (Steps) and can you answer these two questions for me plz

Andrew [12]

Ax²+bx+c=0

The formula to solve for x is:

$x= \frac{-b(+/-) \sqrt{ b^{2}-4ac } }{2a}$

So for 3x²+13x+10=0, plug in the data

x=-13±√13²-4(3)(10) / 2(3)
x=-13±√169-120 / 6
x=-13±√49 /6
x=-13±7 / 6

x=-13+7/6
x=-6/6
x=-1

x=-13-7/ 6
x=-20/6 or 3.3repeating

3x²+14x+15=0
x=-14±√14²-4(3)(15) / 2(3)
x=-14±√196-180 /6
x=-14±√16 / 6

x=-14+4/6
x=-10/6 or 1.6repeating

x=-14-4/6
x=-18/6
x=-3

7 0

3 years ago

Soup is 3 cans for $0.82, so each can of soup is $0.2733. how much does it cost for each one?

Vladimir79 [104]

It would be $0.27 for each can

5 0

3 years ago