Ask question

Ask question

All categories

mrs_skeptik [129]

3 years ago

5

Solve for Z. 3z - 5 = 2z + 2

1 answer:

inna [77]3 years ago

7 0

The answer to your equation is z = 7

You might be interested in

The density of hydrogen is 0.0000899 .

Bumek [7]

0.00009 might help I’m not sure about the possible answers but that one seems reasonable

6 0

3 years ago

Read 2 more answers

Which number is divisible by 5 <br> A 44<br> B 125<br> C 116<br> D 114

Ray Of Light [21]

Answer:

B

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

A mother shares a box of 20 chocolates in the same ratio as her children's ages . Calculate how many chocolates each child gets

Anni [7]

So a box of 20 chocolates is divided between a group of people based on age So the ratio for Sibusiso, Dorris, and Freddy is 2:3:5 Add these together, and you get 10 So to find how much each person gets, divide total amount of chocolates by total age So (20÷10) = 2 So multiply each age term by 2 and you have your answer (4:6:10) So Sibusiso gets 4, Dorris gets 6 and Freddy gets 10

8 0

3 years ago

A lake polluted by bacteria is treated with an antibacterial chemical. Aftertdays, thenumber N of bacteria per milliliter of wat

jeyben [28]

Answer:

N(1)=50 is a minimum

N(15)=4391.7 is a maximum

Step-by-step explanation:

<u>Extrema values of functions </u>

If the first and second derivative of a function f exists, then f'(a)=0 will produce values for a called critical points. If a is a critical point and f''(a) is negative, then x=a is a local maximum, if f''(a) is positive, then x=a is a local minimum.

We are given a function (corrected)

$N(t) = 20(t^2-lnt^2)+ 30$

$N(t) = 20(t^2-2lnt)+ 30$

(a)

First, we take its derivative

$N'(t) = 20(2t-\frac{2}{t})$

Solve N'(t)=0

$20(2t-\frac{2}{t})=0$

Simplifying

$2t^2-2=0$

Solving for t

$t=1\ ,t=-1$

Only t=1 belongs to the valid interval $1\leqslant t\leqslant 15$

Taking the second derivative

$N''(t) = 20(2+\frac{2}{t^2})$

Which is always positive, so t=1 is a minimum

(b)

$N(1)=20(1^2-2ln1)+ 30$

N(1)=50 is a minimum

(c) Since no local maximum can be found, we test for the endpoints. t=1 was already determined as a minimum, we take t=15

(d)

$N(15)=20(15^2-2ln15)+ 30$

N(15)=4391.7 is a maximum

7 0

3 years ago

Madelyn is knitting a scarf for a holiday gift. So far, it's 4.5 centimeters long. She plans to

olchik [2.2K]

In order to make the question true the length should be 45cm not 4.5 cm , Solving accordingly

Let nights be x
Rate of change=5cm

The equation is

y=45-5x

As 45 is base amount we used it as y intercect

7 0

2 years ago