All categories

3 years ago

Examine the following diagram:

Mathematics

2 answers:

enyata [817]3 years ago

7 0

Answer:

Step-by-step explanation:

Tanya [424]3 years ago

5 0

Answer:

Step-by-step explanation:

i just took it on edge

You might be interested in

Use the inequality 18 < -3(4x - 2) Solve the inequality for x. Show your work.

IrinaVladis [17]

Answer:

x > -1

Step-by-step explanation:

Simplify the inequality using the distributive property (multiply the term outside the bracket with each number inside the bracket). Then, isolate 'x' by performing the reverse operations for every number that's on the same side as 'x'. (Reverse operations 'cancel out' a number.)

18 < -3(4x - 2) Expand this to simplify

18 < (-3)(4x) - (-3)(2) Multiply -3 with 4x and -2

18 < -12x + 6 Start isolating 'x'

18 - 6 < -12x + 6 - 6 Subtract 6 from both sides

18 - 6 < -12x '+ 6' is cancelled out on the right side

12 < -12x Subtracted 6 from 18 on the left side

12/-12 < -12x/-12 Divide both sides by -12

12/-12 < x 'x' is isolated. Simplify left side

-1 < x Answer

x > -1 Standard formatting puts variable on the left side

6 0

3 years ago

The lengths of a rectangle have been measured to the nearest tenth of a centimetre they are 87.3cm and 51.8cm what is the upper

vagabundo [1.1K]

Answer: Area (upper bound) = 4527.7056 cm²

Perimeter (lower bound) = 278 cm

Step-by-step explanation:

The length and width of the rectangle have been ROUNDED to the nearest tenth. Let's calculate what their actual measurements could be:

LENGTH: rounded to 87.3, actual is between 87.25 and 87.34

87.25 is the lowest number it could be that would round it UP to 87.3

87.34 is the highest number it could be that would round DOWN to 87.3

WIDTH: rounded to 51.8, actual is between 51.75 and 51.84

51.75 is the lowest number it could be that would round it UP to 51.8

51.84 is the highest number it could be that would round DOWN to 51.8

To find the Area of the upper bound, multiply the highest possible length and the highest possible width:

A = 87.34 × 51.84 = 4527.7056

To find the Perimeter of the lower bound, calculate the perimeter using the lowest possible length and the lowest possible width:

P = 2(87.25 + 51.75) = 278

8 0

3 years ago

Read 2 more answers

If log2 5 = k, determine an expression for log32 5 in terms of k.

lukranit [14]

Answer:

$log_3_2(5)=\frac{1}{5} k$

Step-by-step explanation:

Let's start by using change of base property:

$log_b(x)=\frac{log_a(x)}{log_a(b)}$

So, for $log_2(5)$

$log_2(5)=k=\frac{log(5)}{log(2)}\hspace{10}(1)$

Now, using change of base for $log_3_2(5)$

$log_3_2(5)=\frac{log(5)}{log(32)}$

You can express $32$ as:

$2^5$

Using reduction of power property:

$log_z(x^y)=ylog_z(x)$

$log(32)=log(2^5)=5log(2)$

Therefore:

$log_3_2(5)=\frac{log(5)}{5*log(2)}=\frac{1}{5} \frac{log(5)}{log(2)}\hspace{10}(2)$

As you can see the only difference between (1) and (2) is the coefficient $\frac{1}{5}$ :

So:

$\frac{log(5)}{log(2)} =k\\$

$log_3_2(5)=\frac{1}{5} \frac{log(5)}{log(2)} =\frac{1}{5} k$

6 0

3 years ago

Given f(x)=1/x+5 and g(x)=x-2

vladimir1956 [14]

For this case we have the following functions:

$f (x) = \frac {1} {x + 5}\\g (x) = x-2$

We must find f (g (x)). So:

$f (g (x)) = \frac {1} {(x-2) +5} = \frac {1} {x-2 + 5} = \frac {1} {x + 3}$

Finally we have to:

$f (g (x)) = \frac {1} {x + 3}$

For the function to be defined the denominator must be different from 0. That is x different from -3.

Answer:

Option B

3 0

3 years ago

It takes 22 hours for a hosepipe with a flow of 12 litres per minute to fill a swimming pool. How long will it take if its flow

Monica [59]

Answer:

t2 = 52.8 hours

it will take 52.8 hours if its flow is reduced to 5 litres per minute.

Step-by-step explanation:

The volume of the swimming pool can be written as;

Volume V = flow rate × time

V = R×t

V = R1×t1 = R2×t2

R1×t1 = R2×t2

t2 = R1×t1/R2 ........1

Given;

Flow rate R1 = 12 litres per minute

Flow rate R2 = 5 litres per minute

time t1 = 22 hours

Substituting the given values into equation 1;

t2 = R1×t1/R2 = 12×22/5

t2 = 52.8 hours

it will take 52.8 hours if its flow is reduced to 5 litres per minute.

3 0

3 years ago