I need help on this one please :)

A rectangular box has length x and width 3. The volume of the box is given by y = 3x(8 – x). The greatest x-intercept of the gra

CaHeK987 [17]

Consider rectangular box with

length x units (x≥0);
width 3 units;
height (8-x) units (8-x≥0, then x≤8).

The volume of the rectangular box can be calculated as

$V_{box}=\text{length}\cdot \text{width}\cdot \text{height}.$

In your case,

$V_{box}=3\cdot x\cdot (8-x).$

Note that maximal possible value of the height can be 8 units (when x=0 - minimal possible length) and the minimal possible height can be 0 units (when x=8 - maximal possible length).

From the attached graph you can see that the greatest x-intercept is x=8, then the height will be minimal and lenght will be maximal.

Then the volume will be V=0 (minimal).

Answer: correct choices are B (the maximum possible length), C (the minimum possible height)

7 0

3 years ago

Which number should be added to both sides of the equation to complete the square x^2-10x=7

mixer [17]

Answer:

25

Step-by-step explanation:

The number that needs to be added is the square of half the x-coefficient:

(-10/2)^2 = 25

_____

Adding that gives ...

(x -5)^2 = 32

x = 5 ± 4√2

5 0

4 years ago

Simplify this expression. 2(x - 3) + 4x

Natasha2012 [34]

Answer:

Step-by-step explanation:

2x-6+4x=6x-6

Hope this helps!

6 0

3 years ago

Read 2 more answers

find the area of the trapezium whose parallel sides are 25 cm and 13 cm The Other sides of a Trapezium are 15 cm and 15 CM

Snezhnost [94]

$\huge\underline{\red{A}\blue{n}\pink{s}\purple{w}\orange{e}\green{r} -}$

Given - A trapezium ABCD with non parallel sides of measure 15 cm each ! along , the parallel sides are of measure 13 cm and 25 cm

To find - Area of trapezium

Refer the figure attached ~

In the given figure ,

AB = 25 cm

BC = AD = 15 cm

CD = 13 cm

Construction -

$draw \: CE \: \parallel \: AD \: \\ and \: CD \: \perp \: AE$

Now , we can clearly see that AECD is a parallelogram !

$\therefore$ AE = CD = 13 cm

Now ,

$AB = AE + BE \\\implies \: BE =AB - AE \\ \implies \: BE = 25 - 13 \\ \implies \: BE = 12 \: cm$

Now , In ∆ BCE ,

$semi \: perimeter \: (s) = \frac{15 + 15 + 12}{2} \\ \\ \implies \: s = \frac{42}{2} = 21 \: cm$

Now , by Heron's formula

$area \: of \: \triangle \: BCE = \sqrt{s(s - a)(s - b)(s - c)} \\ \implies \sqrt{21(21 - 15)(21 - 15)(21 - 12)} \\ \implies \: 21 \times 6 \times 6 \times 9 \\ \implies \: 12 \sqrt{21} \: cm {}^{2}$

Also ,

$area \: of \: \triangle \: = \frac{1}{2} \times base \times height \\ \\\implies 18 \sqrt{21} = \: \frac{1}{\cancel2} \times \cancel12 \times height \\ \\ \implies \: 18 \sqrt{21} = 6 \times height \\ \\ \implies \: height = \frac{\cancel{18} \sqrt{21} }{ \cancel 6} \\ \\ \implies \: height = 3 \sqrt{21} \: cm {}^{2}$

Since we've obtained the height now , we can easily find out the area of trapezium !

$Area \: of \: trapezium = \frac{1}{2} \times(sum \: of \:parallel \: sides) \times height \\ \\ \implies \: \frac{1}{2} \times (25 + 13) \times 3 \sqrt{21} \\ \\ \implies \: \frac{1}{\cancel2} \times \cancel{38 }\times 3 \sqrt{21} \\ \\ \implies \: 19 \times 3 \sqrt{21} \: cm {}^{2} \\ \\ \implies \: 57 \sqrt{21} \: cm {}^{2}$

hope helpful :D

6 0

2 years ago

Someone, please help ill give brainliest this is due later today!! and 100 points!!

Vladimir79 [104]

Answer:

I think the answer is 5

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers