All categories

2 years ago

X^3+y^3=???????????????

Mathematics

1 answer:

omeli [17]2 years ago

6 0

Answer:

Well since we don't know what X or Y is, then we can't answer this question.

Step-by-step explanation:

You might be interested in

A kangaroo hops 2 kilometers in 3 minutes at this rate a. how long does it take the kangaroo to travel 5 kilometers b. how far d

spin [16.1K]

Answer:

7.5

Step-by-step explanation:

8 0

3 years ago

2. You plan to replace the carpeting in the room shown. What is the

RSB [31]

Answer:

$198\ ft^2$

Step-by-step explanation:

see the attached figure to better understand the problem

we know that the area of the room is equal to the area of a rectangle plus the area of a trapezoid

step 1

Find out the area of the rectangle

$A_1=(18)(9)=162\ ft^2$

step 2

Find out the area of the trapezoid

$A_2=\frac{1}{2} (9+3)(6)=36\ ft^2$

step 3

Sum the areas

$A=A_1+A_2$

substitute the values

$A=162++36=198\ ft^2$

5 0

3 years ago

Read 2 more answers

The manager of a fleet of automobiles is testing two brands of radial tires and assigns one tire of each brand at random to the

Darya [45]

Answer:

Brand 1 Brand 2 Difference

37734 35202 2532

45299 41635 3664

36240 35500 740

32100 31950 150

37210 38015 −805

48360 47800 560

38200 37810 390

33500 33215 285

Sum of difference = 2532+ 3664+740+150 −805+ 560 +390 +285 = 7516

Mean = $d=\frac{7516}{8}$

Mean = $d=939.5$

a) d= 939.5

$\text{Sample Standard deviation, s} = \sqrt{\dfrac{(x-\bar{x})^2}{n-1}}$

$=\sqrt{\dfrac{(2532-939.5)^2+(3664-939.5)^2+(740-939.5)^2 ...+(285-939.5)^2}{8-1}}$

=1441.21

b)SD= 1441.21

c)Calculate a 99% two-sided confidence interval on the difference in mean life.

confidence level =99%

significance level =α= 0.01

Degree of freedom = n-1 = 8-1 =7

So, $t_{\frac{\alpha}{2}}=3.499$

Formula for confidence interval $= \left( \bar{X} \pm t_{\frac{\alpha}{2}} \times \frac{s}{\sqrt{n}} \right)$

Substitute the values

confidence interval $= 939.5 \pm 3.499 \times \frac{1441.21}{\sqrt{8}} \right)$

confidence interval $= 939.5 - 3.499 \times \frac{1441.21}{\sqrt{8}} \right)$ to $= 939.5 + 3.499 \times \frac{1441.21}{\sqrt{8}} \right)$

Confidence interval $−843.396\$ to $2722.396$

8 0

2 years ago

I'm really struggling with this question. Please help!

gavmur [86]

I got 23 u might want to check it tho

3 0

3 years ago

(I need the answer right now please) ΔA'B'C' was constructed using ΔABC and line segment EH. 2 triangles are shown. Line E H is

irina1246 [14]

Answer:

The true statements are

1. BD = DB'

3. m∠EFA = 90°

4. The line of reflection, EH, is the perpendicular bisector of BB', AA', and

CC'

Step-by-step explanation:

* Lets explain how to solve the problem

- Reflection is flipping an object over the line of reflection.

- The object and its image have the same shape and size, but the

figures are in opposite directions from the line of reflection

- The objects appear as if they are mirror reflections, with right and left

reversed

- The line of reflection is a perpendicular bisector for all lines joining

points on the figure with their corresponding images

- Look to the attached figure for more understand

* Lets solve the problem

- ΔA'B'C' was constructed using ΔABC and line segment EH, where

EH is the reflection line

- D is the mid-point of BB'

- F is the mid-point of AA'

- G is the mid-point of CC'

* Lets find from the answer the true statements

1. BD = DB'

∵ D is the mid point of BB'

- Point D divides BB' into two equal parts

∴ BD = DB' ⇒ True

2. DF = FG

- It depends on the size of the sides and angles of the triangle

∵ We can't prove that

∴ DF = FG ⇒ Not true

3. m∠EFA = 90°

∵ The line of reflection ⊥ the lines joining the points with their

corresponding images

∴ EH ⊥ AA' and bisect it at F

∴ m∠EFA = 90° ⇒ True

4. The line of reflection, EH, is the perpendicular bisector of BB',

AA', and CC'

- Yes the line of reflection is perpendicular bisectors of them

∴ The line of reflection, EH, is the perpendicular bisector of BB',

AA', and CC' ⇒ True

5. ΔABC is not congruent to ΔA'B'C'

∵ In reflection the object and its image have the same shape and size

∴ Δ ABC is congruent to Δ A'B'C'

∴ ΔABC is not congruent to ΔA'B'C' ⇒ Not true

4 0

3 years ago

Read 2 more answers