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juin [17]
3 years ago
8

Which could be the dimensions of a rectangular prism whose surface area is greater than 140 square feet? Select three options. 6

feet by 2 feet by 3 feet, 6 feet by 5 feet by 4 feet, 7 feet by 6 feet by 4 feet, 8 feet by 3 feet by 7 feet, 8 feet by 4 feet by 3 feet​
Mathematics
1 answer:
Murljashka [212]3 years ago
7 0

Answer:

the second one, ((6 x 5) x 4) +((5 x 4) x 2) = 160

the third one, ((7 x 6) x 4) + (6 x 4) x 2) = 216

the fourth one, ((8 x 4) x 4) + ((4 x 3) x 2) = 152

Step-by-step explanation:

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I need help with my little brothers homework?
erik [133]
I only did the first part but maybe later I will solve the rest too. If there are any errors but tell me so I can update the pic.

7 0
3 years ago
Read 2 more answers
Complete the square x^2-10=16<br><br>​
Hatshy [7]

Answer:

(x - 5)² = 41

Step-by-step explanation:

* Lets revise the completing square form

- the form x² ± bx + c is a completing square if it can be put in the form

 (x ± h)² , where b = 2h and c = h²

# The completing square is x² ± bx + c = (x ± h)²

# Remember c must be positive because it is = h²

* Lets use this form to solve the problem

∵ x² - 10x = 16

- Lets equate 2h by -10

∵ 2h = -10 ⇒ divide both sides by 2

∴ h = -5

∴ h² = (-5)² = 25

∵ c = h²

∴ c = 25

- The completing square is x² - 10x + 25

∵ The equation is x² - 10x = 16

- We will add 25 and subtract 25 to the equation to make the

 completing square without change the terms of the equation

∴ x² - 10x + 25 - 25 = 16

∴ (x² - 10x + 25) - 25 = 16 ⇒ add 25 to both sides

∴ (x² - 10x + 25) = 41

* Use the rule of the completing square above

- Let (x² - 10x + 25) = (x - 5)²

∴ (x - 5)² = 41

3 0
3 years ago
Can you tell me the answers, if you do thanks
zloy xaker [14]

7.

3 \frac{8}{11}

8.

4 \frac{31}{42}

9.

7 \frac{11}{24}

10.

2 \frac{79}{90}

5 0
3 years ago
The Oxy coordinate plane for two parallel lines a and a' has the equations 2x - 3y-1 = 0 and 2x - 3y + 5 = 0. respectively. Whic
Andru [333]

Answer:

Remember that a vector translation can be written as:

T(a, b)

And if we apply this to a random point, (x, y), the translation gives:

T(a, b)(x, y) = (x + a, y + b)

now, remember that a general line can be written as:

y = m*x + s

Then a point of that line can be written as: (x, m*x + s)

Then if we apply the translation to a point in the line, we get:

T(a, b)(x, m*x + s) = (x + a, m*x + s + b)

Here we have two lines:

2x - 3y - 1 = 0

2x - 3y + 5 = 0

First, let's rewrite both of these in the slope-intercept form:

y = (2/3)*x - 1/3

y = (2/3)*x + 5/3

Now let's assume that we apply a translation to the first line, that has points of the form (x,  (2/3)*x - 1/3), such that we want to get points of the form:

(x, (2/3)*x + 5/3).

Then we must have:

T(a, b)(x,  (2/3)*x - 1/3) = (x + a,  (2/3)*x - 1/3 + b) = (x, (2/3)*x + 5/3).

Then we need to solve:

(x + a,  (2/3)*x - 1/3 + b) = (x, (2/3)*x + 5/3).

This means that:

x + a = x

(2/3)*x - 1/3 + b = (2/3)*x + 5/3

From the first equation, we can see that a = 0

Now we can solve the second one to find the value of b.

(2/3)*x - 1/3 + b = (2/3)*x + 5/3

subtracting (2/3)*x in both sides, we get:

-1/3 + b = 5/3

b = 5/3 + 1/3

b = 6/3 = 2

b = 2

Then the vector translation is:

T(0, 2)

So it moves the whole line 2 units upwards.

7 0
3 years ago
Find the non-extraneous solutions of the square root of the quantity x plus 6 minus 5 equals quantity x plus 1.
sdas [7]
The radical equation is \sqrt{x+6}-5=x+1.


i) We first isolate the square root, adding 5 to both sides of the equation: 
       
                                        \sqrt{x+6}=x+6.

ii) Here let's substitute x+6 with t. Doing so we have:
 
                                         \sqrt{t}=t.

Squaring both sides, we get:
                       
                                               t=t^2

iii) Collecting the variables on the same side, and factorizing t we have:
 
                                               t(t-1)=0, which yields

                              t=0    or       t=1.

Now we solve for x in x+6=t:

x+6=0 ⇒x=-6       and x+6=1⇒x=-5.


iv) Now we check these values in the original equation  \sqrt{x+6}=x+6 :

a) \sqrt{-6+6}=-6+6. ⇒ 0=0 ; Correct.


b) \sqrt{-5+6}=-5+6. ⇒ 1=1 ; Correct.



Answer: <span>x = −6 and x = −5 </span>
5 0
3 years ago
Read 2 more answers
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