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3 years ago

ASAP HELP PLEASE:

Mathematics

1 answer:

stealth61 [152]3 years ago

4 0

We know that

[Surface area ]=[the area of 4 triangles]+[the area of square base]

[the area of square base]=8*8--------> 64 ft²

[the area of 1 triangle]=(1/2)*b*h
b=8 ft
How the angle is 45°-------> h=b/2-------> h=4 ft

then
[the area of 1 triangle]=(1/2)*8*4------> 16 ft²
and
[the area of 4 triangles]=4*16--------> 64 ft²

[Surface area ]=[64 ft²]+[64 ft²]----------> 128 ft²

the answer is 128 ft²

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Step-by-step explanation:

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The function f(x)=(x-5)^2 +2 is not one-to-one. Identify a restricted domain that makes the function one-to-one, and find the in

Anon25 [30]

We have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

We know that vertex of this quadratic function occurs at (5,2).

Further, we know that range of this function is $[2,\infty)$ .

If we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

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3 years ago

Can you please help me. 6 = 4x + 9y slove for y

kow [346]

Answer:

Step-by-step explanation:

↔Start of with your 'solution section' to make sure you have the right things↔

↔Solution 1 ➡ 4(x) + 9(y) = 16

↔Solution 2 ➡ 6(x) - 9(y) = 8

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▪Looking for the value of the 'x'-intercept ⬇▪

◾4 (3) (y) / 2 + 4/ 3 + 9(y) = 16

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◾ 45(y) = 32

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◾Looking for value of the 'y'-intercept ◾

◾45(y) = 32

◾y-intercept : 45

◾x = 3(y)/2+4/3 which is translated to this, am i correct? ➡ (y) = 32/45

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↪ Now, we have finally found both of their intercept ⬇

↪x-intercept: 12

↪y-intercept: 5

↪Or you can even say the intercepts in other words. Here's what I am talking about if in case you don't know what I am saying

↪x-intercept: 32✅

↪y-intercept✅

6 0

3 years ago

Can someone help me?

Lana71 [14]

Answer:

I'm not sure from my answer but I'm going to try

A = { a^2 - 4 = 0 }

A = { a^2 = 0+ 4 }

A = { a^2 = 4}

Take root of both sides

A = {a = 2}

A = {2}

B = { -2, -1, 0, 1, 2}

C = { 1, 2, 3, 4, 5, 6,7}

i) A\B

= {2} \ {-2,-1,0,1,2}

= ⌀ ( "null set" )

ii) (A\B) \C

= ( {2} \ {-2,-1,0,1,2} ) \ C

= ⌀ \ C

= ⌀ \ { 1, 2, 3, 4, 5, 6,7}

= ⌀ ( "null set" )

iii) (A ∪ B) \ C

= ( { 2 } ∪ { -2, -1, 0, 1, 2} ) \ C

= { -2, -1, 0, 1, 2} \C

= { -2, -1, 0, 1, 2} \ { 1, 2, 3, 4, 5, 6,7}

= { -2, -1, 0 }

iv) ( A\B) ∩ ( A\C)

= ( { 2 } \ {-2,-1,0,1,2} ) ∩ ( { 2 } \ { { 1, 2, 3, 4, 5, 6,7} )

= ⌀ ∩ ⌀

= ⌀

v) ( A ∩ B ) \ C

= ( { 2 } ∩ { -2, -1, 0, 1, 2 } ) \ C

= { 2 } \ { 1, 2, 3, 4, 5, 6, 7 }

= ⌀ "null set"

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