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3 years ago

A rectangle has sides measuring (3x + 2) units and (5x + 8) units.

Mathematics

1 answer:

Nataliya [291]3 years ago

4 0

Answer:

a.(3x+2)(5x+8)

Step-by-step explanation:

area of rectangle formula- length times width

so l= 3x+2 and w= 5x+8

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3 years ago

F(x)=-3x+21,find x when f(x)=6

balu736 [363]

Answer:

x=5

Step-by-step explanation:

F(x)=-3x+21,

f(x)=6 = -3x+21

6 = -3x+21

Subtract 21 from each side

6-21 = -3x+21-21

-15 = -3x

Divide each side by -3

-15/-3 = -3x/-3

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3 years ago

Which of the following is an example of parallel lines?

RSB [31]

Answer: C

Explanation: they will never touch

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3 years ago

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A function g(x) has x-intercepts at (StartFraction 1 Over 2 EndFraction, 0) and (6, 0). Which could be g(x)? g(x) = 2(x + 1)(x +

Tcecarenko [31]

Answer:

$g(x) = (x-6)(2x-1)$

Step-by-step explanation:

$g(x)$ intercepts the x-axis at these 2 points:

$(6,0) ;(1/2,0)$

⇒ 6 and 1/2 are roots ie; if you insert $x=6$ or $x=1/2$ into the equation of g(x) you will obtain a 0.

$g(6) = g(1/2) = 0$

now in order for 0 to appear we should have $x-6$

now in order for 0 to appear we should have $x-1/2$

but $x-1/2$ doesn't appear in any of these, but its multiple of 2 is there:

$2(x-1/2) = 2x-1$

Therefore the function;

$g(x) = (x-6)(2x-1)$

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3 years ago

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Use mathematical induction to prove that for each integer n ≥ 4, 5^n ≥ 2 2^n+1 + 100

anastassius [24]

The given Statement which we have to prove using mathematical induction is

$5^n\geq 2*2^{n+1}+100$

for , n≥4.

⇒For, n=4

LHS

$=5^4\\\\5*5*5*5\\\\=625\\\\\text{RHS}=2.2^{4+1}+100\\\\=64+100\\\\=164$

LHS >RHS

Hence this statement is true for, n=4.

⇒Suppose this statement is true for, n=k.

$5^k\geq 2*2^{k+1}+100$

-------------------------------------------(1)

Now, we will prove that , this statement is true for, n=k+1.

$5^{k+1}\geq 2*2^{k+1+1}+100\\\\5^{k+1}\geq 2^{k+3}+100$

LHS

$5^{k+1}=5^k*5\\\\5^k*5\geq 5 \times(2*2^{k+1}+100)----\text{Using 1}\\\\5^k*5\geq (3+2) \times(2*2^{k+1}+100)\\\\ 5^k*5\geq 3\times (2^{k+2}+100)+2 \times(2*2^{k+1}+100)\\\\5^k*5\geq 3\times(2^{k+2}+100)+(2^{k+3}+200)\\\\5^{k+1}\geq (2^{k+3}+100)+3\times2^{k+2}+400\\\\5^{k+1}\geq (2^{k+3}+100)+\text{Any number}\\\\5^{k+1}\geq (2^{k+3}+100)$

Hence this Statement is true for , n=k+1, whenever it is true for, n=k.

Hence Proved.

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3 years ago