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3 years ago

What is the perimeter of the triangle? units

Mathematics

1 answer:

skelet666 [1.2K]3 years ago

8 0

Answer:

Perimeter = 30

Step-by-step explanation:

hypotenuse = $\sqrt{5^2+12^2}$

hypotenuse = $\sqrt{25+144}$

hypotenuse = $\sqrt{169}$

hypotenuse = 13

Perimeter = 5+12+13

Perimeter = 30

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A high rise building has a sink on the fourth floor located 2-feet above the floor. Each floor is 10-feet above the other. Calcu

kap26 [50]

The pressure at a certain height is the pressure at sea level plus the hydrostatic pressure.

P = 80 psi + ΔP
where
ΔP = ρgh
ρ is the density of water equal to 62.4 lbm/ft³
ΔP = (62.4 lbm/ft³)(1 lbf/lbm)(4*10 + 2 ft)(1 ft/12 in)² = 18.2 psi

Thus,
P = 80 + 18.2
P = 98.2 psi

8 0

4 years ago

Geometric Sequence S = 1.0011892 + ... + 1.0012 + 1.001 + 1

leva [86]

Answer:

$S_{1893} =5632.98$

Step-by-step explanation:

The correct form of the question is:

$S = 1.001^{1892} + ... + 1.001^2 + 1.001 + 1$

Required

Solve for Sum of the sequence

The above sequence represents sum of Geometric Sequence and will be solved using:

$S_n = \frac{a(1 - r^n)}{1 - r}$

But first, we need to get the number of terms in the sequence using:

$T_n = ar^{n-1}$

Where

$a = First\ Term$

$a = 1.001^{1892}$

$r = common\ ratio$

$r = \frac{1}{1.001}$

$T_n = Last\ Term$

$T_n = 1$

So, we have:

$T_n = ar^{n-1}$

$1 = 1.001^{1892} * (\frac{1}{1.001})^{n-1}$

Apply law of indices:

$1 = 1.001^{1892} * (1.001^{-1})^{n-1}$

$1 = 1.001^{1892} * (1.001)^{-n+1}$

Apply law of indices:

$1 = 1.001^{1892-n+1}$

$1 = 1.001^{1892+1-n}$

$1 = 1.001^{1893-n}$

Represent 1 as $1.001^0$

$1.001^0 = 1.001^{1893-n}$

They have the same base:

So, we have

$0 = 1893-n$

Solve for n

$n = 1893$

So, there are 1893 terms in the sequence given.

Solving further:

$S_n = \frac{a(1 - r^n)}{1 - r}$

Where

$a = 1.001^{1892}$

$r = \frac{1}{1.001}$

$n = 1893$

So, we have:

$S_{1893} =\frac{1.001^{1892} *(1 -\frac{1}{1.001}^{1893})}{1 -\frac{1}{1.001} }$

$S_{1893} =\frac{1.001^{1892} *(1 -\frac{1}{1.001}^{1893})}{\frac{1.001 -1}{1.001} }$

$S_{1893} =\frac{1.001^{1892} *(1 -\frac{1}{1.001}^{1893})}{\frac{0.001}{1.001} }$

$S_{1893} =\frac{1.001^{1892} *(1 -\frac{1}{1.001^{1893}})}{\frac{0.001}{1.001} }$

Simplify the numerator

$S_{1893} =\frac{1.001^{1892} -\frac{1.001^{1892}}{1.001^{1893}}}{\frac{0.001}{1.001} }$

$S_{1893} =\frac{1.001^{1892} -1.001^{1892-1893}}{\frac{0.001}{1.001} }$

$S_{1893} =\frac{1.001^{1892} -1.001^{-1}}{\frac{0.001}{1.001} }$

$S_{1893} =(1.001^{1892} -1.001^{-1})/({\frac{0.001}{1.001} })$

$S_{1893} =(1.001^{1892} -1.001^{-1})*{\frac{1.001}{0.001}}$

$S_{1893} =\frac{(1.001^{1892} -1.001^{-1}) * 1.001}{0.001}$

Open Bracket

$S_{1893} =\frac{1.001^{1892}* 1.001 -1.001^{-1}* 1.001 }{0.001}$

$S_{1893} =\frac{1.001^{1892+1} -1.001^{-1+1}}{0.001}$

$S_{1893} =\frac{1.001^{1893} -1.001^{0}}{0.001}$

$S_{1893} =\frac{1.001^{1893} -1}{0.001}$

$S_{1893} =5632.97970294$

Hence, the sum of the sequence is:

$S_{1893} =5632.98$ ----- approximated

4 0

3 years ago

I need to know the distance between the red river

zysi [14]

Answer:

B) 1.5

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

In 5 years, Dad will be three times as old as his daughter Jill will be then. If the sum of their present ages is 50, how old ar

Nataliya [291]

The answer is b.3x + 15 = 55 - x

x - Jill's present age
y - dad's present age

In 5 years, Dad will be three times as old as his daughter Jill will be then:
y + 5 = 3(x + 5)

The sum of their present ages is 50:
x + y = 50

We have the system of two equations now:
y + 5 = 3(x + 5)
x + y = 50

Let's rearrange the second equation:
If: x + y = 50
Then: y = 50 - x

Now, substitute y from the second equation (y = 50 - x) into the first one:
y + 5 = 3(x + 5)
50 - x + 5 = 3(x + 5)
50 + 5 - x = 3*x + 3*5
55 - x = 3x + 15

Rearrange it a bit:
3x + 15 = 55 - x
Therefore, the correct choice is b.

5 0

3 years ago

Will the value of y ever equal 0 y=3^x

Evgesh-ka [11]

No x- intercept/ zero/ 0

6 0

3 years ago