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fiasKO [112]
3 years ago
15

how high up on a building will 15 foot ladder reach if the foot of the ladder is placed 5 feet from the building? (be sure to in

clude a diagram with your answer)

Mathematics
1 answer:
mezya [45]3 years ago
3 0
Use the Pythagorean theorem:
5^2+x^2=15^2 \\
25+x^2=225 \\
x^2=225-25 \\
x^2=200 \\
x=\sqrt{200} \\
x=\sqrt{100 \times 2} \\
x=10\sqrt{2} \\
x \approx 14.14

The ladder will reach approximately 14.14 feet (exactly 10√2 feet) up on the building.

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Answer:

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Step-by-step explanation:

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8 0
1 year ago
A geometric sequence is defined by the general term tn = 75(5n), where n ∈N and n ≥ 1. What is the recursive formula of the sequ
andreyandreev [35.5K]
The correct answer is C) t₁ = 375, t_n=5t_{n-1}.

From the general form,
t_n=75(5)^n, we must work backward to find t₁.

The general form is derived from the explicit form, which is
t_n=t_1(r)^{n-1}.  We can see that r = 5; 5 has the exponent, so that is what is multiplied by every time. This gives us

t_n=t_1(5)^{n-1}

Using the products of exponents, we can "split up" the exponent:
t_n=t_1(5)^n(5)^{-1}

We know that 5⁻¹ = 1/5, so this gives us
t_n=t_1(\frac{1}{5})(5)^n&#10;\\&#10;\\=\frac{t_1}{5}(5)^n

Comparing this to our general form, we see that
\frac{t_1}{5}=75

Multiplying by 5 on both sides, we get that
t₁ = 75*5 = 375

The recursive formula for a geometric sequence is given by
t_n=t_{n-1}(r), while we must state what t₁ is; this gives us

t_1=375; t_n=t_{n-1}(5)

3 0
3 years ago
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fgiga [73]

Answer:

2 3/10, 2.59, 24

Step-by-step explanation:

4 0
3 years ago
Which applies the power of a product rule to simplify (5t)^3?
frozen [14]
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5 0
3 years ago
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PLEASE ANSWER ASAP NEED HELP REALLY BAD CALLING OUT TO ALL MATH PROS WHO CAN HELP ME PLEASE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
sergeinik [125]

Answer:

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Step-by-step explanation:

4 0
3 years ago
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