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3 years ago

What the nearest whole for 80.356 + 46.585

Mathematics

2 answers:

IrinaK [193]3 years ago

7 0

It comes up to 126.941, which can be transferred to 127.

zysi [14]3 years ago

4 0

80.356+46.585=126.941

round that to 127

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2 years ago

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If DE≅EF, CD = v + 45, and CF = 4v, what is the value of v?

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Answer:

<h2>since DE≅EF</h2>

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A drone travels 180 kilometers in the same time a bird travels 120 kilometers. If the drone's speed is 20 km/hr faster than the

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2 years ago

A sequence is defined by f(0) = -20, f(n) = f(n-1) - 5 forn > 1.

bulgar [2K]

Answer:

1. Proved down

2. proved down

3. f(10) = -20 - 5 - 5 - 5 - 5 - 5 - 5 - 5 - 5 - 5 - 5

Step-by-step explanation:

Let us explain how to solve the question

∵ f(0) = -20, f(n) = f(n - 1) - 5 for n > 1

→ That means we have an arithmetic sequence with constant

difference -5 and first term -20

1. → f(1) means we need to find the second term, which equal the

term - 5

∵ f(1) means n = 1

∴ f(1) = f(1 - 1) - 5

∴ f(1) = f(0) - 5

∵ f(0) = -20

∴ f(1) = -20 - 5 → Proved

2. → f(3) means we need to find the third term, which equal the

second term - 5

∵ f(3) means n = 3

∴ f(3) = f(3 - 1) - 5

∴ f(3) = f(2) - 5

→ f(2) = f(1) - 5

∵ f(1) = -20 - 5

∴ f(2) = [-20 - 5] - 5 = -20 - 5 - 5

∴ f(3) = [-20 - 5 - 5] - 5

∴ f(3) = -20 - 5 - 5 - 5 → Proved

3. → From 1 and 2 we notice that the number of -5 is equal to n,

at n = 1 there is one (-5), when n= 3 there are three (-5)

∵ n = 10

∴ There are ten (-5)

∴ f(10) = -20 - 5(10)

∴ f(10) = -20 - 5 - 5 - 5 - 5 - 5 - 5 - 5 - 5 - 5 - 5 → Proved

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3 years ago

A right-angled triangle has shorter side lengths exactly c^2-b^2 and 2bc units respectively, where b and c are positive real num

yKpoI14uk [10]

Answer: hypotenuse = $c^{2} + b^{2}$

Step-by-step explanation: Pythagorean theorem states that square of hypotenuse (h) equals the sum of squares of each side ( $s_{1},s_{2}$ ) of the right triangle, .i.e.:

$h^{2} = s_{1}^{2} + s_{2}^{2}$

In this question:

$s_{1}$ = $c^{2}-b^{2}$

$s_{2} =$ 2bc

Substituing and taking square root to find hypotenuse:

$h=\sqrt{(c^{2}-b^{2})^{2}+(2bc)^{2}}$

Calculating:

$h=\sqrt{c^{4}+b^{4}-2b^{2}c^{2}+(4b^{2}c^{2})}$

$h=\sqrt{c^{4}+b^{4}+2b^{2}c^{2}}$

$c^{4}+b^{4}+2b^{2}c^{2}$ = $(c^{2}+b^{2})^{2}$ , then:

$h=\sqrt{(c^{2}+b^{2})^{2}}$

$h=(c^{2}+b^{2})$

Hypotenuse for the right-angled triangle is $h=(c^{2}+b^{2})$ units

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3 years ago