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3 years ago

A rocket was launched and its height, h, in metres, above the ground after time,

Mathematics

1 answer:

Marianna [84]3 years ago

6 0

Short answer b 7m

Very simple just do the math mostly multiplying

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Jet001 [13]

Answer:

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3 years ago

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2 years ago

Find the solution of the following system of linear equations. x+y=6 -2x+y=0

andre [41]

Hope it helped you.

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2 years ago

Find the sum of the first six terms of a geometric series in which a1 = 4 and r = -3: A. 729 B. -729 C. -728 D. 728 plz help

natta225 [31]

Answer:

Step-by-step explanation:

The sum to n terms of a geometric sequence is

$S_{n}$ = $\frac{a_{1}(r^n-1) }{r-1}$

Here a₁ = 4 and r = - 3, thus

$S_{6}$ = $\frac{4((-3)^6-1)}{-3-1}$

= $\frac{4(729-1)}{-4}$ = - (729 - 1) = - 728 → C

6 0

2 years ago

Given right triangle ABC with altitude BD drawn to hypotenuse AC. If AC =16 and DC=5 what is the length of BC in the simplest ra

Nana76 [90]

The length of BC is $4 \sqrt{5}$ .

Solution:

Given ABC is a right triangle.

AC is the hypotenuse and BD is the altitude.

AB and BC are legs of the triangle ABC.

AC = 16 and DC = 5

<u>Leg rule of geometric mean theorem:</u>

$\frac{\text { hypotenuse }}{\text { leg }}=\frac{\text { leg }}{\text { part }}$

$\Rightarrow \frac{AC}{BC}=\frac{BC}{DC}$

$\Rightarrow \frac{16}{x}=\frac{x}{5}$

Do cross multiplication.

$\Rightarrow 16\times 5 = x\times x$

$\Rightarrow 80= x^2$

$\Rightarrow 16\times 5= x^2$

Taking square root on both sides.

$\Rightarrow \sqrt{16\times 5} = \sqrt{x^2}$

$\Rightarrow \sqrt{4^2\times 5} = \sqrt{x^2}$

square and square roots get canceled, we get

$\Rightarrow 4\sqrt{ 5} = x$

The length of BC is $4 \sqrt{5}$ .

7 0

2 years ago