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3 years ago

14

Ben made a huge number, writing out the consecutive natural numbers from 1 to 500:123 ... 10111213 ... 499500. Lusine erased fir

st 500 digits. With what digit did the remaining number begin?

1 answer:

Aleksandr-060686 [28]3 years ago

4 0

123 :D hope I helped

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Which of the following is equivalent??

Naddik [55]

The answer is b because it is equivalent.

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3 years ago

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Four less than the square of a number

matrenka [14]

Answer: x² - 4

<u>Four less than</u> the <u>square of a number</u>

↓ ↓

(Take away 4) x²

Four less than the square of a number = x² - 4

6 0

3 years ago

Another one <br><br> Question <br><br> b+11=4 <br> b=?

steposvetlana [31]

Answer:

b = -7 please mark brainliest :))))

Step-by-step explanation:

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3 years ago

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The height h (in feet) of an object dropped from a ledge after x seconds can be modeled by h(x)=−16x2+36 . The object is dropped

kakasveta [241]

Check the picture below.

$\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(t) = -16t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}&\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&\\ \qquad \textit{at "t" seconds} \end{cases}$

so the object hits the ground when h(x) = 0, hmmm how long did it take to hit the ground the first time anyway?

$\bf h(x)=-16x^2+36\implies \stackrel{h(x)}{0}=-16x^2+36\implies 16x^2=36 \\\\\\ x^2=\cfrac{36}{16}\implies x^2 = \cfrac{9}{4}\implies x=\sqrt{\cfrac{9}{4}}\implies x=\cfrac{\sqrt{9}}{\sqrt{4}}\implies x = \cfrac{3}{2}~~\textit{seconds}$

now, we know the 2nd time around it hit the ground, h(x) = 0, but it took less time, it took 0.5 or 1/2 second less, well, the first time it took 3/2, if we subtract 1/2 from it, we get 3/2 - 1/2 = 2/2 = 1, so it took only 1 second this time then, meaning x = 1.

$\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(x) = -16x^2+v_ox+h_o \quad \begin{cases} v_o=\textit{initial velocity}&0\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&0\\ \qquad \textit{at "t" seconds}\\ x=\textit{seconds}&1 \end{cases} \\\\\\ 0=-16(1)^2+0x+h_o\implies 0=-16+h_o\implies 16=h_o \\\\[-0.35em] ~\dotfill\\\\ ~\hfill h(x) = -16x^2+16~\hfill$

quick info:

in case you're wondering what's that pesky -16x² doing there, is gravity's pull in ft/s².

4 0

3 years ago

Please how to answer.

vivado [14]

Answer:

Blue Rectangle: <u>135 mm²</u>

Blue Triangle: <u>45 mm²</u>

<em>Not sure if you need this but the </em>Total Square: <u>225 mm²</u>

Step-by-step explanation:

Area of a rectangle: <u>length x width</u>

Area of a triangle: $\frac{1}{2}$ <u> x base x height</u>

First find the area of the blue rectangle.

Length = 15 mm

Width = 9 mm

Area = 135 mm²

Now find the area of the blue triangle:

Base = 6 mm (because the bottom is 15 total and you subtract the 9)

Height = 15 mm

Area = 45 mm²

Hope it helps!

6 0

1 year ago