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3 years ago

ITS A ANGLE QUESTON PLS HELP

Mathematics

2 answers:

Svetlanka [38]3 years ago

4 0

This is a right angles triangle

(6x6)/2

<h2>Hope this helps :></h2><h2 /><h2>From Cambridge</h2><h2 /><h2>Please give brainliest</h2>

luda_lava [24]3 years ago

4 0

Answer:

18 cm²

Step-by-step explanation:

The area (A) of a triangle is calculated as

A = $\frac{1}{2}$ bh ( b is the base and h the perpendicular height )

Here b = 6 and h = 6 , then

A = $\frac{1}{2}$ × 6 × 6 = 3 × 6 = 18 cm²

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Bret needs 450 packs of paper for next week. He has 284 packs of paper. How many more packs of paper dose bret need?

gulaghasi [49]

Answer:

166 packs

Step-by-step explanation:

Data obtained from the question include:

Total pack of paper needed by Bret 450 packs

Bret currently has = 284 packs

The remaining packs of paper needed by Bret = 450 — 284 = 166 packs

5 0

3 years ago

Read 2 more answers

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Archy [21]

(a) If the particle's position (measured with some unit) at time <em>t</em> is given by <em>s(t)</em>, where

$s(t) = \dfrac{5t}{t^2+11}\,\mathrm{units}$

then the velocity at time <em>t</em>, <em>v(t)</em>, is given by the derivative of <em>s(t)</em>,

$v(t) = \dfrac{\mathrm ds}{\mathrm dt} = \dfrac{5(t^2+11)-5t(2t)}{(t^2+11)^2} = \boxed{\dfrac{-5t^2+55}{(t^2+11)^2}\,\dfrac{\rm units}{\rm s}}$

(b) The velocity after 3 seconds is

$v(3) = \dfrac{-5\cdot3^2+55}{(3^2+11)^2} = \dfrac{1}{40}\dfrac{\rm units}{\rm s} = \boxed{0.025\dfrac{\rm units}{\rm s}}$

$\dfrac{-5t^2+55}{(t^2+11)^2} = 0 \implies -5t^2+55 = 0 \implies t^2 = 11 \implies t=\pm\sqrt{11}\,\mathrm s \imples t \approx \boxed{3.317\,\mathrm s}$

(d) The particle is moving in the positive direction when its position is increasing, or equivalently when its velocity is positive:

$\dfrac{-5t^2+55}{(t^2+11)^2} > 0 \implies -5t^2+55>0 \implies -5t^2>-55 \implies t^2 < 11 \implies |t|$

In interval notation, this happens for <em>t</em> in the interval (0, √11) or approximately (0, 3.317) s.

(e) The total distance traveled is given by the definite integral,

$\displaystyle \int_0^8 |v(t)|\,\mathrm dt$

By definition of absolute value, we have

$|v(t)| = \begin{cases}v(t) & \text{if }v(t)\ge0 \\ -v(t) & \text{if }v(t)$

In part (d), we've shown that <em>v(t)</em> > 0 when -√11 < <em>t</em> < √11, so we split up the integral at <em>t</em> = √11 as

$\displaystyle \int_0^8 |v(t)|\,\mathrm dt = \int_0^{\sqrt{11}}v(t)\,\mathrm dt - \int_{\sqrt{11}}^8 v(t)\,\mathrm dt$

and by the fundamental theorem of calculus, since we know <em>v(t)</em> is the derivative of <em>s(t)</em>, this reduces to

$s(\sqrt{11})-s(0) - s(8) + s(\sqrt{11)) = 2s(\sqrt{11})-s(0)-s(8) = \dfrac5{\sqrt{11}}-0 - \dfrac8{15} \approx 0.974\,\mathrm{units}$

7 0

3 years ago

What is .65% of 160? Show your work

vladimir2022 [97]

0.65/100 = x/160 Ok, until now?

You then find the cross product: (0.65)(160) = 100x

104 = 100x 104/100 = 1.04

So x = 1.04 and it is also 0.65% of 160

Hope this helps!

7 0

3 years ago

T<br>this is a 7.5 trig ratios solving missing sides assignment

Vsevolod [243]

Answer: the awnser is b 22.2.

Step-by-step explanation: have a nice day

4 0

3 years ago

Can u someone help me ? Plz

NARA [144]

What is the question?

6 0

4 years ago