Which statement best explains whether △PQR is congruent to △XYZ?

Which number has an absolute value greater than 4?

pav-90 [236]

Answer:

0

Step-by-step explanation:

Sorry if this doesn't help much and if its wrong, but maybe you could mark this as brainly? Have a wonderful day!

5 0

2 years ago

Hank the hiker would like to go on a hike. He only has four hours available today for the hike. The expression 1/2d + 1/2e is us

alisha [4.7K]

Answer:

No. Hank wont be able to complete the hike.

Step-by-step explanation:

Time Available for hike = 4 hours

The expression which represents the time Hank needs to complete the hike is:

$\text{time}=\frac{1}{2}d+\frac{1}{2}e$

d is the distance in miles of the hiking trail. Since, Hank would like to hike 6 miles, d = 6

e is the elevation gain in thousands. Since, the trail hank will hike on has an elevation gain of 3000, so e = 3

Using these values in the expression, we get:

$\text{Time}=\frac{1}{2}(6)+\frac{1}{2}(3)=4.5$

This means Hank should allow him 4.5 hours or the hike. Since he has only 4 hours available, he wont be able to complete the hike. So the answer to this question is No.

6 0

2 years ago

I will give brainliest if correct!!! Please show work so I know how to do it. :)

yawa3891 [41]

Answer:

.2625

Step-by-step explanation:

logb(3) = 0.5646, logb(4) = 0.7124, and logb(5) = 0.8271

logb(5/3)

We know that logx ( y/z) = logx (y) - logz (z)

logb ( 5) - logb(3)

0.8271 - 0.5646

.2625

5 0

3 years ago

Read 2 more answers

A small rocket is fired from a launch pad 10 m above the ground with an initial velocity left angle 250 comma 450 comma 500 righ

jonny [76]

Let $\vec r(t),\vec v(t),\vec a(t)$ denote the rocket's position, velocity, and acceleration vectors at time $t$ .

We're given its initial position

$\vec r(0)=\langle0,0,10\rangle\,\mathrm m$

and velocity

$\vec v(0)=\langle250,450,500\rangle\dfrac{\rm m}{\rm s}$

Immediately after launch, the rocket is subject to gravity, so its acceleration is

$\vec a(t)=\langle0,2.5,-g\rangle\dfrac{\rm m}{\mathrm s^2}$

where $g=9.8\frac{\rm m}{\mathrm s^2}$ .

a. We can obtain the velocity and position vectors by respectively integrating the acceleration and velocity functions. By the fundamental theorem of calculus,

$\vec v(t)=\left(\vec v(0)+\displaystyle\int_0^t\vec a(u)\,\mathrm du\right)\dfrac{\rm m}{\rm s}$

$\vec v(t)=\left(\langle250,450,500\rangle+\langle0,2.5u,-gu\rangle\bigg|_0^t\right)\dfrac{\rm m}{\rm s}$

(the integral of 0 is a constant, but it ultimately doesn't matter in this case)

$\boxed{\vec v(t)=\langle250,450+2.5t,500-gt\rangle\dfrac{\rm m}{\rm s}}$

and

$\vec r(t)=\left(\vec r(0)+\displaystyle\int_0^t\vec v(u)\,\mathrm du\right)\,\rm m$

$\vec r(t)=\left(\langle0,0,10\rangle+\left\langle250u,450u+1.25u^2,500u-\dfrac g2u^2\right\rangle\bigg|_0^t\right)\,\rm m$

$\boxed{\vec r(t)=\left\langle250t,450t+1.25t^2,10+500t-\dfrac g2t^2\right\rangle\,\rm m}$

b. The rocket stays in the air for as long as it takes until $z=0$ , where $z$ is the $z$ -component of the position vector.

$10+500t-\dfrac g2t^2=0\implies t\approx102\,\rm s$

The range of the rocket is the distance between the rocket's final position and the origin (0, 0, 0):

$\boxed{\|\vec r(102\,\mathrm s)\|\approx64,233\,\rm m}$

c. The rocket reaches its maximum height when its vertical velocity (the $z$ -component) is 0, at which point we have

$-\left(500\dfrac{\rm m}{\rm s}\right)^2=-2g(z_{\rm max}-10\,\mathrm m)$

$\implies\boxed{z_{\rm max}=125,010\,\rm m}$

7 0

2 years ago

A paperclip manufacturer puts 75 paperclips in each box. Every hour the factory produces 114,375 paperclips. How many boxes does

scZoUnD [109]

The answer is A 1,525 paperclips
because 114,375 divided by 75 is 1,525

3 0

2 years ago