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3 years ago

Need help with geo please 20 points

Mathematics

1 answer:

Serjik [45]3 years ago

6 0

Answer:

Step-by-step explanation:

Option B is correct - angle ABE is less than 90 degrees, therefore it is acute, and angle DBC is greater than 90 degrees, therefore it is obtuse.

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Which expression does not belong with the other three? Explain your reasoning. −58−34 −34+58 −58+(−34) −34−58 This expression eq

Svetach [21]

Answer:

The expression which doesn't belong to the other three is (-34 + 58).

Step-by-step explanation:

Given:

Four expressions.

We have to find the odd one out.

The expressions are :

$-58-34$
$-34+58$
$-58+(-34)$
$-34-58$

We know that with same signs (operator) in addition/subtraction of an expression we we add the values and the answer remains in negative.

But with different signs on numbers we will subtract the numbers and the greater number's sign will be carried forward.

So,

$-58-34$ $= -92$
$-34+58$ $=24$
$-58+(-34)$ $=-58-34 =-92$
$-34-58$ $=-92$

We can see that the second expression is not equal to the other three.

(-34 + 58) is the odd one out of the four expressions.

6 0

3 years ago

Pablo is running a 15.5-kilometer race.

Montano1993 [528]

It would be answer D

5 0

3 years ago

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

3 years ago

List the potential rational zeros of the polynomial function. Do not find the zeros.f(x) = x5 - 3x2 + 3x + 14

uysha [10]

$f(x) = x^5 - 3x^2 + 3x + 14\\\\14\to\{\pm1;\ \pm2;\ \pm7;\ \pm14\}\ and\ \left\{\pm\dfrac{1}{14};\ \pm\dfrac{1}{7}; \pm\dfrac{1}{2}\right\}$

4 0

3 years ago

The radius AND height of a cylinder are each doubled. What is the new volume?

Naya [18.7K]

The volume of a cylinder is given by

$V = \pi hr^2$

Let $h\mapsto 2h$ and $r\mapsto 2r$

The new volume becomes

$V = \pi(2h)(2r)^2 = \pi\cdot 2h \cdot 4r^2 = 8\pi hr^2$

So, the new volume is 8 times the original volume. In fact, volume scales linearly with the height and quadratically with the radius.

7 0

3 years ago