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3 years ago

Please help! I've attached an image of the math question

Mathematics

2 answers:

Rashid [163]3 years ago

5 0

Answer:

Step-by-step explanation:

a_sh-v [17]3 years ago

4 0

the answer should be D

Step-by-step explanation:

when looking this up it seems foil and multiplying binomial go hand and hand but this more closely resembles D

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Who made the telescope

kaheart [24]

Answer:

Hans Lipperhey made the telescope

6 0

3 years ago

Read 2 more answers

Find x . Please and thank you

Mila [183]

Answer:

x = 6

Step-by-step explanation:

In similar triangles, corresponding sides are in same ratio

$\frac{x+3}{3}= \frac{6}{2}\\\\ \frac{x+3}{3}=3\\\\x+3 = 3*3\\\\x+3 = 9\\\\x = 9-3\\\\x = 6$

8 0

2 years ago

Você é um comandante de uma espaçonave. Sua missão é chegar até Alfa Centauro em cinco anos. A distância do Sol até Alfa Centaur

Deffense [45]

Answer:

The spaceship will get to Alpha Centaur in time

Step-by-step explanation:

<u><em>The complete question in English is</em></u>

You're the commander of a spaceship. Your mission is to reach Alpha Centaur in five years. The distance from the sun to Alpha Centaur is 2.5 x 10^13 miles. The distance from Earth to Sun is approximately 9.3 x 10^7

miles. Your spaceship can travel at the speed of light. You know that light can travel a distance of 5.88 x 10^12 miles in a year. Can you get to Alpha Centaur in time?

step 1

Find the time it takes for the spaceship to travel from Earth to the Sun

The distance from the Earth to Sun is equal to

$9.3*10^7\ miles$

The spaceship can travel at the speed of light

The light can travel a distance of $5.88*10^{12}\ miles$ in a year

using a proportion

$\frac{5.88*10^{12}\ miles}{1\ year}=\frac{9.3*10^7\ miles}{x}\\\\x=(9.3*10^7)/5.88*10^{12}\\\\x=1.58*10^{-5} \ years$

This time is very small in years

Convert to minutes

$1.58*10^{-5} \ years=1.58*10^{-5} (365)(24)(60)=8.3\ min$

step 2

Find the time it takes for the spaceship to travel from the Sun to Alpha Centaur

The distance from the the Sun to Alpha Centaur is equal to

$2.5*10^{13}\ miles$

The spaceship can travel at the speed of light

The light can travel a distance of $5.88*10^{12}\ miles$ in a year

using a proportion

$\frac{5.88*10^{12}\ miles}{1\ year}=\frac{2.5*10^{13}\ miles}{x}\\\\x=(2.5*10^{13})/5.88*10^{12}\\\\x=4.25\ years$

$4.25\ years< 5\ years$

therefore

The spaceship will get to Alpha Centaur in time

3 0

2 years ago

The graph of the continuous function g, the derivative of the function f, is shown above. The function g is piecewise linear for

4vir4ik [10]

A) $g=f'$ is continuous, so $f$ is also continuous. This means if we were to integrate $g$ , the same constant of integration would apply across its entire domain. Over $0$ , we have $g(x)=2x$ . This means that

$f_{0$

For $f$ to be continuous, we need the limit as $x\to1^-$ to match $f(1)=3$ . This means we must have

$\displaystyle\lim_{x\to1}x^2+C=1+C=3\implies C=2$

Now, over $x$ , we have $g(x)=-3$ , so $f_{x$ , which means $f(-5)=17$ .

b) Integrating over [1, 3] is easy; it's just the area of a 2x2 square. So,

$\displaystyle\int_1^6g(x)=4+\int_3^62(x-4)^2\,\mathrm dx=4+6=10$

c) $f$ is increasing when $f'=g>0$ , and concave upward when $f''=g'>0$ , i.e. when $g$ is also increasing.

We have $g>0$ over the intervals $0$ and $x>4$ . We can additionally see that $g'>0$ only on $0$ and $x>4$ .

d) Inflection points occur when $f''=g'=0$ , and at such a point, to either side the sign of the second derivative $f''=g'$ changes. We see this happening at $x=4$ , for which $g'=0$ , and to the left of $x=4$ we have $g$ decreasing, then increasing along the other side.

We also have $g'=0$ along the interval $-1$ , but even if we were to allow an entire interval as a "site of inflection", we can see that $g'>0$ to either side, so concavity would not change.

5 0

2 years ago

How do I do it and show my work along with it?

Rudiy27

Be kind and say please,that is being rude when you don't say it. Well, how much times does 6 go into 34? 5 times right. So then put 30 under the 34. You get 4. So then bring down the 5. How much times does 6 go into 45. 7 right. So then put 42 under 45. You get 3. Now add a decimal point after the 5 and after the 7. Add zero after the 5 . Bring down the zero. How much times does 6 go into 30? 5 times, so you put 30 under 30 and get zero. If you have any more questions,please leave them below in the comments.

8 0

3 years ago