All categories

2 years ago

Can someone answer this and show me how u got it cause I’m kind of confused on this one

Mathematics

1 answer:

balu736 [363]2 years ago

4 0

Answer:

26 1/32

Step-by-step explanation:

<em>Doing good so far</em>

To make an exponent positive, you need to make the number a fraction and move it to the opposite of what it is in <em>(so 4^-3 = 1/4^3)</em>

25 + 1 + (1/(4^3) + 1/(2^6))

26 + (1/64 + 1/64)

26 + 2/64

26 + 1/32

= 26 1/32

You might be interested in

It has been reported that men are more likely than women to participate in online auctions. In a recent survey, 65% of repondent

kupik [55]

Answer:

Hence, the Female and not participated is $\frac{28}{100}=0.28\%$

Step-by-step explanation:

Let Total respondents $=100$

Participated in online action $(P)=65$

Men $(m)=45$ and Women $(F)=100-45=55$

Male and participated $(M\; \text{and}\;P)=38$

Female and participated $(F\; \text{and}\;P)=65-38=27$

Male and not participated is $45-38=7$

Female and not participated is $55-27=28$

Therefore, the Female and not participated is $\frac{28}{100}=0.28\%$

Hence, the correct option is $D$ .

5 0

3 years ago

There are 16 boys and 19 girls in a class. The teacher calls the

san4es73 [151]

Answer:

1/12/2121 precent

Step-by-step explanation:

6 0

3 years ago

A rocket is fired with an initial vertical velocity of 40 m/s from a launch pad 45 m high, and its height is given by the formul

N76 [4]

Answer:

1) 125 meters

2) For 9 seconds.

Step-by-step explanation:

The rocket's height is given by the formula $-5t^2+40t+45$ .

Notice that this is a quadratic.

Part 1)

Since this is a quadratic, the highest height the rocket goes will be the vertex of our quadratic. Remember that the vertex of a quadratic in standard form is:

$(-\frac{b}{2a}, f(-\frac{b}{2a}))$

So, let's identify our coefficients. The standard quadratic form is:

$ax^2+bx+c$

Therefore, in this case, our a is -5, b is 40, and c is 45.

So, let's find the x-coordinate of our vertex. Substitute 40 for b and -5 for a. This yields:

$x=\frac{-(40)}{2(-5)}$

Multiply and reduce. So:

$x=-40/-10=4$

Now, substitute 4 for t to find the height. So:

$-5(4)^2+40(4)+45$

Evaluate:

$=-5(16)+40(4)+45$

Multiply and add:

$=-80+160+45=125\text{ meters}$

Therefore, the highest the rocket goes up is 125 meters.

Part 2)

To find out for how much time the rocket is in the air, we can think about after how many seconds after launch will the rocket land.

If the rocket lands, the height h will be 0. So, we can set our expression equal to 0 and solve for t:

$-5t^2+40t+45=0$

First, let's divide everything by -5. This yields:

$t^2-8t-9=0$

We can factor:

$(t+1)(t-9)=0$

Zero Product Property:

$t+1=0\text{ or } t-9=0$

Solve for t:

$t=-1\text{ or } t=9$

In this case, since t is our time in seconds, -1 seconds does not make sense. So, we can remove -1 from our solution set.

Therefore, 9 seconds after launch, the rocket will touch the ground.

Therefore, the rocket was in the air for 9 seconds.

And we're done!

6 0

3 years ago

Select the graph that represents two quantities in a proportional relationship.

julia-pushkina [17]

Answer:

Step-by-step explanation:

6 0

3 years ago

A circle has the center of (1,-5) and a radius of 5 determine the location of the point (4,-1)

Sliva [168]

"determine the location" or namely, is it inside the circle, outside the circle, or right ON the circle?

well, we know the center is at (1,-5) and it has a radius of 5, so the distance from the center to any point on the circle will just be 5, now if (4,-1) is less than that away, is inside, if more than that is outiside and if it's exactly 5 is right ON the circle.

well, we can check by simply getting the distance from the center to the point (4,-1).

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ \stackrel{center}{(\stackrel{x_1}{1}~,~\stackrel{y_1}{-5})}\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{-1})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d = \sqrt{[4-1]^2+[-1-(-5)]^2}\implies d=\sqrt{(4-1)^2+(-1+5)^2} \\\\\\ d = \sqrt{3^2+4^2}\implies d =\sqrt{9+16}\implies d=\sqrt{25}\implies \stackrel{\textit{right on the circle}}{d = 5}$

5 0

3 years ago