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3 years ago

15

A student is decorating a circular picture frame that has a diameter of 8 inches. How many inches of fabric will the student nee

d to go completely around the outside of the frame?

1 answer:

Akimi4 [234]3 years ago

7 0

Answer:

B

Step-by-step explanation:

Circumference is equal to $\pi d$

$C=\pi d$

$C=\pi \times 8$

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jose jumped 8 1/3 feet. this was 2 2/3 feet farther than lila jumped. how far did lila jump TELL MEH AFTER THIS I GET ON IXL

wariber [46]

Subtract 2 2/3 from 8 1/3

6 1/3

5 0

3 years ago

The ratio of boys to girls in a group is 3:1. If there are 52 more boys than girls, work out how many people there are in total.

UkoKoshka [18]

Answer: 104 (Good chance its wrong)

Step-by-step explanation: if the ratio is 3:1 then there's 3 times as much boys as girls. if there are 52 more boys then that means 52 is twice as much as the number of girls so if we divide is by 2 we get 26. if we add 26 with 52 we get 78 and if we divide 78 by 3 we get 26. so I believe there are 26 girls and 78 boys. 26+78=104

I believe 104 is the answer but there's a good chance i'm wrong.

7 0

3 years ago

An object launched straight up at a speed of 29.4 meters per second has a height, h, in meters of h , t seconds after the object

Nostrana [21]

Answer:

h = 44.06 meters (maximum height)

the time the object takes to complete this whole path is 6 seconds, this is why the time at which the object reaches its maximum height will between 0 and 6 seconds

Step-by-step explanation:

To solve this question, we need to first recognize that this is a constant acceleration problem, specifically, it can be thought of as a projectile motion problem.

Recall, the equations of motion:

$1) v^2 - v_0^2 = 2a(s - s_0)\\2) s = v_0^2 + \frac{1}{2} at^2\\3) v = v_0 + at$

What do we already know?

$v_0 = 29.4 ms^-1$
The launch is straight up
$a = -9.81 ms^-2$ this is the gravitational acceleration $g$

$s_0 = 0 m$ , since our reference point is at s = 0, (the ground)

We can use use the Eq(1):

we know that when any object is launched up, at maximum height its velocity is going to be zero, $v = 0 ms^-2$

$v^2 - v_0^2 = 2a(s - s_0)\\0^2 - (29.4)^2 = 2(-9.81)(s- 0)\\s = 44.06 m$

this is the maximum height!

Why does t have to between zero and six?

We can answer this using a bit visualization, if you think about the second equation

$s = v_0 t - \frac{1}{2}at^2\\ s = 29.4t - 4.905t^2$

this is the equation of the whole trajectory that object makes.

and if you solve this by making $s = 0$ , you will get the times at which the object was at the ground. the times will be 0s and 5.99s.

so the amount of time the object takes to go through this whole path is 6 seconds and this why the object will only reach its maximum height in between this time interval.

hope this helps :)

5 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B%20%7B2%7D%5E%7Bx%20%2B%202%7D%20-%20%20%7B2%7D%5E%7B%20x%20%2B%203%7D%20%7D%7B%2

weeeeeb [17]

I suppose you just have to simplify this expression.

(2ˣ⁺² - 2ˣ⁺³) / (2ˣ⁺¹ - 2ˣ⁺²)

Divide through every term by the lowest power of 2, which would be <em>x</em> + 1 :

… = (2ˣ⁺²/2ˣ⁺¹ - 2ˣ⁺³/2ˣ⁺¹) / (2ˣ⁺¹/2ˣ⁺¹ - 2ˣ⁺²/2ˣ⁺¹)

Recall that <em>n</em>ª / <em>n</em>ᵇ = <em>n</em>ª⁻ᵇ, so that we have

… = (2⁽ˣ⁺²⁾ ⁻ ⁽ˣ⁺¹⁾ - 2⁽ˣ⁺³⁾ ⁻ ⁽ˣ⁺¹⁾) / (2⁽ˣ⁺¹⁾ ⁻ ⁽ˣ⁺¹⁾ - 2⁽ˣ⁺²⁾ ⁻ ⁽ˣ⁺¹⁾)

… = (2¹ - 2²) / (2⁰ - 2¹)

… = (2 - 4) / (1 - 2)

… = (-2) / (-1)

… = 2

Another way to get the same result: rewrite every term as a multiple of <em>y</em> = 2ˣ :

… = (2²×2ˣ - 2³×2ˣ) / (2×2ˣ - 2²×2ˣ)

… = (4×2ˣ - 8×2ˣ) / (2×2ˣ - 4×2ˣ)

… = (4<em>y</em> - 8<em>y</em>) / (2<em>y</em> - 4<em>y</em>)

… = (-4<em>y</em>) / (-2<em>y</em>)

… = 2

8 0

3 years ago

Angle Bisector Solve for Y

olganol [36]

As <ABC=40

3x-1+34-2x=40

x =40-34+1 = 7

In triangle ABD

3x-1 + 90 +3y + 6= 180

3(7) +3y +95 =180

3y= 180-95-21=64

Y =64/3

4 0

3 years ago