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3 years ago

Find the measure of the indicated arc

Mathematics

1 answer:

Yanka [14]3 years ago

4 0

Answer:

m(arc WXY) = 224°

Step-by-step explanation:

By inscribed angle theorem,,

"Measure of intercepted arc is double of the inscribed angle"

m(arc WXY) = 2[m(∠YCW)]

Here, arc WXY is the intercepted arc and ∠YCW is the inscribed angle.

By substituting the value of inscribed angle,

m(arc WXY) = 2(112°)

= 224°

Therefore, measure of arc WXY is 224°.

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1. If $18,000 is invested at 6% compounded monthly, what is the amount after 7 years?

stira [4]

P = 18,000
R=6%
T=7 years

A = P ( 1 + 1 divided by r) raised to n
Substitute the numbers an u will fin the answer

8 0

2 years ago

2w + 7 = 2(w + 5) - 8

BARSIC [14]

Answer:

no solution

Step-by-step explanation:

2w + 7 = 2(w + 5) - 8

Distribute

2w+7 = 2w+10-8

Combine like terms

2w+7 = 2w+2

Subtract 2w from each side

7 = 2

This is never true so there is no solution

7 0

3 years ago

Read 2 more answers

You are arguing over a cell phone while trailing an unmarked police car by 25 m; both your car and the police car are traveling

BARSIC [14]

Answer:

(a) 15 m

(b) ground speed: 26.14 m/s (94.1 km/h); relative speed 12 m/s

Step-by-step explanation:

We can choose the frame of reference to be the trailing car. We can start counting time from the point the lead car begins deceleration. Then its position (in meters) relative to the trailing car is ...

d(t) = 25 - 5/2t² . . . . . . where 25 m is the initial distance and -5 is the acceleration in m/s².

(a) Then the distance between cars at t=2 is ...

d(2) = 25 - (5/2)·4 = 15 . . . . meters

(b) The <em>relative velocity</em> at 2.4 seconds is ...

d'(t) = -5t

d'(2.4) = -5(2.4) = -12.0 . . . m/s

The closure speed with the police car is 12 m/s.

We need to do a little more work to find the ground speed of the trailing car when the cars bump.

The distance between the cars at t=2.4 seconds is ...

d(2.4) = 25 -(5/2)2.4² = 10.6 . . . . meters

At the closure speed of 12 m/s, it will take an additional ...

10.6 m/(12 m/s) = 0.8833... s = 53/60 s

to close the gap. The speed of the trailing car at that point in time will be the original speed less the deceleration for 0.8833 seconds:

(110 km/h)·(1/3.6 (m/s)/(km/h))-(5 m/s²)(53/60 s)

= 26 5/36 m/s = 94.1 km/h

_____

<em>Comment on following distance</em>

To avoid collision, the trailing car must be trailing by at least the distance it covers in 2.4 seconds, about 73 1/3 meters.

6 0

3 years ago

220 students were asked if they liked chocolate or vanilla ice cream. 140 said they liked chocolate and 120 said they liked vani

Flura [38]

Answer: D. 40

Step-by-step explanation:

Given: Number of students liked vanilla or chocolate ice cream = $n(V\cup C)=220$

Number of students liked chocolate = $n(C)=140$

Number of students liked vanilla= $n(V)=120$

Now, Number of students liked vanilla and chocolate ice cream is given by :-

$n(V\cap C)=n(C)+n(V)-n(V\cup C)\\\\\Rightarrown(V\cap C)=140+120-220\\\\\Rightarrown(V\cap C)=260-220\\\\\Rightarrown(V\cap C)=40$

Number of students liked vanilla and chocolate ice cream= 40

6 0

3 years ago

I just need 30 please

PIT_PIT [208]

In the smaller triangle, let $y$ be the length of the smallest leg. Then the smallest leg in the larger triangle is $14-y$ .

Within the scope of the smaller triangle, we have $y$ such that

$x^2+y^2=13^2\implies y=\sqrt{13^2-x^2}$

Then within the larger the triangle, we would have

$x^2+(14-y)^2=15^2\iff x^2+\left(14-\sqrt{13^2-x^2}\right)^2=15^2$

Now we can solve for $x$ :

$x^2+14^2-28\sqrt{13^2-x^2}+(13^2-x^2)=15^2$

$\implies x=12$

8 0

3 years ago