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Degger [83]
3 years ago
14

The distance from the center of an archery target to the outside edge of the target is 24 in. What is the diameter of the target

?
Mathematics
1 answer:
klemol [59]3 years ago
7 0

Answer:

The diameter of the target is 48\ in

Step-by-step explanation:

we know that

In this problem the  distance from the center of an archery target to the outside edge of the target is equal to the radius of the target

so

r=24\ in

Remember that

The diameter is two times the radius

D=2r

substitute

D=2(24)=48\ in

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Manuel needs to make a total of 60 deliveries this week. So far he has completed 36 of them. What percentage of his total delive
Alenkasestr [34]

Answer:

60%

Step-by-step explanation:

It’s simple

He needs to do 60, he has done 36

Percentage = (36/60) * 100

= 60%

5 0
4 years ago
Trapezoid MNPQ is similar to Trapezoid RSTU. What is the length of the "x" on the misside side NP ? Also, what is the length of
ivolga24 [154]

Answer:

x =  15

y =  25

Step-by-step explanation:

Given

See attachment for MNPQ and RSTU

Required

Find x and y

To solve this question, we make use of equivalent ratios of corresponding side lengths.

The ratio of corresponding sides are:

MN : RS

NP : ST

PQ : TU

MQ : RU

From the attachment, we have:

MN : RS \to 18 : 30

NP : ST \to x : 25

PQ : TU \to 15 : y

To solve for x, we equate MN : RS and NP : ST

18 : 30 = x : 25

Express as fraction

\frac{18 }{ 30 }= \frac{x }{ 25}

Make x the subject

x =  25 * \frac{18 }{ 30 }

x =  \frac{25 * 18 }{ 30 }

x =  \frac{450}{ 30 }

x =  15

To solve for y, we equate MN : RS and PQ : TU

18 : 30 = 15 : y

Express as fraction

\frac{18 }{ 30 }= \frac{15 }{ y}

Make y the subject

y = 15 * \frac{30 }{ 18 }

y =  \frac{15 *30}{ 18 }

y =  \frac{450}{ 18 }

y =  25

8 0
3 years ago
This is for algebra 2 can someone tell me if I’m right and can someone explain how do you get a= ?
algol13

Answers:

a = 2

h = 3

k = -3

You have the correct h and k values. Nice work so far.

==========================================================

Explanation:

(h,k) is the vertex. So (h,k) = (3,-3) meaning h = 3 and k = -3

To get the value of 'a', we need to plug in some point on the red graph. Each point is of the form (x,y). We can pick any point we want that isn't the vertex.

Let's pick (0,3) which is the y intercept. I'm picking this because 0 is an easy number to work with

------------

y = a|x-h| + k

y = a|x-3| + (-3) .... plug in the h,k values

y = a|x-3| - 3

3 = a|0-3| - 3 .... plug in (x,y) = (0,3)

3 = a|-3| - 3

3 = a*3 - 3

3 = 3a - 3

3a-3 = 3

3a = 3+3 ... adding 3 to both sides

3a = 6

a = 6/3 ... divide both sides by 3

a = 2

The value a = 2 indicates that the parent function y = |x| has been stretched vertically by a factor of 2. So it is twice as tall as before. Then it has been shifted to place the vertex at (3,-3) as shown in the graph.

------------

You may be wondering why you can't pick on the vertex for (x,y)

Let's see what happens if we use (x,y) = (3,-3)

y = a|x-3| - 3

-3 = a|3-3| - 3 ... plug in (x,y) = (3,-3)

-3 = a|0| - 3 ... uh oh, we get 0 here

-3 = a*0 - 3

-3 = 0 - 3

-3 = -3

We get a true statement, which is nice, but it doesn't tell us anything about what the value of 'a' is. That 'a' term goes away entirely. So I avoided using x = 3 to prevent x-3 from being 0.

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