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lakkis [162]
3 years ago
8

The length of a rectangle is 15 and its width is w.

Mathematics
2 answers:
Fed [463]3 years ago
6 0

A rectangle with sides 15 and w has perimeter

2p = 2(15+w) = 2w+30

We want this quantity to be at most 50, so it must be less than or equal to 50:

2w+30\leq 50

For the record, this implies that

2w\leq 20 \iff w \leq 10

So, the width can be at most 10.

Svetradugi [14.3K]3 years ago
3 0

Answer:

2x+30< 50   ...answer : A

Step-by-step explanation:

the perimeter is :  P = 2(L+W)

let : L= x   given : L=15

P= 2(x+15)

The perimeter of the rectangle is, at most, 50:   P<50

2(x+15) < 50

2x+30< 50   ...answer : A

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If anyone could help me with this last problem, I would greatly appreciate it.
il63 [147K]

Answer:

From the given information, the value of a is 3 and the measurement of ∠R is 25°

Step-by-step explanation:

For this problem, we have to find the value of a and the measurement of ∠R. We are given some information already in the problem.

<em>ΔJKL ≅ ΔPQR</em>

This means that all of the angles and all of the sides of each triangle are going to be equal to each other.

So, knowing this, let;s find the measurement of ∠R first.

All triangles have a total measurement of 180°. We are already given two angle measurements. We are given that the m∠P is 90° because the small box in the triangle represents a right angle and right angles equal 90°. We are also given that the m∠Q is 65° because ∠Q is equal to ∠K so they have the same measurement. Now, let's set up our equation.

65 + 90 + m∠R = 180

Add 65 to 90.

155 + m∠R = 180

Subtract 155 from 180.

m∠R = 25°

So, the measurement of ∠R is 25°.

Now let's find the value of a.

KL is equal to PQ so we will set up an equation where they are equal to each other.

7a - 10 = 11

Add 10 to 11.

7a = 21

Divide 7 by 21.

a = 3

So, the value of a is 3.

4 0
4 years ago
PLEASE HELP ON THIS<br><br> THX
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Determine whether the three given points are collinear. (3, 2), (4, 6), (0, 8) True: they are collinear False: they are not coll
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Answer:

They are not co-linear.

Step-by-step explanation:

For three point's to be co linear the slopes of the lines connecting them should be same

Mathematically we can write for (x_{1},y_{1}),(x_{2},y_{2}),(x_{3},y_{3})

to be co-linear we should have

\frac{y_{3}-y_{2}}{x_{3}-x_{2}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{y_{3}-y_{1}}{x_{3}-x_{1}}

Applying the given values we obtain

\frac{y_{3}-y_{2}}{x_{3}-x_{2}}=\frac{8-6}{0-4}=-1/2\\\\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{6-2}{4-3}=3\\\\\frac{y_{3}-y_{1}}{x_{3}-x_{1}}=\frac{8-2}{0-3}=-2

As we can see the values are not equal thus the points are not co-linear.

6 0
3 years ago
What is 25 over 6 as a mixed number?
DaniilM [7]

4 1/6 is your answer because 6* 4 = 24 so then there is just one more left over which makes the 1/6
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Steven's mother asked him to buy 750 grams of sugar. At the supermarket, he finds that the sugar packages give the weight in kil
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Answer:

0.75 kilograms

Step-by-step explanation:

All you need to do is move the decimal.

7 0
4 years ago
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