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Gnom [1K]
3 years ago
14

A rectangle is four times as long as it is wide. The perimeter is 40 cm. Find the length of each side.

Mathematics
1 answer:
rjkz [21]3 years ago
6 0

Answer:

W=4 l=16

Step-by-step explanation:

Let w = width

Let l = length

Since the length is 4 times the width, l=4w

Perimeter of rectangle = 2l + 2w

= 2(4w)+2w

=10w

The perimeter of the rectangle =40, so

40=10w and w=4

Plug w into the length equation : l=4w=4(4)=16

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How to solve 10 (g+5)=2(g+9)
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first multiply the outside numbers by both the numbers in the ()

10xg=10g       10x5=50    2xg=2g             2x9=18

then combine like terms

10g+50=2g+18

8g=32

then divide

g=4

4 0
3 years ago
What type of number is 0? Rational, Irrational, Whole number, Integer
gladu [14]
It is a whole number and integer.
4 0
4 years ago
Read 2 more answers
800 divided by 10 to the power 3
Blababa [14]
I think that the answer you are looking for is 0.3

7 0
4 years ago
Please answer this question now
Rashid [163]

Answer:

54

Step-by-step explanation:

To solve problems like this, always recall the "Two-Tangent theorem", which states that two tangents of a circle are congruent if they meet at an external point outside the circle.

The perimeter of the given triangle = IK + KM + MI

IK = IJ + JK = 13

KM = KL + LM = ?

MI = MN + NI ?

Let's find the length of each tangents.

NI = IJ = 5 (tangents from external point I)

JK = IK - IJ = 13 - 5 = 8

JK = KL = 8 (Tangents from external point K)

LM = MN = 14 (Tangents from external point M)

Thus,

IK = IJ + JK = 5 + 8 = 13

KM = KL + LM = 8 + 14 = 22

MI = MN + NI = 14 + 5 = 19

Perimeter = IK + KM + MI = 13 + 22 + 19 = 54

4 0
3 years ago
Let Q(x, y) be the predicate "If x < y then x2 < y2," with domain for both x and y being R, the set of all real numbers.
Levart [38]

Answer:

a) Q(-2,1) is false

b) Q(-5,2) is false

c)Q(3,8) is true

d)Q(9,10) is true

Step-by-step explanation:

Given data is Q(x,y) is predicate that x then x^{2}. where x,y are rational numbers.

a)

when x=-2, y=1

Here -2 that is x  satisfied. Then

(-2)^{2}

4 this is wrong. since 4>1

That is x^{2}>y^{2} Thus Q(x,y) =Q(-2,1)is false.

b)

Assume Q(x,y)=Q(-5,2).

That is x=-5, y=2

Here -5 that is x this condition is satisfied.

Then

(-5)^{2}

25 this is not true. since 25>4.

This is similar to the truth value of part (a).

Since in both x satisfied and x^{2} >y^{2} for both the points.

c)

if Q(x,y)=Q(3,8) that is x=3 and y=8

Here 3 this satisfies the condition x.

Then 3^{2}

9 This also satisfies the condition x^{2}.

Hence Q(3,8) exists and it is true.

d)

Assume Q(x,y)=Q(9,10)

Here 9 satisfies the condition x

Then 9^{2}

81 satisfies the condition x^{2}.

Thus, Q(9,10) point exists and it is true. This satisfies the same values as in part (c)

6 0
3 years ago
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