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2 years ago

The perimeter of a rectangle is 202. The length is 26 more than 4 times the width. Find the dimensions

Mathematics

1 answer:

jek_recluse [69]2 years ago

6 0

Answer:

$P = 2*(26+4y) + 4y$

And if we solve for y we got:

$202= 52 +8y +4y$

$202= 52 +12 y$

And replacing we got:

$150 = 12y$

And we got:

$y = \frac{150}{12}= 12.5$

And for x we got:

$x = 26 +4*12.5 = 76$

So then the length would be 76 and the width we got 12.5

Step-by-step explanation:

For this case we have a rectangle. The perimeter is given by:

$P= 2x+2y$

Where x represent the length and y the the width. We can set the following conditions:

$x = 26 +4y$

And if we replace the conditions we got:

$P = 2*(26+4y) + 4y$

And if we solve for y we got:

$202= 52 +8y +4y$

$202= 52 +12 y$

And replacing we got:

$150 = 12y$

And we got:

$y = \frac{150}{12}= 12.5$

And for x we got:

$x = 26 +4*12.5 = 76$

So then the length would be 76 and the width we got 12.5

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Answer:

74613

Step-by-step explanation:

22C6 = 74613

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2 years ago

0.01234567901 as a fraction

balu736 [363]

Answer: 1234567901 /100000000000

Step-by-step explanation:

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1 year ago

Your school organizes a clothing drive as a fundraiser for your class trip. The school earns $100 for every 400 pounds of donati

noname [10]

Answer:

$550

Step-by-step explanation:

400x=2200, x is the number of groups needed to get 2200 pounds.

If you divide both sides by 400, you'll get x=5.5

Next, multiply 5.5 by $100 because each group gives $100

hope this helps!

4 0

1 year ago

Read 2 more answers

In equilateral ∆ABC with side a, the perpendicular to side AB at point B intersects extension of median AM in point P. What is t

Sergeeva-Olga [200]

Answer:

Perimeter = (2 + √3)·a

Step-by-step explanation:

Given: ΔABC is equilateral and AB = a

The diagram is given below :

AM is a median , PB ⊥ AB , PM = b

Now, by using properties of equilateral triangle, median is perpendicular bisector and each angle is of 60°.

We get, ∠AMB = 90°. So, by linear pair ∠AMB + ∠PMB = 180° ⇒ ∠PMB = 90°. Also, ∠ABC = 60° and ∠ABP = 90° (given) So, ∠PBM = 30°

Since, AM is perpendicular bisector of BC. So,

$MB = \frac{a}{2}$

Now in ΔAMB , By using Pythagoras theorem

$AB^{2}=AM^{2}+MB^{2}\\AM^{2}=AB^{2}-MB^{2}\\AM^{2}=a^{2}-(\frac{a}{2})^{2}\\AM=\frac{\sqrt{3}\cdot a}{2}$

Now, in ΔBMP :

$sin\thinspace 30^{o}=\frac{\text{Perpendicular}}{\text{Hypotenuse}}\\\\sin\thinspace 30^{o}=\frac{\text{MB}}{\text{PB}}\\\\PB=\frac{\text{MB}}{\text{sin 30}}\\\\PB=\frac{\frac{a}{2}}{\frac{1}{2}}\implies PB = a\\\\tan\thinspace 30^{o}=\frac{\text{Perpendicular}}{\text{Base}}\\\\tan\thinspace 30^{o}=\frac{\text{MB}}{\text{PM}}\\\\PM=\frac{\text{MB}}{\text{tan 30}}\\\\PM=\frac{\frac{a}{2}}{\frac{1}{\sqrt3}}\implies PM=b= \frac{\sqrt{3}\cdot a}{2}$

Perimeter of ABM = AB + PB + PM + AM

$\text{Perimeter = }a+a+b+ \frac{\sqrt{3}\cdot a}{2}\\\\=2\cdot a + \frac{\sqrt{3}\cdot a}{2} +\frac{\sqrt{3}\cdot a}{2}\\\\=2\cdot a +\sqrt{3}\cdot a\\\\=(2+\sqrt3})\cdot a$

Hence, Perimeter of ΔABP = (2 + √3)·a units

3 0

1 year ago

The point (–1, 0.5) lies on the graph of f –1(x) = 2x. Based on this information, which point lies on the graph of f(x) = log2x?

Taya2010 [7]

<span>(0.5, -1)
f^-1(x) and f(x) are inverse functions of each other. That means that for all x, f^-1(f(x)) = x and f(f^-1(x)) = x. So for f^-1(x) we've been given the point (-1, 0.5) which means that we have an x value of -1 and a y value of 0.5. So if we swap the x and y values, we'll have a valid point for function f(x). Therefore an x value of 0.5 and a y value of -1 will work. So the answer is (0.5, -1). Notice that no math has to be done, you just need to know the meaning of inverse functions.</span>

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2 years ago

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