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3 years ago

If the breadth of rectangle is one-fourth of its length and perimeter in60 CM find the length and breadth of rectangle

Mathematics

1 answer:

Gennadij [26K]3 years ago

4 0

Answer:

Step-by-step explanation:

Let length be x

So breadth will be 1/4x

2(x+1/4x)=60

x+1/4x=60/2=30

find lcm and the lcm would be 4 and make denominators equal and add it

4/4x+1/4x=30

5/4x=30

x=30*4/50=6*4=24

Therefore the length is 24/4=6cm and breadth is 24cm.

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A local hamburger shop sold a combined total of 763 hamburgers and cheeseburgers on Saturday. There were 63 more cheeseburgers s

Airida [17]

In order to solve this we'll start by assigning variables to hamburgers and cheeseburgers, since these are what we're trying to find. Lets say x = hamburgers and y = cheeseburgers. So we know two things, we know that x+y= 763 (hamburgers plus cheeseburgers sold equals 763, and we know that y= x+63 (cheeseburgers sold equals 63 more than hamburgers sold). Now we have a system of equations. This can be solved most easily by rearranging each equation to each y, and then set them equal to each other:
x+y=763 -> y=763-x, and we already have y=x+63. Set them equal to each other:
x+63 = 763-x (add x to both sides) -> 2x+63 = 763 (subtract 63 from both sides) -> 2x = 700 (divide both sides by 2) x = 350. So we solved for x, which is hamburgers sold, which is what the question asks for, so your answer is 350 hamburgers were sold on Saturday

6 0

4 years ago

The age of a father is to less than seven times the age of his son in three years the sum oftheir ages will be 52 if the sons pr

vekshin1

Answer:

The correct answer is:

(7s-2)+3+(s+3) = 52, or 8s+4 = 52.

Step-by-step explanation:

Since s is the son's age, "two less than seven times" the son's age would be represented by 7s-2. To represent this in 3 years, we would add 3: (7s-2)+3. In 3 years, the son's age, s, would be represented by s+3. We are told that the sum of these ages will be 52; this gives us (7s-2)+3+(s+3) = 52.

To simplify this, combine like terms. 7s+s = 8s; -2+3+3 = 4. This gives us 8s+4=52.

3 0

3 years ago

A tunnel is in the shape of a parabola. The maximum height is 31 m and it is 13 m wide at the base as shown below.

yan [13]

Using the given values from the problem and the illustration, three points are known which are (0,0), (6.5,-31), (-6.5,-31). The first step in solving this problem is to determine the equation of the parabola.
y = ax²
-31 = a(6.5)²
-31 = 42.25a
a = -31/42.25
a = -124/169

Therefore, the equation of the parabola is y = (-124/169)x². The value 4.5 is then substituted in the equation as x to get the answer which is 16.14 meters.

6 0

3 years ago

Find the remaining trigonometric ratios of θ if csc(θ) = -6 and cos(θ) is positive

VikaD [51]

Now, the cosecant of θ is -6, or namely -6/1.

however, the cosecant is really the hypotenuse/opposite, but the hypotenuse is never negative, since is just a distance unit from the center of the circle, so in the fraction -6/1, the negative must be the 1, or 6/-1 then.

we know the cosine is positive, and we know the opposite side is -1, or negative, the only happens in the IV quadrant, so θ is in the IV quadrant, now

$\bf csc(\theta)=-6\implies csc(\theta)=\cfrac{\stackrel{hypotenuse}{6}}{\stackrel{opposite}{-1}}\impliedby \textit{let's find the \underline{adjacent side}}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
\pm\sqrt{6^2-(-1)^2}=a\implies \pm\sqrt{35}=a\implies \stackrel{IV~quadrant}{+\sqrt{35}=a}$

recall that

$\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad\qquad 
cos(\theta)=\cfrac{adjacent}{hypotenuse}
\\\\\\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}
\qquad \qquad 
% cotangent
cot(\theta)=\cfrac{adjacent}{opposite}
\\\\\\
% cosecant
csc(\theta)=\cfrac{hypotenuse}{opposite}
\qquad \qquad 
% secant
sec(\theta)=\cfrac{hypotenuse}{adjacent}$

therefore, let's just plug that on the remaining ones,

$\bf sin(\theta)=\cfrac{-1}{6}
\qquad\qquad 
cos(\theta)=\cfrac{\sqrt{35}}{6}
\\\\\\
% tangent
tan(\theta)=\cfrac{-1}{\sqrt{35}}
\qquad \qquad 
% cotangent
cot(\theta)=\cfrac{\sqrt{35}}{1}
\\\\\\
sec(\theta)=\cfrac{6}{\sqrt{35}}$

now, let's rationalize the denominator on tangent and secant,

$\bf tan(\theta)=\cfrac{-1}{\sqrt{35}}\implies \cfrac{-1}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{-\sqrt{35}}{(\sqrt{35})^2}\implies -\cfrac{\sqrt{35}}{35}
\\\\\\
sec(\theta)=\cfrac{6}{\sqrt{35}}\implies \cfrac{6}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{6\sqrt{35}}{(\sqrt{35})^2}\implies \cfrac{6\sqrt{35}}{35}$

3 0

4 years ago

An amusement park charges $9.00 for admission $4.00 per ride. Write an equation that gives the cost in dollars as a function of

Gnoma [55]

T = total cost

X= number of rides

T=4.00x+9.00

3 0

3 years ago