All categories

3 years ago

The length of the garden should be 15 feet longer than its width and the area is less than 100 ft^2

Mathematics

2 answers:

Scrat [10]3 years ago

4 0

Answer:

If the perimeter is 200feet (two lengths and two widths) then one width+one length equals 100feet.

If the length is 10 feet longer than the width, then length = 55 feet, and width =45 feet

Therefore the area of the rectangle is 45x55 feet, which equals 2475 sq.ft

Step-by-step explanation:

Hi! Nice to meet you and have a great day

Sholpan [36]3 years ago

4 0

Answer:

If the perimeter is 200feet (two lengths and two widths) then one width+one length equals 100feet.

If the length is 10 feet longer than the width, then length = 55 feet, and width =45 feet

Therefore the area of the rectangle is 45x55 feet, which equals 2475 sq.ft

You might be interested in

(3xy^3)^2 (xy)^6 simplify the expression

solong [7]

9x8y12 it a lot of work to do so I don't want to type how to do the problem

5 0

3 years ago

Help me please.<br><br> Rounding questions.

ehidna [41]

Step-by-step explanation:

a) 6400

b) 4800

c) 32000

hope this answer helps you dear!

3 0

3 years ago

Please help !! ASAP on all questions

Vera_Pavlovna [14]

So i think u subtract how much money the person got and then subtract the answer by his pay.

7 0

3 years ago

Write an equation that has a graph with the shape of y = x^2, but shifted left 3 units.

svetoff [14.1K]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\
% left side templates
\begin{array}{llll}
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}\\\\
--------------------\\\\$

$\bf % template detailing
\bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\$

$\bf \bullet \textit{ vertical shift by }{{ D}}\\
\left. \qquad \right. if\ {{ D}}\textit{ is negative, downwards}\\\\
\left. \qquad \right. if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}$

now, with that template in mind, let's see

$\bf y=x^2\implies 
\begin{array}{llll}
y=(&1x&-0)^2\\
&B&C
\end{array}$

so, just change C to +3, thus C/B is 3/1

6 0

3 years ago

Simplify. y 5 · y 3 ÷ y 2

Gnesinka [82]

Y^5 x y^3 /y^2
y^8/y^2
y^4

3 0

3 years ago