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2 years ago

The length of a rectangular garden is 30 yards more than the width. The perimeter is 300 yards. Find the length and width

Mathematics

1 answer:

andre [41]2 years ago

5 0

Answer:

60 yards

Step-by-step explanation:

300 = 2(30 + w) + 2(w)

300 = 60 +2w + 2w

300 = 60 + 4w

240 = 4w

60 = w

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Step-by-step explanation:

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3 years ago

There are 50 cupcakes and 160 cookies to be sold at the school bake sale. What is the greatest number of packages of baked items

Sav [38]

The greatest number of packages of baked items that can be sold is 10

Solution:

Given that there are 50 cupcakes and 160 cookies to be sold at the school bake sale

To find: greatest number of packages of baked items that can be sold

Each package has the same number of cupcakes and same number of cookies with none left over

So, we have to find the greatest common factor of 50 and 160

The greatest number that is a factor of two (or more) other numbers.

When we find all the factors of two or more numbers, and some factors are the same ("common"), then the largest of those common factors is the Greatest Common Factor.

Greatest common factor of 50 and 160:

The factors of 50 are: 1, 2, 5, 10, 25, 50

The factors of 160 are: 1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 80, 160

Then the greatest common factor is 10

So the greatest number of packages sold is 10

Each package will contain:

Given that there 50 cupcakes and 160 cookies

Number of cupcakes in 1 package:

$\rightarrow \frac{50}{10} = 5$

Number of cookies in 1 package:

$\rightarrow \frac{160}{10} = 16$

So each package has the same number of cupcakes and same number of cookies with none left over

7 0

3 years ago

Which situation can be represented by 3s = 5s - 6 ?

grin007 [14]

Answer:

First one

Step-by-step explanation:

3s=5s-6

8 0

3 years ago

How many extraneous solutions does the equation below have?

aev [14]

Answer:

A solution is said to be extraneous, if it is a zero of the equation, but it does not satisfy the equation,when substituted in the original equation,L.H.S≠R.H.S.

The given equation consisting of variable , m is

$\frac{2 m}{2 m+3} -\frac{2 m}{2 m-3}=1\\\\ 2 m[\frac{1}{2 m+3} -\frac{1}{2 m-3}]=1\\\\ 2 m\times \frac{[2 m-3 -2 m- 3]}{4m^2-9}=1\\\\ -6 \times 2 m=4 m^2 -9\\\\ 4 m^2 +1 2 m -9=0\\\\m=\frac{-12 \pm\sqrt{12^2-4 \times 4 \times (-9)}}{2\times 4}\\\\m=\frac{-12 \pm \sqrt {144+144}}{8}\\\\m=\frac{-12 \pm \sqrt {288}}{8}\\\\m=\frac{-12 \pm 12 \sqrt{2}}{8}\\\\m=\frac{3}{2}\times(-1 \pm \sqrt{2})$

None of the two solution

$m=\frac{3}{2}\times(-1 +\sqrt{2}) and , m=\frac{3}{2}\times(-1 -\sqrt{2})$ , is extraneous.

Here, L.H.S= R.H.S

Option A: 0→ extraneous

8 0

3 years ago

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