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3 years ago

Help!! i don't understand either of these :(

Mathematics

1 answer:

irga5000 [103]3 years ago

5 0

Answer:

19. 15.5x8X3=372ft

C=372ft

20. 15.5x8x3=372ft

C=372ft

Hope it helped

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I will give brainliest!!!!!

user100 [1]

The sum of their age is "s" years? Please fix the typo. Then I might be able to help :D

4 0

3 years ago

Read 2 more answers

the length of a rectagle is 5 in longer than its width. if the perimeter of the rectangle is 58 in, find its length and width

dolphi86 [110]

Answer:

Length = 17 inches

Width = 12 inches

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Step-by-step explanation:

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As it is given that, the length of a rectangle is 5 in longer than its width and the perimeter of the rectangle is 58 in and we are to find the length and width of the rectangle. So,

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Let us assume the width of the rectangle as x inches and therefore, the length will be (x + 5) inches .

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Now, <u>According to the Question :</u>

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${\longrightarrow \qquad { \pmb{\frak {2 ( Length + Breadth )= Perimeter_{(Rectangle)} }}}}$

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${\longrightarrow \qquad { {\sf{2 ( x + 5 + x )= 58 }}}}$

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${\longrightarrow \qquad { {\sf{2 ( 2x + 5 )= 58 }}}}$

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${\longrightarrow \qquad { {\sf{ 4x + 10= 58 }}}}$

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${\longrightarrow \qquad { {\sf{ 4x = 58 - 10}}}}$

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${\longrightarrow \qquad { {\sf{ 4x = 48}}}}$

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${\longrightarrow \qquad { {\sf{ x = \dfrac{48}{4} }}}}$

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${\longrightarrow \qquad{ \underline{ \boxed { \pmb{\mathfrak {x = 12}} }}} }\: \: \bigstar$

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Therefore,

The width of the rectangle is 12 inches .

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Now, We are to find the length of the rectangle:

${\longrightarrow \qquad{ { \frak{\pmb{Length = x + 5 }}}}}$

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${\longrightarrow \qquad{ { \frak{\pmb{Length = 12 + 5 }}}}}$

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${\longrightarrow \qquad{ { \frak{\pmb{Length = 17}}}}}$

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Therefore,

The length of the rectangle is 17 inches .

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8 0

2 years ago

Read 2 more answers

X and y are both bigger than 30. x and y have HCF 21 and LCM 1617. Find x and y.

Sunny_sXe [5.5K]

The values of x and y are 99 and 343

The HCF of numbers is the highest common factors of the numbers

The LCM of numbers is the lowest common multiple of the numbers

The HCF and the LCM are given as:

$HCF = 21$

$LCM = 1617$

Multiply the HCF and the LCM

$HCF \times LCM = 21 \times 1617$

$HCF \times LCM = 33957$

The product of the numbers x and y equals the product of the HCF and the LCM.

So, we have:

$x \times y = 33957$

<h3>Prime Factors</h3>

Express 33957 as a prime factor

$x \times y = 3^2 \times 7^3 \times 11$

Rewrite the equation as

$x \times y = 99 \times 7^3$

$x \times y = 99 \times 343$

By comparison:

$x = 99$

$y= 343$

Hence, the values of x and y are 99 and 343