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3 years ago

!!I NEED AN ANSWER QUICKLY!!

Mathematics

2 answers:

Arisa [49]3 years ago

8 0

Answer:

Step-by-step explanation:

The volume (V) of the tent is calculated as

V = Al ( A is the area of the triangular end and l is its length )

A = $\frac{1}{2}$ bh ( b is the base and h the height )

Here b = 5 and h = 6 , so

A = $\frac{1}{2}$ × 5 × 6 = 15 ft² , then

V = 15 × 11 = 165 ft³ → A

Katyanochek1 [597]3 years ago

7 0

Answer:

Step-by-step explanation:

Ans:A.

6×5÷2×11=165

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First five problems

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Answer:

1. B

3. C

4. B

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Step-by-step explanation:

Sry... I dont think I'm able to do #2 bc I'm only in Algebra I but there u go ;)

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3 years ago

In the spinner below, what is the probability of landing on 3?

WINSTONCH [101]

The spinner is divided into 10 equal parts. Since only one part is labeled 3, the answer is 1/10.

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4 years ago

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Find the equation of ellipse passing throgh (1,4) and (-3,2)

irinina [24]

Answer:

$\displaystyle \frac{ {3x}^{2} }{ 35 } + \frac{{2y}^{2} }{ 35 } = 1$

Step-by-step explanation:

we want to figure out the ellipse equation which passes through (1,4) and (-3,2)

the standard form of ellipse equation is given by:

$\displaystyle \frac{(x - h {)}^{2} }{ {a}^{2} } + \frac{(y - k {)}^{2} }{ {b}^{2} } = 1$

where:

(h,k) is the centre
a is the horizontal redius
b is the vertical radius

since the centre of the equation is not mentioned, we'd assume it (0,0) therefore our equation will be:

$\displaystyle \frac{ {x}^{2} }{ {a}^{2} } + \frac{{y}^{2} }{ {b}^{2} } = 1$

substituting the value of x and y from the point (1,4),we'd acquire:

$\displaystyle \frac{ 1}{ {a}^{2} } + \frac{16}{ {b}^{2} } = 1$

similarly using the point (-3,2), we'd obtain:

$\displaystyle \frac{ 9}{ {a}^{2} } + \frac{4 }{ {b}^{2} } = 1$

let 1/a² and 1/b² be q and p respectively and transform the equation:

$\displaystyle \begin{cases} q + 16p = 1 \\ 9q + 4p = 1 \end{cases}$

solving the system of linear equation will yield:

$\displaystyle \begin{cases} q = \dfrac{3}{35} \\ \\ p = \dfrac{2}{35} \end{cases}$

substitute back:

$\displaystyle \begin{cases} \dfrac{1}{ {a}^{2} } = \dfrac{3}{35} \\ \\ \dfrac{1}{ {b}^{2} } = \dfrac{2}{35} \end{cases}$

divide both equation by 1 which yields:

$\displaystyle \begin{cases} {a}^{2} = \dfrac{35}{ 3} \\ \\ {b}^{2} = \dfrac{35}{2} \end{cases}$

substitute the value of a² and b² in the ellipse equation , thus:

$\displaystyle \frac{ {x}^{2} }{ \dfrac{35}{3} } + \frac{{y}^{2} }{ \dfrac{35}{2} } = 1$

simplify complex fraction:

$\displaystyle \frac{ {3x}^{2} }{ 35 } + \frac{{2y}^{2} }{ 35 } = 1$

and we're done!

(refer the attachment as well)

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Answer:

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Step-by-step explanation:

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3 years ago

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JulsSmile [24]

Answer:

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Step-by-step explanation:

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