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2 years ago

13

Solve for the value of R. SOMEONE PLEASE HELP ILL GIVE YOU POINTS AND BRAINLIST ANSWER !!!

1 answer:

lord [1]2 years ago

7 0

Step-by-step explanation:

$(3r + 4) + (8r - 2) = 90$

$3r + 4 + 8r - 2= 90$

$11r = 88$

$r = \frac{88}{11} \\$

$r = 8$

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Plss help I need answers asap

elena55 [62]

Answer:

52

Step-by-step explanation:

Let total plants = x

grape tomato: 1/5 of x=18; (1/5)*x=18; x=18*5=90

other plants other than grape tomatoes and cherry tomatoes = 90-18-20=52

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3 years ago

Whats the answer to 10 + 5( x - 3) -7x when simplified

DedPeter [7]

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3 years ago

Read 2 more answers

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Korvikt [17]

Answer:

25%

Step-by-step explanation:

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3 years ago

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(10 points) Person A leaves a $16 tip for a meal that costs $123. Person B leaves a tip of $12 for a meal that costs $89. Who is

miskamm [114]

Answer:

person b because the person has $89 so the person can get any meal

Step-by-step explanation:

3 0

3 years ago

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 <u>(</u><u>1</u><u>,</u><u>4</u><u>)</u><u> </u>and <u>(</u><u>-</u><u>3</u><u>,</u><u>2</u><u>)</u>

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)

8 0

3 years ago