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3 years ago

Help me please will mark brainliest

Mathematics

1 answer:

weqwewe [10]3 years ago

3 0

Answer:

Interest earned: 100

Step-by-step explanation:

I=prt

100 = 2000 x 0.05 x 1

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What is 9.03×10^2 in standard form

yawa3891 [41]

9.03 * 10² =
9.03 * 100 =
903

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BlackZzzverrR [31]

Answer:

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3 years ago

How many breaths does it take jesse to fill a balloon?

Aleksandr [31]

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3 years ago

A grasshopper jumps off a tree stump. The height, in feet, of the grasshopper above the ground after t seconds is modeled by the

Monica [59]

The height, in feet, of the grasshopper above the ground after t seconds is modeled by the function , which is

$h(t) =-t^2+\frac{4}{3}t+ \frac{1}{4}$

When the grasshopper land on the ground,

$h(t)=0$

that is,

$-t^2+\frac{4}{3}t+ \frac{1}{4}=0$

Multiplying whole equation by 12 to get rid of denominators 3 and 4

$-12t^2+16t +3=0$

$(3-2t)(6t+1)=0 \\ 3-2t=0 , 6t+1=0 \\ t= \frac{3}{2} , \frac{-1}{6}$

And time cant be negative, so the correct option is

$t = \frac{3}{2}=1.5 seconds$

Correct option is C .

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4 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 (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)

8 0

3 years ago