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3 years ago

11

Can you guys help me out with this? Thanks! <3

1 answer:

Tamiku [17]3 years ago

6 0

Answer:

Step-by-step explanation:

1.Jenna's interest was $480

2. 7 years Mike has saved & eared interest of $280.

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Answer question 3-8 plz

telo118 [61]

Work is shown (^_^)
Did #9-10 also

3) 3p -2 = -29 Isolate 3
3p= -29+2
3p= -27 Divide by 3
p = -9

4) 1- r = -5 Isolate r
-r= -5-1
-r= -6 Multiply by -1
r= 6

5) k-10/2 = -7 Multiply by 2
k-10= 2(-7)
k-10= -14
k=-14+10
k=-4

See attached file for #6-10
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7 0

3 years ago

Use a transformation to solve the equation. b + 0.6 = –1

Nina [5.8K]

You answer should be :

-1.6

Hope that helped!

8 0

3 years ago

The perimeter of the rectangle is (14,19,28,45) cm. If the rectangle is dilated by a scale factor of 6 to create a new rectangle

Lorico [155]

Ok ok so 28 because 9+9 is 18 and then 5+5 is 10 so you add them up and then 28

6 0

3 years ago

Read 2 more answers

Only the correct answer please

11111nata11111 [884]

A. because he had ridden less than 140 so 5s >140

3 0

3 years ago

Read 2 more answers

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