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OLga [1]
3 years ago
13

I need help with equivalent ratio

Mathematics
2 answers:
Vedmedyk [2.9K]3 years ago
8 0

Answer:

What is the question i can help!

Step-by-step explanation:

♡♡♡♡♡♡♡♡♡

goldfiish [28.3K]3 years ago
5 0

Answer:

what is the question

Step-by-step explanation:

if you r asking what it is than;

it is something that can be simplified and broke down into a simple fraction i think

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Jamie puts $500 into a savings account that has an interest rate of 1% how much will be in her account after 6 years?
IgorLugansk [536]

Step-by-step explanation:

interest:

500 \times \frac{1}{100}  \times 6 = 30

Total=500+30=530

4 0
3 years ago
A right-angled triangle has shorter side lengths exactly c^2-b^2 and 2bc units respectively, where b and c are positive real num
yKpoI14uk [10]

Answer: hypotenuse = c^{2} + b^{2}

Step-by-step explanation: Pythagorean theorem states that square of hypotenuse (h) equals the sum of squares of each side (s_{1},s_{2}) of the right triangle, .i.e.:

h^{2} = s_{1}^{2} + s_{2}^{2}

In this question:

s_{1} = c^{2}-b^{2}

s_{2} = 2bc

Substituing and taking square root to find hypotenuse:

h=\sqrt{(c^{2}-b^{2})^{2}+(2bc)^{2}}

Calculating:

h=\sqrt{c^{4}+b^{4}-2b^{2}c^{2}+(4b^{2}c^{2})}

h=\sqrt{c^{4}+b^{4}+2b^{2}c^{2}}

c^{4}+b^{4}+2b^{2}c^{2} = (c^{2}+b^{2})^{2}, then:

h=\sqrt{(c^{2}+b^{2})^{2}}

h=(c^{2}+b^{2})

Hypotenuse for the right-angled triangle is h=(c^{2}+b^{2}) units

3 0
3 years ago
Is √8 a rational or irrational number
yawa3891 [41]

Answer:

irrational

Step-by-step explanation:

irrational because there will still be a 2 under the root when you simplify to 2√2

3 0
3 years ago
The expression 1662 models the distance in feet that an object falls during t seconds after being dropped. What distance will an
Lunna [17]
100 center meters I think that should be right
6 0
3 years ago
A parabola can be drawn given a focus of (-11, -2) and a directrix of x= -3
fgiga [73]

Check the picture below, so the parabola looks more or less like that.

now, the vertex is half-way between the focus point and the directrix, so that puts it where you see it in the picture, and the horizontal parabola is opening to the left-hand-side, meaning that the distance "P" is negative.

\textit{horizontal parabola vertex form with focus point distance} \\\\ 4p(x- h)=(y- k)^2 \qquad \begin{cases} \stackrel{vertex}{(h,k)}\qquad \stackrel{focus~point}{(h+p,k)}\qquad \stackrel{directrix}{x=h-p}\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix}\\\\ \stackrel{"p"~is~negative}{op ens~\supset}\qquad \stackrel{"p"~is~positive}{op ens~\subset} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}

\begin{cases} h=-7\\ k=-2\\ p=-4 \end{cases}\implies 4(-4)[x-(-7)]~~ = ~~[y-(-2)]^2 \\\\\\ -16(x+7)=(y+2)^2\implies x+7=-\cfrac{(y+2)^2}{16}\implies x=-\cfrac{1}{16}(y+2)^2-7

8 0
2 years ago
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