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Aloiza [94]
3 years ago
13

Which equations are equivalent to a/b = c/d? Choose more than one answer.

Mathematics
1 answer:
Kruka [31]3 years ago
3 0

Answer:

1/3= 5/15

Step-by-step explanation:

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a local school raised 5687.50 for the terry fox foundation. Hua raised 0.8% of this total. how much money did Hua raise?
zlopas [31]
You multiply the $5687.50 by 0.8 to get the total of $4550.
So Hua raised $4550 out of the $5687.50.
6 0
3 years ago
The graph below represents which of the following functions?
erica [24]

This is a more difficult one to explain. Let's start with the line to the right.  It has a slope (m) of \frac{-2}{2} = -1 and a y-intercept of +3 (continue the line to see where it crosses the y-axis). So, the equation of that line is: y = -x + 3 when x ≥ 2. This can also be written as y = 3 - x when x ≥ 2.  The only option that has that equation is A.

Answer: A


7 0
3 years ago
Read 2 more answers
The sum of five times a number and 6 more than the number is the same as the difference between -18 and twice the number. What i
goblinko [34]
N = -3
5n + n + 6 = -18 - 2n
6n + 6 = -18 - 2n
-6 to both sides
6n = -24 - 2n
+2n to both sides
8n = -24
divide 8 to both sides
n = -3

4 0
3 years ago
Read 2 more answers
The height h (in feet) of an object dropped from a ledge after x seconds can be modeled by h(x)=−16x2+36 . The object is dropped
kakasveta [241]

Check the picture below.

\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(t) = -16t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}&\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&\\ \qquad \textit{at "t" seconds} \end{cases}

so the object hits the ground when h(x) = 0, hmmm how long did it take to hit the ground the first time anyway?

\bf h(x)=-16x^2+36\implies \stackrel{h(x)}{0}=-16x^2+36\implies 16x^2=36 \\\\\\ x^2=\cfrac{36}{16}\implies x^2 = \cfrac{9}{4}\implies x=\sqrt{\cfrac{9}{4}}\implies x=\cfrac{\sqrt{9}}{\sqrt{4}}\implies x = \cfrac{3}{2}~~\textit{seconds}

now, we know the 2nd time around it hit the ground, h(x) = 0, but it took less time, it took 0.5 or 1/2 second less, well, the first time it took 3/2, if we subtract 1/2 from it, we get 3/2 - 1/2  = 2/2 = 1, so it took only 1 second this time then, meaning x = 1.

\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(x) = -16x^2+v_ox+h_o \quad \begin{cases} v_o=\textit{initial velocity}&0\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&0\\ \qquad \textit{at "t" seconds}\\ x=\textit{seconds}&1 \end{cases} \\\\\\ 0=-16(1)^2+0x+h_o\implies 0=-16+h_o\implies 16=h_o \\\\[-0.35em] ~\dotfill\\\\ ~\hfill h(x) = -16x^2+16~\hfill

quick info:

in case you're wondering what's that pesky -16x² doing there, is gravity's pull in ft/s².

4 0
3 years ago
In an Animal shelter, the ratio of dogs to cats is 5 to 3. There are 25 dogs, how many cats are there?
LekaFEV [45]

Answer: 25:15

Step-by-step explanation: in the ratio 5:3 the 5 is dogs and the 3 is cats you multiply the 5 (aka the dogs) by 5 to get 25 (the amount of dogs in the shelter) when you multiply 3 by 5 you get the amount of cats which is 15, a good way to remember is what you do to one side you have to do to the other. I hope this helps.

7 0
3 years ago
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