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3 years ago

How to do the lesson 4 7th grade

Mathematics

1 answer:

wlad13 [49]3 years ago

4 0

For the first one you would put 1,2,3,4 in the top then in the bottom you would put 15,30,45,60

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how many circles can be drawn passing through three non collinear points and how many circles can be drawn passing through three

MrRissso [65]

Answer:

Only one circle can be drawn through three non nonlinear points, No circle can be drawn through three nonlinear points

Step-by-step explanation:

A circle has a curve. As long as the three non nonlinear points both have the same amount of distance from the center of the circle it can be a circle drawn. No circle can be drawn through three points on the line.

7 0

2 years ago

Write .33 as a simple form

anyanavicka [17]

.33 in its simplest form is 1/3 I think

3 0

3 years ago

Question

leva [86]

Answer:

1800 dollars

Step-by-step explanation:

We can set up an equation here to represent the exponential growth in Terry's bank account.

$y = 1600(1.04)^x$

We can plug in 3 for x since x is the # of years the account is receiving compound interest on.

$y = 1600(1.04)^3$

$1.04^3 = 1.124864\\1.124864*1600= 1800$

8 0

2 years ago

If Sally can paint a house in 4 hours, and John can paint the same house in 6 hour, how long will it take for both of them to pa

astraxan [27]

Sally can paint a house in 4 hours; Which means in 1 hour she can paint 1/4th of the house.

John can paint the same house in 6 hours. Which means in 1 hour she can paint 1/6th of the house.

Together in 1 hour they can paint: - 1/4 + 1/6 = 5/12th of the house.

Total Hours for painting the house together will be 12/5 = 2.4 Hours.

So the answer is 2 hours and 24 minutes

Please mark as brainliest

8 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