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3 years ago

Help me cuz yah gurl gotta test UwU

Mathematics

1 answer:

Hitman42 [59]3 years ago

4 0

Answer:

got u

Step-by-step explanation:

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Mr. Lewis designed a circular walkway in

NARA [144]

Answer:

The diameter is equal to the circumference divided by the number pi

Step-by-step explanation:

we know that

The circumference of the circular walkway is equal to

$C=\pi D$

where

D is the diameter

we have

$D=6\ ft$

substitute

$C=\pi (6)$

$C=6\pi\ ft$

Divide the diameter by the circumference

$\frac{D}{C}=\frac{6}{6\pi}$

$\frac{D}{C}=\frac{1}{\pi}$

$D=\frac{C}{\pi}$

The diameter is equal to the circumference divided by the number pi

8 0

3 years ago

How is the product of two and -5 shown on a number line

tatuchka [14]

The correct answer is D

3 0

3 years ago

Read 2 more answers

the the circle shape part of the necklace above has a diameter of 5 mm the circumfrence of the circle-shaped part of each of the

tankabanditka [31]

7.855 mm maybe the answer unless rounded to the tenths place to be 7.9 mm.

8 0

4 years ago

Read 2 more answers

D)) The ratio of the monthly income and expenditure of Sunayana is 5 : 3. If she

Gnom [1K]

Answer:

Sunayana's income is ₹25000 and expenditures are ₹15000.

Step-by-step explanation:

Let i denote the monthly income and e denote the expenditures.

We know that the ratio of the monthly income to expenditure is 5:3. So, we can write the following proportion:

$\frac{i}{e}=\frac{5}{3}$

Let's multiply both sides by e. This yields:

$i=\frac{5}{3}e$

We know that when the income is <em>increased</em> by 5000 and the expenditures are <em>decreased </em>by 3000, the new ratio is 5:2. So, we can write the following proportion:

$\frac{i+5000}{e-3000}=\frac{5}{2}$

Let's multiply both sides by $(e-3000)$ :

$i+5000=\frac{5}{2}(e-3000)$

Since we know that $i=\frac{5}{3}e$ , substitute:

$\frac{5}{3}e+5000=\frac{5}{2}(e-3000)$

So, let's solve for the expenditures. Distribute the right:

$\frac{5}{3}e+5000=\frac{5}{2}e-7500$

Subtract $\frac{5}{2}e$ from both sides:

$-\frac{5}{6}e+5000=-7500$

Subtract 5000 from both sides:

$-\frac{5}{6}e=-12500$

Multiply both sides by -6/5. So, the expenditures are:

$e=\text{Rs }15000$

We can use the original ratio to find Sunayana's income:

$i=\frac{5}{3}e$

Substitute 15000 for e. Evaluate:

$i=\frac{5}{3}(15000)=\text{Rs }25000$

So, Sunayana's income is ₹25000 and expenditures are ₹15000.

And we're done!

Edit: Wrong currency, sorry about that!

3 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