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3 years ago

Logarithm

ign="absmiddle" class="latex-formula">

Mathematics

1 answer:

Genrish500 [490]3 years ago

8 0

Answer:

log(53.75) ≈ 1.73

Step-by-step explanation:

Not sure if this was a question but I just put it into my calculator.

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HELP!<br> What is the missing term? What is the rule? 8 9/10, 7 7/10, 6 1/2 , _________

KengaRu [80]

Answer:

5 3/10

Step-by-step explanation:

You subtract 1 1/5 from the number (n).

(In equation) x = n - 1 1/5

5 0

3 years ago

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Brady made a scale drawing of a rectangular swimming pool on a coordinate grid. The points (-20, 25), (30, 25), (30, -10) and (-

djverab [1.8K]

Answer:

Length = 50 units

width = 35 units

Step-by-step explanation:

Let A, B, C and D be the corner of the pools.

Given:

The points of the corners are.

$A(x_{1}, y_{1}})=(-20, 25)$

$B(x_{2}, y_{2}})=(30, 25)$

$C(x_{3}, y_{3}})=(30, -10)$

$D(x_{4}, y_{4}})=(-20, -10)$

We need to find the dimension of the pools.

Solution:

Using distance formula of the two points.

$d(A,B)=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ ----------(1)

For point AB

Substitute points A(30, 25) and B(30, 25) in above equation.

$AB=\sqrt{(30-(-20))^{2}+(25-25)^{2}}$

$AB=\sqrt{(30+20)^{2}}$

$AB=\sqrt{(50)^{2}$

AB = 50 units

Similarly for point BC

Substitute points B(-20, 25) and C(30, -10) in equation 1.

$d(B,C)=\sqrt{(x_{3}-x_{2})^{2}+(y_{3}-y_{2})^{2}}$

$BC=\sqrt{(30-30)^{2}+((-10)-25)^{2}}$

$BC=\sqrt{(-35)^{2}}$

BC = 35 units

Similarly for point DC

Substitute points D(-20, -10) and C(30, -10) in equation 1.

$d(D,C)=\sqrt{(x_{3}-x_{4})^{2}+(y_{3}-y_{4})^{2}}$

$DC=\sqrt{(30-(-20))^{2}+(-10-(-10))^{2}}$

$DC=\sqrt{(30+20)^{2}}$

$DC=\sqrt{(50)^{2}}$

DC = 50 units

Similarly for segment AD

Substitute points A(-20, 25) and D(-20, -10) in equation 1.

$d(A,D)=\sqrt{(x_{4}-x_{1})^{2}+(y_{4}-y_{1})^{2}}$

$AD=\sqrt{(-20-(-20))^{2}+(-10-25)^{2}}$

$AD=\sqrt{(-20+20)^{2}+(-35)^{2}}$

$AD=\sqrt{(-35)^{2}}$

AD = 35 units

Therefore, the dimension of the rectangular swimming pool are.

Length = 50 units

width = 35 units

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3 years ago

Divide 2/8 by 9/18 . Input your answer as a reduced fraction.\

Katen [24]

$\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{\sf{\frac{2}{8}\div\frac{9}{18} }} \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf Simplify \ \frac{2}{8} \ to \ \frac{1}{4}. \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{\sf{\frac{1}{4}\div\frac{18}{9} }} \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf Use \ this \ rule: a\div \frac{b}{c}=a\times \frac{c}{b}a \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{\sf{\frac{1}{4}\times\frac{18}{9} }} \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf Use \ this \ rule: \frac{a}{b} \times \frac{c}d{}=\frac{ac}{bd}. \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{\sf{\frac{1\times18}{4\times9} \ \to \ \ Multiply }} \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{\sf{ \frac{18}{36}= }}\boxed{\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{\sf{\frac{1}{2} }} \end{gathered}$}} \end{gathered}$}$

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2 years ago

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Please help me with this math question!

aivan3 [116]

If you do what part B says for the first two equations in part A, it will answer part A.
Part C: You plug in coordinates on a table and see which points match and which points don't fit in the equations

6 0

4 years ago

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The height meters of a mass projected vertically upwards at a time seconds s=ut-0.5gt².

Sergeu [11.5K]

9514 1404 393

Answer:

t ≈ 0.590 s (ascent); 5.532 s (descent)

Step-by-step explanation:

We are interested in the values of t when s=16.

s = 30t -4.9t²

4.9t^2 -30t +16 = 0 . . . . . substitute 16 for s; put in standard form

The quadratic formula can be used to find the solutions:

t = (-(-30) ±√((-30)² -4(4.9)(16)))/(2(4.9))

t = (30 ±√586.4)/9.8 ≈ 0.59023, 5.53221 . . . . seconds after launch

a) It will take 0.590 seconds to reach 16 m height initially.

b) It will take 5.532 seconds to return to 16 m height on descent.

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3 years ago