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3 years ago

What is equivalent to logbase2n=4?

Mathematics

1 answer:

Wittaler [7]3 years ago

5 0

I'm pretty sure its B (:

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Graph the function:<br> y = 5(3)

jonny [76]

Answer:

A straight horizontal line, with no slope, across the graph at 15 on the y-axis

Step-by-step explanation:

y = 5(3)

y = 15

4 0

2 years ago

En una competición participan 200 atletas, 2/5 son españoles, 1/3 del resto son franceses, 35 son alemanes y el resto son ingles

Bas_tet [7]

Answer:

7/40

Step-by-step explanation:

La fraccion de Alemanes son 35/200=7/40

6 0

2 years ago

Which equation describes a line that passes through (-5,-8) and is perpendicular to y= 5/4x + 10

lara31 [8.8K]

Answer: y = -4/5x - 12

1. First find slope. Since you are trying to find a line perpendicular to y = 5/4x + 10, the slope would be the negative reciprocal of 5/4x.

slope = - 4/5x

2. plug in (-5,-8) and the slope of -4/5 to the point slope form equation
(y-y2) = slope (x-x2)

= y+8=-4/5(x+5)

3. solve and simply y+8=-4/5(x+5)

y = -4/5x - 12 is your answer

8 0

2 years ago

What sum will amount to rupees 4000 in 3 years at 6% p.a. compound interest

natali 33 [55]

Answer:

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{purple}{Given:}}}}}}}\end{gathered}$

⇢ Principle = Rs.4000
⇢ Rate = 6%
⇢ Time = 3 year

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{purple}{To Find:}}}}}}}\end{gathered}$

⇢ Amount

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{purple}{Using Formula:}}}}}}}\end{gathered}$

${\dag{\underline{\boxed{\sf{Amount ={P{\bigg(1 + \dfrac{R}{100}{\bigg)}^{T}}}}}}}}$

$\dag{\underline{\boxed{\sf{Compound \: Interest = Amount- Principle }}}}$

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{purple}{Solution:}}}}}}}\end{gathered}$

${\bigstar \:{\underline{\pmb{\frak{\red{Firstly,Finding \: the \: Amount }}}}}}$

$\quad {:\implies{\sf{Amount = \bf{P{\bigg(1 + \dfrac{R}{100}{\bigg)}^{T}}}}}}$

Substituting the values

$\quad {:\implies{\sf{Amount = \bf{4000{\bigg(1 + \dfrac{6}{100}{\bigg)}^{3}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4000{\bigg(1 \times 100 + \dfrac{6}{100}{\bigg)}^{3}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4000{\bigg( \dfrac{100 + 6}{100}{\bigg)}^{3}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4000{\bigg( \dfrac{106}{100}{\bigg)}^{3}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4000{\bigg({\cancel{\dfrac{106}{100}}{\bigg)}}^{3}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4000{\bigg( \dfrac{53}{50}{\bigg)}^{3}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4000{\bigg( \dfrac{53}{50} \times \dfrac{53}{50} \times \dfrac{53}{50}{\bigg)}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4000{\bigg( \dfrac{148877}{125000}{\bigg)}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4000 \times \dfrac{148877}{125000}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4{\cancel{000}} \times \dfrac{148877}{125{\cancel{000}}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{\dfrac{148877 \times 4}{125}}}}}$

$\quad {:\implies{\sf{Amount = \bf{\dfrac{595508}{125}}}}}$

$\quad {:\implies{\sf{Amount = \bf{\cancel{\dfrac{595508}{125}}}}}}$

$\quad {:\implies{\sf{Amount = \bf{4764.064}}}}$

$\begin{gathered} \dag{\boxed{\textsf{\textbf{\underline{\color{green}{Amount = {Rs.4764.064}}}}}}}\end{gathered}$

Hence, The Amount is Rs.4764.064

$\begin{gathered}\end{gathered}$

${\bigstar \:{\underline{\pmb{\frak{\red{ Now,Finding \: The \: Compound \: Interest }}}}}}$

$\quad{: \implies{\sf{Compound \: Interest = \bf{Amount- Principle }}}}$

Substituting the values

$\quad{: \implies{\sf{Compound \: Interest = \bf{4764.064- 4000 }}}}$

$\quad{: \implies{\sf{Compound \: Interest =\bf{764.064}}}}$

$\begin{gathered} \dag{\boxed{\textsf{\textbf{\underline{\color{green}{Compound Interest = Rs.764.064}}}}}}\end{gathered}$

Henceforth,The Compound Interest is Rs.764064

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{purple}{Learn More:}}}}}}}\end{gathered}$

$\begin{gathered}\begin{gathered}\begin{gathered} \dag \: \underline{\bf{More \: Useful \: Formula}}\\ {\boxed{\begin{array}{cc}\dashrightarrow {\sf{Amount = Principle + Interest}} \\ \\ \dashrightarrow \sf{ P=Amount - Interest }\\ \\ \dashrightarrow \sf{ S.I = \dfrac{P \times R \times T}{100}} \\ \\ \dashrightarrow \sf{P = \dfrac{Interest \times 100 }{Time \times Rate}} \\ \\ \dashrightarrow \sf{P = \dfrac{Amount\times 100 }{100 + (Time \times Rate)}} \\ \end{array}}}\end{gathered}\end{gathered}\end{gathered}$

8 0

2 years ago

A car braked with a constant deceleration of 16ft/s2, producing skid marks measuring 200 feet before coming to a stop. How fast

Llana [10]

Answer:the car was traveling at a speed of 80 ft/s when the brakes were first applied.

Step-by-step explanation:

The car braked with a constant deceleration of 16ft/s^2. This is a negative acceleration. Therefore,

a = - 16ft/s^2

While decelerating, the car produced skid marks measuring 200 feet before coming to a stop.

This means that it travelled a distance,

s = 200 feet

We want to determine how fast the car was traveling (in ft/s) when the brakes were first applied. This is the car's initial velocity, u.

Since the car came to a stop, its final velocity, v = 0

Applying Newton's equation of motion,

v^2 = u^2 + 2as

0 = u^2 - 2 × 16 × 200

u^2 = 6400

u = √6400

u = 80 ft/s

7 0

2 years ago