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3 years ago

An item on sale costs 80% of the original price. If the original price was $95, what is the sale price?

Mathematics

2 answers:

WITCHER [35]3 years ago

7 0

Answer:

19$

Step-by-step explanation:

den301095 [7]3 years ago

3 0

Answer:

$\frac{selling \: price \: - cost \: price}{cost \: price} \times 100 \\ \frac{sp - 95}{95} \times 100 = 80 \\ open \: the \: bracket \: \\ \frac{100sp - 9500}{95} = 80 \\ 100sp - 9500 = 7600 \\ 100sp = 7600 + 9500 \\ 100sp = 17100 \\ sp = 171$

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The mean salary if of 5 employees is $35900. the median is $37000. the mode is $382000. If the median payed employee gets a $310

mr Goodwill [35]

Answer:

Step-by-step explanation:

New median:40100

New mode:385100

5 0

3 years ago

This isosceles triangle has two sides of equal length , a, that are longer than the length of the base , b. The perimeter of the

Sholpan [36]

We are give the equation of the perimeter of the triangle as follows:
2a + b = 15.7
where b represents the base.

Now, if we want to calculate the length of the base, all we have to do is isolate the b in one side of the equation as follows:
b = 15.7 - 2a

We know that a = 6.3 cm, therefore, the length of the base can be calculated as follows:
b = 15.7 - 2(6.3) = 3.1 cm

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3 years ago

Find the area of the equilateral triangle

uranmaximum [27]

Answer:

Step-by-step explanation:

Suppose that the side length is s. Imagine that one side is level. Draw a vertical line through the top vertex perpendicular to the base (level side). Label this vertical line "h." Then h and s are related as follows:

√3 h

sin 60 degrees = ------- = --------

2 s

s√3

and so h = the height of the triangle = -----------

The area of this triangle is (1/2)(base)(height), which here is:

(1/2)(s/2)( s√3 /2), or

(s^2)√3

A = ---------------

8 0

3 years ago

Read 2 more answers

Please show me how to solve the initial value problem <br> y'=tanx y(pi/4)=3

AlexFokin [52]

The ODE is separable, i.e. you can write

$\dfrac{\mathrm dy}{\mathrm dx}=\tan x\iff\mathrm dy=\tan x\,\mathrm dx$

Integrating both sides gives the general solution.

$\displaystyle\int\mathrm dy=\int\tan x\,\mathrm dx$
$y=-\ln|\cos x|+C$

Given that $y\left(\dfrac\pi4\right)=3$ , we have

$3=-\ln\left|\cos\dfrac\pi4\right|+C$
$3=-\ln\dfrac1{\sqrt2}+C$
$3-\ln\sqrt2=C$

and so the particular solution to the IVP is

$y=-\ln|\cos x|+3-\ln\sqrt2$
$y=3-\ln|\sqrt2\cos x|$

4 0

3 years ago

S_A_V [24]

The corect answer for this question

6 0

2 years ago