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3 years ago

The price of an item has been reduced by 65% the original price was $95

Mathematics

1 answer:

EastWind [94]3 years ago

6 0

Difference is 61.75 and the answer is 33.25 for the total price after reduction

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Given that rectangle MNOP~ rectangle STUV~, what is the length of ___<br> TU

IrinaK [193]

A parallelogram in which adjacent sides are perpendicular to each other is called a rectangle. The value of x or the length of TU is 22.5.

<h3>What is a rectangle?</h3>

That parallelogram in which adjacent sides are perpendicular to each other is called a rectangle. A rectangle is always a parallelogram and a quadrilateral but the reverse statement may or may not be true.

Since all the rectangles are similar, therefore, the corresponding sides of the rectangle will be in ratio. Therefore,

$\dfrac{\text{Length of STUV}}{\text{Length of MNOP}} = \dfrac{9}{6}$

Similarly, the breadth will be,

$\dfrac{\text{Width of STUV}}{\text{Width of MNOP}} = \dfrac{\text{Length of STUV}}{\text{Length of MNOP}}\\\\\\\dfrac{\text{Width of STUV}}{\text{Width of MNOP}} = \dfrac{9}{6}\\\\\\\dfrac{x}{15}=\dfrac96\\\\\\x = \dfrac{9 \times 15}{6} = 22.5$

Hence, the value of x or the length of TU is 22.5.

Learn more about Rectangle:

brainly.com/question/15019502

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5 0

2 years ago

Work out the area of this circle.

exis [7]

We know that

Area of circle = πr²

=> π × (6)²

=> π × 36

=> 36π mm²

3 0

3 years ago

Read 2 more answers

I need help I'm so confused

borishaifa [10]

A = L x W = 8 x 17 = 136
answer
C. 136 sq. units

8 0

3 years ago

Read 2 more answers

In quadratic drag problem, the deceleration is proportional to the square of velocity

Mars2501 [29]

Part A

Given that $a= \frac{dv}{dt} =-kv^2$

Then,

$\int dv= -kv^2\int dt \\ \\ \Rightarrow v(t)=-kv^2t+c$

For $v(0)=v_0$ , then

$v(0)=-kv^2(0)+c=v_0 \\ \\ \Rightarrow c=v_0$

Thus, $v(t)=-kv(t)^2t+v_0$

For $v(t)= \frac{1}{2} v_0$ , we have

$\frac{1}{2} v_0=-k\left( \frac{1}{2} v_0\right)^2t+v_0 \\ \\ \Rightarrow \frac{1}{4} kv_0^2t=v_0- \frac{1}{2} v_0= \frac{1}{2} v_0 \\ \\ \Rightarrow kv_0t=2 \\ \\ \Rightarrow t= \frac{2}{kv_0}$

Part B

Recall that from part A,

$v(t)= \frac{dx}{dt} =-kv^2t+v_0 \\ \\ \Rightarrow dx=-kv^2tdt+v_0dt \\ \\ \Rightarrow\int dx=-kv^2\int tdt+v_0\int dt+a \\ \\ \Rightarrow x=- \frac{1}{2} kv^2t^2+v_0t+a$

Now, at initial position, t = 0 and $v=v_0$ , thus we have

$x=a$

and when the velocity drops to half its value, $v= \frac{1}{2} v_0$ and $t= \frac{2}{kv_0}$

Thus,

$x=- \frac{1}{2} k\left( \frac{1}{2} v_0\right)^2\left( \frac{2}{kv_0} \right)^2+v_0\left( \frac{2}{kv_0} \right)+a \\ \\ =- \frac{1}{2k} + \frac{2}{k} +a$

Thus, the distance the particle moved from its initial position to when its velocity drops to half its initial value is given by

$- \frac{1}{2k} + \frac{2}{k} +a-a \\ \\ = \frac{2}{k} - \frac{1}{2k} = \frac{3}{2k}$

7 0

3 years ago

I have to do Oder of operation and I need to show my work for all 10 questions so can someone help me?

Feliz [49]

Answer:

Order of Operations:

Parenthesis

Exponents

Multiplication

Division

Addition

Subtraction

Step-by-step explanation:

3 0

3 years ago