All categories

3 years ago

Dan bought a stereo at a 15% discount. the original price was $350.00. how much money did Dan save?

Mathematics

2 answers:

Lemur [1.5K]3 years ago

7 0

Answer:

$52.5 saved. And he paid $297.50

Step-by-step explanation:

First finding the amount of money of the 15% discount

350.00 × .15 = $52.5

Then the the original price minus the dicount

350.00 - 52.50 = $297.50

siniylev [52]3 years ago

3 0

Answer:

he saved 52.5$

Step-by-step explanation:

350.00*15/100

=52.5

You might be interested in

Solve: 7.85 + q ≤ -18.15 *

Ilia_Sergeevich [38]

Answer:

q≤-26

Step-by-step explanation:

7.85+q≤-18.15

-7.85 -7.85

q≤-26

7 0

3 years ago

Real answers only pls! no links either!

MaRussiya [10]

Answer:

A. 4/8 + 2/4 =1 B.5/8 + 1/4 =0.875

C.6/8 + 3/4 =1.5 D.7/8 + 2/4 =1.375

3 0

2 years ago

What is the estimate of 979

saw5 [17]

I'm pretty sure that it would be 1,000...

4 0

3 years ago

Solve! Image attached! Geometry, dumb/incorrect answers will get account, and answer reported, please answer all the way through

neonofarm [45]

Answer:

+) m∠1 = m∠2 (vertical angles theorem)

+) m∠2 and ∠3 are supplementary (because a//b)

+)m∠3 = m∠4 (alternate interior angles theorem, c//d)

=> ∠1 and ∠4 are supplementary (substitution property of equality)

Step-by-step explanation:

5 0

3 years ago

A curve is given by y=(x-a)√(x-b) for x≥b, where a and b are constants, cuts the x axis at A where x=b+1. Show that the gradient

ankoles [38]

<u>Answer:</u>

A curve is given by y=(x-a)√(x-b) for x≥b. The gradient of the curve at A is 1.

<u>Solution:</u>

We need to show that the gradient of the curve at A is 1

Here given that ,

$y=(x-a) \sqrt{(x-b)}$ --- equation 1

Also, according to question at point A (b+1,0)

So curve at point A will, put the value of x and y

$0=(b+1-a) \sqrt{(b+1-b)}$

0=b+1-c --- equation 2

According to multiple rule of Differentiation,

$y^{\prime}=u^{\prime} y+y^{\prime} u$

so, we get

${u}^{\prime}=1$

$v^{\prime}=\frac{1}{2} \sqrt{(x-b)}$

$y^{\prime}=1 \times \sqrt{(x-b)}+(x-a) \times \frac{1}{2} \sqrt{(x-b)}$

By putting value of point A and putting value of eq 2 we get

$y^{\prime}=\sqrt{(b+1-b)}+(b+1-a) \times \frac{1}{2} \sqrt{(b+1-b)}$

$y^{\prime}=\frac{d y}{d x}=1$

Hence proved that the gradient of the curve at A is 1.

7 0

2 years ago