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lianna [129]
3 years ago
14

If Ben bikes 3 miles and then bikes 1,000 yards back toward his starting point, how many yards is Ben from his starting point?

Mathematics
1 answer:
WINSTONCH [101]3 years ago
7 0
1 mile = 1760 yds...so 3 miles is (3 * 1760) = 5280 yds

so if he bikes 5280 yds...and then bikes 1000 yds back....then he would be (5280 - 1000) = 4280 yds from the starting point 
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Kyle's car repair costs $1000. He has 2 coupons which could save him $80 off of $1000, or 15% off of $1000. EXPLAIN WHICH COUPON
never [62]
Answer:

Kyle would save more using his 15% off of $1000 coupon.

Explanation:

15 ÷ 100 • 1000 = 150

1000 - 150 = $850

$150 is the amount the 15% coupon took off.

$850 is the amount Kyle will pay.

vs.

1000 - 80 = $920
6 0
3 years ago
When y=7.5 x=2 what is the value of y when x is 2 1/4
swat32
<h2>Answer:</h2>

The value of y is 8.4375

<h2>Step-by-step explanation:</h2><h3>Known :</h3>
  • y = 7.5
  • x = 2

<h3>Asked :</h3>
  • The value of y when x is 2.25

<h3>Solution :</h3>

2/7.5 = 2.25/y

Do a cross multiplication,

2/7.5 = 2.25/y

=> 20/75 = 2.25/y

=> 75 . 2.25 = 20y

=> 168.75 = 20y

Reverse the equation,

168.75 = 20y

=> 20y = 168.75

Find the value of y,

20y = 168.75

=> y = 8.4375

<h3>Conclusion :</h3>

y = 8.4375

7 0
2 years ago
The Gordon family plans to buy a TV. One TV has a purchase price of $330 and an estimated yearly operating cost of $14. The othe
Ann [662]
Answer:
The second tv option (cheaper by $1)

Step by step explanation:
To find how much the first tv will cost all together complete the equation:
$330 + $14 * 8 = $442
Second tv cost:
$369 + $9 * 8 = $441

Hope this helps
Pls give me the brainliest
5 0
3 years ago
Consider the function g(x) = (x-e)^3e^-(x-e). Find all critical points and points of inflection (x, g(x)) of the function g.
Elden [556K]

Answer:

The answer is "cirtical\  points \ (x,g(x))\equiv  (e,0),(e+3,\frac{27}{e^3})"

Step-by-step explanation:

Given:

g(x) = (x-e)^3e^{-(x-e)}

Find critical points:

g(x) = (x-e)^3e^{(e-x)}

differentiate the value with respect of x:

\to g'(x)= (x-e)^3 \frac{d}{dx}e^{e-r} +e^{e-r}  \frac{d}{dx}(x-e)^3=(x-e)^2 e^{(e-x)} [-x+e+3]

critical points g'(x)=0

\to (x-e)^2 e^{(e-x)} [e+3-x]=0\\\\\to e^{(e-x)}\neq 0 \\\\\to (x-e)^2=0\\\\ \to [e+3-x]=0\\\\\to x=e\\\\\to x=e+3\\\\\to x= e,e+3

So,

The critical points of (x,g(x))\equiv  (e,0),(e+3,\frac{27}{e^3})

7 0
3 years ago
Part A
Liono4ka [1.6K]

A

Because i did the math and i had this question before

3 0
3 years ago
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