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3 years ago

11

Which technique is most appropriate to solve this equation?

1 answer:

Free_Kalibri [48]3 years ago

4 0

quadratic formula

Step-by-step explanation:

x(x-9)+4(x-9)

x²-9x+4x-36

x²-5x-36

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Arte-miy333 [17]

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if you have a bad credit score it can increase your debt ratio

Step-by-step explanation:

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3 years ago

Assume that a simple random sample has been selected from a normally distributed population and test the given claim. Identify t

CaHeK987 [17]

Answer:

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i was thinking the same thing thank you

4 0

3 years ago

Ray has found that his new car gets 31 miles per gallon on the highwayand 26 miles per gallon in the city. He recently drove 285

Mkey [24]

Answer:

$Highway = 155\ miles$

$City = 130\ miles$

Step-by-step explanation:

Given

$Highway = 31mi/gallon$

$City = 26mi/gallon$

$Total\ Miles = 285$

$Total\ Gallons = 10$

Required

Determine the number of miles driven on the highway and on the city

Represent the gallons used on highway with h and on city with c.

So, we have:

$c + h = 10$ ---- gallons used

and

$31h + 26c = 285$ --- distance travelled

In the first equation, make c the subject

$c = 10 - h$

Substitute 10 - h for c in the second equation

$31h + 26c = 285$

$31h + 26(10 - h) = 285$

Open bracket

$31h + 260 - 26h = 285$

Collect like terms

$31h - 26h = 285 - 260$

$5h = 25$

Make h the subject

$h = \frac{25}{5}$

$h = 5$

Substitute 5 for h in $c = 10 - h$

$c = 10 - 5$

$c = 5$

If on the highway, he travels 31 miles per gallon, then his distance on the highway is:

$Highway = 31 * 5$

$Highway = 155\ miles$

If in the highway, he travels 26 miles per gallon, then his distance on the highway is:

$City = 26 * 5$

$City = 130\ miles$

8 0

3 years ago

josiah is taking his friends to the movies each ticket cost $10, and popcorn is $5 a bag. there is a 3$ service fee for the enti

musickatia [10]

X = total amount paid
y = number of tickets
z = number of bags

your equation:
x = 10y + 5z + 3

hope this helps

4 0

4 years ago

Read 2 more answers

Lim x--> 0 (e^x(sinx)(tax))/x^2

Tom [10]

Make use of the known limit,

$\displaystyle\lim_{x\to0}\frac{\sin x}x=1$

We have

$\displaystyle\lim_{x\to0}\frac{e^x\sin x\tan x}{x^2}=\left(\lim_{x\to0}\frac{e^x}{\cos x}\right)\left(\lim_{x\to0}\frac{\sin^2x}{x^2}\right)$

since $\tan x=\dfrac{\sin x}{\cos x}$ , and the limit of a product is the same as the product of limits.

$\dfrac{e^x}{\cos x}$ is continuous at $x=0$ , and $\dfrac{e^0}{\cos 0}=1$ . The remaining limit is also 1, since

$\displaystyle\lim_{x\to0}\frac{\sin^2x}{x^2}=\left(\lim_{x\to0}\frac{\sin x}x\right)^2=1^2=1$

so the overall limit is 1.

4 0

3 years ago