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3 years ago

10

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1 answer:

zvonat [6]3 years ago

3 0

-2(3y = -x/2 + 2)
+ y = -x + 9
---------------------
-5y = 0 + 5

y = -1
x = 10

Answer A

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Instructions:Drag the tiles to the correct boxes to complete the pairs. Match each polynomial function with one of its factors.

Svetach [21]

We have the following functions:
f (x) = x3 - 3x2 - 13x + 15
f (x) = x4 + 3x3 - 8x2 + 5x - 25
f (x) = x3 - 2x2 - x + 2
f (x) = -x3 + 13x - 12
Factoring we have:
f (x) = x3 - 3x2 - 13x + 15 -------> x + 3
f (x) = x4 + 3x3 - 8x2 + 5x - 25 -> x + 5
f (x) = x3 - 2x2 - x + 2 ------------> x - 2
f (x) = -x3 + 13x - 12 -------------> x + 4

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4 years ago

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it's a shame 1 hour and 45 minutes to pick 3 1/2 pounds of raspberry at what unit rate did she pick raspberries

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Answer:2 pounds per hour

Step-by-step explanation:

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3 years ago

The price of a pair of shoes increases from $64 to $66. What is the percent increase to the nearest percent? The percent increas

Strike441 [17]

Answer:

Percent symbol 2 I think

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3 years ago

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A steady wind blows a kite due west. The kite's height above ground from horizontal position x = 0 to x = 80 ft is given by y =

GaryK [48]

Answer:

The distance travelled by the kite is 122.8 ft ( approx )

Step-by-step explanation:

Here, the given function,

$y=150-\frac{1}{40}(x-50)^2$

Differentiating with respect to x,

$y'=-\frac{1}{20}(x-50)$

∵ arc length of a curve is,

$L=\int_{a}^{b} \sqrt{1+y'^2}dx$

Where, y shows the height of the curve for a ≤ x ≤ b,

Thus, the arc length of the given curve is,

$L=\int_{0}^{80} \sqrt{1+(-\frac{1}{20}(x-50)^2}dx$

Put $-\frac{1}{20}(x-50)=tan\theta$

$\implies -dx=-20 sec^2\theta d\theta$

$\implies L=-20\int_{0}^{80} \sqrt{1+tan^2\theta}sec^2\theta d\theta$

$=-20\int_{0}^{80} (sec \theta ) sec^2\theta d\theta$

$=-20\int_{0}^{80} (sec \theta ) sec^2\theta d\theta$

By integration by parts,

$=|-\frac{20}{2}(sec \theta tan\theta +ln|sec\theta +tan\theta |) |^{x=80}_{x=0}$

If x = 80, $tan \theta = -\frac{1}{20}(30-50)=\frac{3}{2}$

$sec \theta = \frac{\sqrt{13}}{2}$

$\implies \theta = \frac{1}{20}(0-50)=\frac{5}{2}$

$sec \theta = \frac{\sqrt{29}}{2}$

Thus, the length of the curve is,

$=-10(\frac{\sqrt{13}}{2}(-\frac{3}{2}) +ln|\frac{13}{2}-\frac{3}{2}|) + 10(\frac{5\sqrt{29}}{4} + ln |\frac{29}{2} + \frac{5}{2} |)$

$\approx 122.8\text{ feet}$

8 0

3 years ago

Two savings accounts earn simple interest. Account A has a beginning balance of $500 and grows by $25 per year. Account B has a

GalinKa [24]

Answer:

After 25 years both the accounts will have the same balance.

Step-by-step explanation:

Let t years, both the account will have the same balance.

As account A has a beginning balance of $500 and grows by $25 per year.

So, the total balance of account A after t years = 500+25t

Similarly, account B has a beginning balance of $750 and grows by $15 per year.

So, the total balance of account A after t years = 750+15t

As both the accounts has the same balance, so

500+25t= 750+15t

25t-15t=750-500

10t=250

t=250/10

t=25 years

Hence, after 25 years both the accounts will have the same balance.

5 0

3 years ago