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2 years ago

5x-6y=3 7y=2x+8 Help me out please

Mathematics

1 answer:

zaharov [31]2 years ago

3 0

Answer:

A) 5x-6y=3

B) 7y=2x+8

B) -2x + 7y = 8 then multiplying "B" by 2.5

B) -5x +17.5y = 20 then adding this to A

A) 5x -6y = 3

11.5y = 23

y = 2

x = 3

Step-by-step explanation:

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pantera1 [17]

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2 years ago

Tony was Challenged to a half-court shooting competition. For every half-court shot that he makes, he will earn 20 points. For e

Mkey [24]

Answer:

Tony make 4 half-court shots and misses 16 half-court shots.

Step-by-step explanation:

Given:

Points earned for every successful, half-court shots = $20$

Points deducted for missing the shots = $5$ or $-5$

Total number of shots in the game = $20$

Final points earned by Tony = $0$

According to the question:

Let the number of successful half-court shot be $x$

And the number of shots missed be $20-x$

So,

In terms of points we can re-write it as:

⇒ $20(x)-5(20-x) =0$

⇒ $20x-100+5x=0$

⇒ $25x-100=0$

⇒ $25x-100+100=100$

⇒ $25x=100$

⇒ $\frac{25x}{25} =\frac{100}{25}$

⇒ $x=\frac{100}{25}$

⇒ $x=4$

Tony make 'x' shots that is 4 and the number of shot he missed is (20-x) =(20-4)=16

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2 years ago

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ohaa [14]

Answer:

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Step-by-step explanation:

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2 years ago

My brain is rotting and so is my mental state

geniusboy [140]

Answer:

I'd recommend the photo math app

Step-by-step explanation:

it's easy to use and it can help

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2 years ago

Read 2 more answers

Determine the values of the constants b and c so that the function given below is differentiable. f(x)={2xbx2+cxx≤1x>1

Lera25 [3.4K]

Assuming the function is

$f(x)=\begin{cases}2x&\text{for }x\le1\\bx^2+cx&\text{for }x>1\end{cases}$

For $f(x)$ to be differentiable, it necessarily has to be continuous. For this condition to be met, we need

$\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)=f(1)$
$\iff\displaystyle\lim_{x\to1}2x=\lim_{x\to1}(bx^2+cx)$
$\iff2=b+c$

For the derivative to exist, the one-sided limits of the derivative must also exist and be equal. We have

$f'(x)=\begin{cases}2&\text{for }x1\end{cases}$

$\displaystyle\lim_{x\to1^-}2=\lim_{x\to1^+}(2bx+c)$
$\iff2=2b+c$

Now we solve for $b$ and $c$ :

$\begin{cases}b+c=2\\2b+c=2\end{cases}\implies b=0,c=2$

5 0

3 years ago