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2 years ago

Their corresponding helpppopooequivalent expressions, Help help

Mathematics

2 answers:

Helga [31]2 years ago

7 0

YOU PLAY ROBLÓX I PLAY TOOOOOO

Harrizon [31]2 years ago

4 0

I think it’s the 4th option lmk if I’m wrong

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Here is a sketch of y=x^2+bx+c The curve intersects

atroni [7]

Answer:

y = 18 and x = -2

Step-by-step explanation:

y = x^2+bx+c To find the turning point, or vertex, of this parabola, we need to work out the values of the coefficients b and c. We are given two different solutions of the equation. First, (2, 0). Second, (0, -14). So we have a value (-14) for c. We can substitute that into our first equation to find b. We can now plug in our values for b and c into the equation to get its standard form. To find the vertex, we can convert this equation to vertex form by completing the square. Thus, the vertex is (4.5, –6.25). We can confirm the solution graphically Plugging in (2,0) :

y=x2+bx+c

0=(2)^2+b(2)+c

y=4+2b+c

-2b=4+c

b=-2+2c

Plugging in (0,−14) :

y=x2+bx+c

−14=(0)2+b(0)+c

−16=0+b+c

b=16−c

Now that we have two equations isolated for b , we can simply use substitution and solve for c . y=x2+bx+c 16 + 2 = y y = 18 and x = -2

4 0

2 years ago

The length of the rectangle is x cm. The length of a diagonal of the rectangle is 8 crn. The perimeter of the rectangle is 20 cm

klemol [59]

Answer:

$x^2-10x+18=0$

a = -10 and b = 18.

Step-by-step explanation:

Let w represent the width of the rectangle.

We are given that the perimeter of the rectangle is 20 cm, this means that:

$20=2(x + w)$

Let's put w in terms of x. Divide both sides by two:

$10=x+w$

And solve for w:

$w=10-x$

So, the rectangle measures x by (10 - x) cm.

According to the Pythagorean Theorem:

$a^2+b^2=c^2$

a and b are the legs and c is the hypotenuse.

Substitute x for a, w for b, and 8 for c:

$x^2+w^2=8^2$

Simplify and substitute:

$x^2+(10-x)^2=64$

Square:

$x^2+(100-20x+x^2)=64$

Isolate the equation. So:

$2x^2-20x+36=0$

Since the leading coefficient is one, divide both sides by two:

$x^2-10x+18=0$

Therefore, a = -10 and b = 18.

6 0

2 years ago

For the following telescoping series, find a formula for the nth term of the sequence of partial sums

gtnhenbr [62]

I'm guessing the sum is supposed to be

$\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}$

Split the summand into partial fractions:

$\dfrac1{(5k-1)(5k+4)}=\dfrac a{5k-1}+\dfrac b{5k+4}$

$1=a(5k+4)+b(5k-1)$

If $k=-\frac45$ , then

$1=b(-4-1)\implies b=-\frac15$

If $k=\frac15$ , then

$1=a(1+4)\implies a=\frac15$

This means

$\dfrac{10}{(5k-1)(5k+4)}=\dfrac2{5k-1}-\dfrac2{5k+4}$

Consider the $n$ th partial sum of the series:

$S_n=2\left(\dfrac14-\dfrac19\right)+2\left(\dfrac19-\dfrac1{14}\right)+2\left(\dfrac1{14}-\dfrac1{19}\right)+\cdots+2\left(\dfrac1{5n-1}-\dfrac1{5n+4}\right)$