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3 years ago

Can someone please help me on this?

Mathematics

1 answer:

Shtirlitz [24]3 years ago

7 0

Answer:

x = 7.5

Step-by-step explanation:

4.5 (1/3) = 1.5

5 (1.5) = 7.5

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Find a parabola with equation y = ax2 + bx + c that has slope 5 at x = 1, slope −11 at x = −1, and passes through the point (2,

azamat

By "slope" I assume you mean slope of the tangent line to the parabola.

For any given value of x, the slope of the tangent to the parabola is equal to the derivative of y :

$y=ax^2+bx+c\implies y'=2ax+b$

The slope at x = 1 is 5:

$2a+b=5$

The slope at x = -1 is -11:

$-2a+b=-11$

We can already solve for a and b :

$\begin{cases}2a+b=5\\-2a+b=-11\end{cases}\implies 2b=-6\implies b=-3$

$2a-3=5\implies 2a=8\implies a=4$

Finally, the parabola passes through the point (2, 18); that is, the quadratic takes on a value of 18 when x = 2:

$4a+2b+c=18\implies2(2a+b)+c=10+c=18\implies c=8$

So the parabola has equation

$\boxed{y=4x^2-3x+8}$

6 0

3 years ago

What is bigger 7/8 or 14/15

GalinKa [24]

14/15 is .93333333333333333 and 7/8 is .875 so 14/15 is
bigger

8 0

4 years ago

Solve the given inequality :

tigry1 [53]

Answer:

x < -5 or x = 1 or 2 < x < 3 or x > 3

Step-by-step explanation:

Given rational inequality:

$\dfrac{(x-1)^2(x-2)^3}{(x^2-5x+6)^2(x+5)}\geq 0$

$\textsf{Factor }(x^2-5x+6):$

$\implies x^2-2x-3x+6$

$\implies x(x-2)-3(x-2)$

$\implies (x-3)(x-2)$

Therefore:

$\dfrac{(x-1)^2(x-2)^3}{(x-3)^2(x-2)^2(x+5)}\geq 0$

Find the roots by solving f(x) = 0 (set the numerator to zero):

$\implies (x-1)^2(x-2)^3=0$

$\implies (x-1)^2=0\implies x=1$

$\implies (x-2)^3=0 \implies x=2$

Find the restrictions by solving f(x) = undefined (set the denominator to zero):

$\implies (x-3)^2(x-2)^2(x+5)=0$

$\implies (x-3)^2=0 \implies x=3$

$\implies (x-2)^2=0 \implies x=2$

$\implies (x+5)=0 \implies x=-5$

Create a sign chart, using closed dots for the roots and open dots for the restrictions (see attached).

Choose a test value for each region, including one to the left of all the critical values and one to the right of all the critical values.

Test values: -6, 0, 1.5, 2.5, 4

For each test value, determine if the function is positive or negative:

$f(-6)=\dfrac{(-6-1)^2(-6-2)^3}{(-6-3)^2(-6-2)^2(-6+5)}=\dfrac{(+)(-)}{(+)(+)(-)}=+$

$f(0)=\dfrac{(0-1)^2(0-2)^3}{(0-3)^2(0-2)^2(0+5)}=\dfrac{(+)(-)}{(+)(+)(+)}=-$

$f(1.5)=\dfrac{(1.5-1)^2(1.5-2)^3}{(1.5-3)^2(1.5-2)^2(1.5+5)}=\dfrac{(+)(-)}{(+)(+)(+)}=-$

$f(2.5)=\dfrac{(2.5-1)^2(2.5-2)^3}{(2.5-3)^2(2.5-2)^2(2.5+5)}=\dfrac{(+)(+)}{(+)(+)(+)}=+$

$f(4)=\dfrac{(4-1)^2(4-2)^3}{(4-3)^2(4-2)^2(4+5)}=\dfrac{(+)(+)}{(+)(+)(+)}=+$

Record the results on the sign chart for each region (see attached).

As we need to find the values for which f(x) ≥ 0, shade the appropriate regions (zero or positive) on the sign chart (see attached).

Therefore, the solution set is:

x < -5 or x = 1 or 2 < x < 3 or x > 3

As interval notation:

$(- \infty,-5) \cup x=1 \cup (2,3) \cup(3,\infty)$

4 0

2 years ago

El valor que falta en la ecuacion 3+×=10

grandymaker [24]

3 + x = 10
x = 10 -3
x = 7

5 0

3 years ago

An isosceles triangle has an angle that measures 116.6°. Which other angles could be in that isosceles triangle? Choose all that

katrin2010 [14]

D is wrong. It can't have 2 angles equal to or greater than 90.

2x + 116.6 = 180
2x = 180 - 116.6
2x = 63.4
x = 63.4/2
x = 31.7 Two of this size are the only possible answers.

6 0

4 years ago