Can someone help me with this

When asked to factor the x^2-121 a student gives the answer (x-11)(x-11) what is wron with this answer?

Art [367]

Given expression is x² - 121

This is called Difference of Squared terms, we have a formula that is given as :-

a² - b² = (a - b) · (a + b)

Now using the above formula in the given expression, we get :-

x² - 121

x² - (11)²

here a = x and b = 11

x² - (11)² = (x - 11) · (x + 11)

but it says that student gave the answer as (x - 11) · (x - 11).

So, student's answer should be (x - 11) · (x + 11) instead of product of two (x - 11) terms.

7 0

3 years ago

Read 2 more answers

Find the quotient of (11 ^ 18)/(11 ^ 3)

astraxan [27]

Answer:

The first choice, $11^{15}$ .

Step-by-step explanation:

The numerator, $11^{18}$ , is equivalent to eighteen " $11$ " multiplied together:

$11^{18} = \underbrace{11 \times 11 \times 11 \times 11 \times \cdots \times 11 \times 11}_{\text{$18$ in total}}$ .

On the other hand, the denominator, $11^{3}$ , is equivalent to three " $11$ " multiplied with one another:

$11^{3} = 11 \times 11 \times 11$ .

Dividing $11^{18}$ by $11^{3}$ would eliminate three " $11$ " from the numerator:

$\begin{aligned}& \frac{11^{18}}{11^{3}} \\ =\; & \frac{(11 \times 11 \times 11) \times \overbrace{11 \times \cdots \times 11}^{\text{$(18 - 3)$ in total}}}{(11 \times 11 \times 11)} \\ =\; & \overbrace{11 \times \cdots \times 11}^{\text{$15$ in total}} \\ =\; & 11^{15}\end{aligned}$ .

In general, dividing an expression by $y^{b}$ ( $y \ne 0$ ) is equivalent to multiplying that expression by $y^{-b}$ .

For example, in this question, dividing $11^{18}$ by $11^{3}$ would be equivalent to multiplying $11^{18}\!$ by $11^{-3}$ . In other words:

$\begin{aligned}& \frac{11^{18}}{11^{3}} \\ =\; & 11^{18} \times 11^{-3} \\ =\; & 11^{18 - 3} \\ =\; & 11^{15}\end{aligned}$ .

6 0

3 years ago

A student stands 20 m away from the foot

Ostrovityanka [42]

Answer:

Height of tree is $\approx$ 15 m.

Step-by-step explanation:

Given that student is 20 m away from the foot of tree.

and table is 1.5 m above the ground.

The angle of elevation is: 34°28'

Please refer to the attached image. The given situation can be mapped to a right angled triangle as shown in the image.

AB = CP = 20 m

CA = PB = 1.5 m

$\angle C$ = 34°28' = 34.46°

To find TB = ?

we can use trigonometric function tangent to find TP in right angled $\triangle TPC$

$tan \theta = \dfrac{Perpendicular}{Base}\\tan C= \dfrac{PT}{PC}\\\Rightarrow tan 34.46^\circ = \dfrac{PT}{20}\\\Rightarrow PT = 20 \times 0.686 \\\Rightarrow PT = 13.72\ m$

Now, adding PB to TP will give us the height of tree, TB

Now, height of tree TB = TP + PB

TB = 13.72 + 1.5 = 15.22 $\approx$ 15 m

7 0

4 years ago

How many real number solutions does the equation have 0=5x^2+2x-12

Harman [31]

Rearrange the equation to standard form of a quadratic equation (ax^2+bx+c=0) by switching sides: 5x^2+2x-12=0. Now, use the quadratic equation formula to solve. You should come out with x_1=sqrt61-1/5 and x_2=-1+sqrt61/5. Thus, your answer is B, or two solutions.

7 0

3 years ago

Read 2 more answers

Let f(x)=x^3-x-1

alexira [117]

$f(x) = x^3 - x - 1$

To find the gradient of the tangent, we must first differentiate the function.

$f'(x) = \frac{d}{dx}(x^3 - x - 1) = 3x^2 - 1$

The gradient at x = 0 is given by evaluating f'(0).

$f'(0) = 3(0)^2 - 1 = -1$

The derivative of the function at this point is negative, which tells us the function is decreasing at that point.

The tangent to the line is a straight line, so we will have a linear equation of the form y = mx + c. We know the gradient, m, is equal to -1, so

$y = -x + c$

Now we need to substitute a point on the tangent into this equation to find c. We know a point when x = 0 lies on here. To find the y-coordinate of this point we need to evaluate f(0).

$f(0) = (0)^3 - (0) - 1 = -1$

So the point (0, -1) lies on the tangent. Substituting into the tangent equation:

$y = -x + c \\\\ -1 = -(0) + c \\\\ -1 = c \\\\ \text{Equation of tangent is } \boxed{y = -x - 1}$

6 0

3 years ago