All categories

2 years ago

Bidmass Working out brackets

Mathematics

1 answer:

andreyandreev [35.5K]2 years ago

8 0

Answer:

27 hope it helps you,.................

You might be interested in

Click here please. The problems are in picture format

padilas [110]

First picture:

You got the first two right (good job!), so I'll just tackle the third one. If you have to compute $f(x-4)$ , it means that you have to substitute $x-4$ in place of $x$ . Differently from the first two points, this will generate a new function, rather than a specific value:

$f(x) = \sqrt{x+4}+2 \implies f(x-4) = \sqrt{(x-4)+4}+2 = \sqrt{x}+2$

Second picture:

Given the point $(x,y)$ highlighted in the picture, you can deduce that the base is $2x$ units long (since it spans from $(-x,0)$ to $(x,0)$ ) and the height is $y$ units long (because it spans from $(x,0)$ to $(x,y)$ ). So, the area of the rectangle is the multiplication between base and height:

$A = 2xy$

But we know that $y=f(x)=\sqrt{36-x^2}$ , so we have

$A = 2x\sqrt{36-x^2}$

The domain of this function is given by the domain of the square root: we want its argument to be non, negative, so we have

$36-x^2 \geq 0 \iff x^2 \leq 36 \iff -6 \leq x \leq 6$

But since the problem is symmetric, the answer is

$0 \leq x \leq 6$

You can only see the answer $0 < x < 6$ because, if you choose $x=0$ or $x=6$ , the rectangle degenerates to a segment, and your exercise doesn't like this scenario, apparently

4 0

3 years ago

**WILL MARK BRAINLIEST AND THANK YOU**

omeli [17]

A is correct because f(1)=3^1+1 so f(1)=4 and g(x)=4 so f(x)=g(x)

B is also correct because the ordered pair we get is (1,4) and this is in quadrant 1

C is incorrect

D is incorrect

E is incorrect

Ans: A and B are correct

5 0

3 years ago

1-116 1.2.6

ankoles [38]

Answer:

A. 30 adults; B. 24 not dolls

Step-by-step explanation:

$\begin{array}{rcl}\text{No. of adults} & = & \text{No. of visitors} \times \text{Fraction who are adults}\\& = & \text{100 visitors} \times \dfrac{\text{3 adults}}{\text{10 visitors}} \\& = & \textbf{30 adults}\\\end{array}$

If five-eighths of the prizes were dolls, then three-eighths of the prizes were not dolls.

$\begin{array}{rcl}\text{No. not dolls} & = & \text{No. of prizes} \times \text{Fraction not dolls}\\& = & \text{64 prizes} \times \dfrac{\text{3 not dolls}}{\text{8 prizes}} \\& = & \textbf{24 not dolls}\\\end{array}$