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1 year ago

I neee help plss i have so lil time to hand it in

Mathematics

1 answer:

SashulF [63]1 year ago

5 0

The Solution.

The given equation is

$\begin{gathered} y=|x|+7 \\ \text{Where x=11} \end{gathered}$

Substituting 11 for x, we can get the value of y.

$\begin{gathered} y=|11|+7 \\ y=11+7 \\ y=18 \end{gathered}$

Hence, the value of y is 18.

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Beth wants to plant a garden at the back of her house. She has 32m of fencing. The area that can be enclosed is modelled by the

Maksim231197 [3]

Answer:

The maximum area that can be obtained by the garden is 128 meters squared.

Step-by-step explanation:

A represents area and we want to know the maximum.

$A(x)=-2x^2+32x$ is a parabola. To find the maximum of a parabola, you need to find it's vertex. The y-coordinate of the vertex will give us the maximum area.

To do this we will need to first find the x-coordinate of our vertex.

$x=\frac{-b}{2a}{/tex] will give us the x-coordinate of the vertex.Compare [tex]-2x^2+32x$ to $ax^2+bx+c$ then $a=-2,b=32,c=0[tex].So the x-coordinate is [tex]\frac{-(32)}{2(-2)}=\frac{-32}{-4}=8$ .

To find the y that corresponds use the equation that relates y and x.

$y=-2x^2+32x$

$y=-2(8)^2+32(8)$

$y=-2(64)+32(8)$

$y=-128+256$

$y=128$

The maximum area that can be obtained by the garden is 128 meters squared.

8 0

3 years ago

What is the range of the function f(x) = –2|x + 1|?

beks73 [17]

Answer:

Range $\rightarrow$ all real numbers less than or equal to 0 $\rightarrow$ ( - ∞ , 0 ]

Step-by-step explanation:

For visual understanding a graph of the function is attached with the answer.

For calculating the range of any modulus function you need to know that if modulus is there across any function then the output will be always positive.

For example: x has a range of ( - ∞ , + ∞ ) but |x| has a range of [ 0 , + ∞ ). Similarly range of |x + 1| is [ 0 , + ∞ ).

If you multiply the modulus function with a negative sign then the output will always be negative.

For example: Range of |x| is [ 0 , + ∞ ) but range of -|x| is ( - ∞ , 0 ]. Similarly range of -|x + 1| is ( - ∞ , 0 ]

Range in this case won't be affected on multiplying a positive constant with the modulus function.

Therefore the range of f(x) = -2|x + 1| will be ( - ∞ , 0 ].

(NOTE : [a,b] means all the numbers between 'a' and 'b' including 'a' and 'b'.

(a,b) means all the numbers between 'a' and 'b' excluding 'a' and 'b'.

(a,b] means all the numbers between 'a' and 'b' including only 'b' not 'a'.

[a,b) means all the numbers between 'a' and 'b' including only 'a' not 'b'.

{a,b} means only 'a' and 'b'.

{a,b] or (a,b} doesn't mean anything. )

6 0

3 years ago

Read 2 more answers

Pick correct graph from multiple choice options. A.B.C.D

Bezzdna [24]

Answer:

The answer would be C.

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

PLEASE ANSWER ASAP!!!!

Arlecino [84]

We'll use the notation $\textrm{Arcsin } x$ for the principal value and $\arcsin x$ for all the values:

$\arcsin x = \textrm{Arcsin } x + 2\pi k \textrm{ or } (\pi - \textrm{Arcsin }x) + 2\pi k \quad \textrm{ integer } k$

Part I

$\sin(\theta/2) = \frac 1 2$

$\theta/2 = \arcsin \frac 1 2$

Part II. In our range we can write

$\sin(\theta/2) = \sin(\frac \pi 6)$

$\theta/2 = \frac \pi 6 \textrm{ or } \theta /2 = \pi - \frac \pi 6 = \frac{5\pi}{6}$

Part III.

$\theta = \frac \pi 3 \textrm{ or } \theta = \frac{5\pi}{3}$

Part IV.

$\theta/2 = \frac \pi 6 + 2\pi k \textrm{ or } \theta = \frac{5 \pi}{6} + 2\pi k \quad \textrm{integer } k$

$\theta = \frac \pi 3 + 4 \pi k \textrm{ or } \theta = \frac{5 \pi}{3} + 4\pi k \quad \textrm{integer } k$

5 0

3 years ago

HELP ASAP

Alexxandr [17]

Answer: A − 9/7

Step-by-step explanation:

Hi, to answer this question we have to convert all the numbers into decimal form.

-3 1/3 = - (3x3+1)/3 =-10/3 = - 3.3334

-4/5 = -0.8

Since he number must greater than −3 1/3 but less than − 4/5.

−3 1/3 < x < − 4/5.

- 3.3334 < x < -0.8

The correct option is A = -9/7 , because in decimal form is equal to -1.28-

- 3.3334 < -1.28 < -0.8

5 0

3 years ago