I don't understand this please help me ill really appreciate it :)

Plot and connect the points A(6,-7), B(1,-7), C(1,-4), D(3,-2), E(7,-2), F(7,-4), and find the length of DE

strojnjashka [21]

The length is 4 units...
They share the same value of y so the length would be | 7 - 3 | = 4

8 0

2 years ago

Write an algebraic expression that represents each word expression. You want to store an equal number of books on each of the 5

Butoxors [25]

Answer: $\dfrac{b}{5}$

Step-by-step explanation:

Given: Let b = Total number of books.

Total shelves = 5

Number of books on each shelf = (Total number of books) ÷ (Total shelves)

= b ÷ 5

$=\dfrac{b}{5}$

The required expression for the number of books on each shelf = $\dfrac{b}{5}$

3 0

2 years ago

How do you write 36,985?

Karo-lina-s [1.5K]

Thirty-six thousand, nine hundred eighty-five

6 0

2 years ago

I need to know the lengths of MN, NP, and PM

andrezito [222]

Answer:

MN = 12 units

NP = 16 units

PM = 20 units

Step-by-step explanation:

You can find MN and NP by counting on the graph. To find PM you will need to use the formula a^2 + b^2 = c^2 ( ^ stands for exponent)

MN and NP will be a and b.

Example: 12^2 + 16^2 = c^2

Simplify.

Example:

144 + 256 = c^2

144 + 256 = 400

Then find the square root of 400 which is 20.

To check your work put all of the sides into the formula and it should be true.

8 0

2 years ago

Read 2 more answers

A ball is thrown in the air from a platform that is 96 feet above ground level with an initial vertical velocity of 32 feet per

pishuonlain [190]

Answer:

y = -16 (x - 1)^2 + 112

The object lands on the ground in approximately 3.6s

Explanation:

The equation given is that of a parabola.

Now the maximum (local) point of a parabola is the vertex. Therefore, if we want to rewrite our function in the form that would be used to find the maximum height, then that form must be the vertex form of a parabola.

The vertex form of a parabola is

$y=a(t-h)^2+k$

where (h, k) is the vertex.

The only question is, what is the vertex for our function h(t)?

Remember that if we have an equation of the form

$y=ax^2+bx+c$

then the x-coordinate of the vertex is

$h=-\frac{b}{2a}$

Now in our case b = 32 and a = -16; therefore,

$h=\frac{-32}{2(16)}=1$

We've found the value of the x-coordinate of the vertex. What about the y-coordinate? To get the y-coordinate, we put x = 1 into h(t) and get

$k=-16(1)+32(1)+96=112$

Hence, the y-coordindate is k = 112.

Therefore, the vertex of the parabola is (1, 112).

With the coordinates of the vertex in hand, we now write the equation of the parabola in vertex form.

$h(t)=a(t-1)^2+112$

The only problem is that we don't know what the value of a is. How do we find a?

Note that the point (0, 96) lies on the parabola. In other words,

$h(0)=-16(0)^2+32(0)+96=96$

Therefore, the vertex form of the parabola must also contain the point (0, 96).

Putting in t = 0, h = 96 into the vertex form gives

$96=a(0-1)^2+112$ $96=a+112$

subtracting 112 from both sides gives

$a=-16$

With the value of a in hand, we can finally write the equation of the parabola on vertex form.

$\boxed{h\mleft(t\mright)=-16\left(t-1\right)^2+112.}$

Now when does the object hit the ground? In other words, for what value of t is h(t) = 0? To find out we just have to solve the following for t.

$h(t)=0.$

We could either use h(t) = -16t^2 + 32t + 96 or the h(t) = -16(t - 1)^2 + 112 for the above equation. But it turns out, the vertex form is more convenient.

Thus we solve,