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Serggg [28]
3 years ago
15

What is the representation of the third element in an array called a?

Mathematics
1 answer:
tekilochka [14]3 years ago
5 0
Aluminum
It's the second element in 3A and the third in the third period.

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6–[(−3)+(−7)]–[(−1)–(−5)–(−8)] help me
stellarik [79]

Answer:

4

Copy it, paste it on a new tab, bada Bing Bada Boom, google, the thing that gets me past school.

5 0
3 years ago
What do the differences between the points (as shown on the graph) represent?
Mamont248 [21]
The slope of the graph

14-7/4-2=7/2=3.5
7 0
2 years ago
Complete the factorization of 3x2 – 5x + 2.
Ad libitum [116K]
Multiply the first coefficient (3) and the last term (2) to get 3*2 = 6

We need to find two numbers that

a) multiply to 6
AND
b) add to -5 (middle coefficient)

Such two numbers are -3 and -2
-3 times -2 = 6
-3 plus -2 = -5

So we can break up the -5x into -3x-2x and then factor by grouping

--------------

3x^2-5x+2

3x^2-3x-2x+2

(3x^2-3x)+(-2x+2)

3x(x-1)+(-2x+2)

3x(x-1)-2(x-1)

(3x-2)(x-1)

So the two factors are (3x-2) and (x-1)

The answers are A and C
8 0
4 years ago
Read 2 more answers
NEED ANSWERS PLEASE!!
joja [24]

(A)  4 sec the ball is in the air.

(B) Height of the ball = 49 ft.

(C) Yes, the ball is at its maximum height at 1.5 seconds.

Solution:

Given data:

h(t)=-16t^2+48t+64

Initial velocity = 48 ft/s

Height = 64 ft

(A)  -16t^2+48t+64=0

a = –16, b = 48, c = 64

We can solve it by using a quadratic formula,

$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

$\Rightarrow t=\frac{-48 \pm \sqrt{(48)^{2}-4 \times(-16)(64)}}{2(-16)}

$\Rightarrow t=\frac{-48 \pm \sqrt{2304+4096}}{-32}

$\Rightarrow t=\frac{-48 \pm \sqrt{6400}}{-32}

$\Rightarrow t=\frac{-48 \pm 80}{-32}

$\Rightarrow t=\frac{-48 + 80}{-32},\frac{-48 - 80}{-32}

$\Rightarrow t=-1,t=4

Time cannot be in negative. So neglect t = –1

t = 4 sec

Hence, 4 sec the ball is in the air.

(B) When t = 1.5 sec,

h(1.5)=-16(1.5)^2+48(1.5)+64

h(1.5) = 49 ft

(C) The maximum height occurs at the average of zeros.

Average = \frac{(-1+4)}{2}=1.5 sec

Yes, the ball is at its maximum height at 1.5 seconds.

8 0
4 years ago
For the circle whose diameter has endpoints of (-3, 6) and (5, -2), write the equation in standard form.
dsp73

Answer:

B

Step-by-step explanation:

We are given a circle whose diameter has endpoints (-3, 6) and (5, -2).

And we want to equation of the circle in standard form.

First, let's determine the center of the circle. Since we are given the diameter, the center will be the midpoint of the diameter. The midpoint is given by:

\displaystyle M=\Big(\frac{x_1+x_2}{2},\frac{y_1+y_1}{2}\Big)

By substitution:

\displaystyle M=\Big(\frac{(-3)+(5)}{2},\frac{(6)+(-2)}{2}\Big)

Evaluate:

\displaystyle M=(1, 2)

Thus, the center of our circle is (1, 2).

Next, we need to find the radius of our circle. We can use the distance formula to find the diameter, and then divide that by two to find the radius. The distance formula is given by:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2

Let (-3, 6) be (x₁, y₁) and (5, -2) be (x₂, y₂). Substitute:

d=\sqrt{(-3-5)^2+(6-(-2))^2}

Evaluate:

\begin{aligned} d&= \sqrt{(-8)^2+(8)^2}\\&=\sqrt{64+64}\\&=\sqrt{2(64)}\\&=8\sqrt{2}}\end{aligned}

Therefore, the radius will be:

\displaystyle r=\frac{8\sqrt{2}}{2}=4\sqrt{2}

The equation for a circle is given by:

(x-h)^2+(y-k)^2=r^2

Where (h, k) is the center.

By substituting everything in, we acquire:

(x-(1))^2+(y-(2))^2=(4\sqrt{2})^2

Simplify:

(x-1)^2+(y-2)^2=32

Therefore, our answer is B.

7 0
3 years ago
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