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3 years ago

Find the magnitude of AB. A(-2, 6), B(1, 10) O A 2 OB. ✔️15 O C.5 OD ✔️2

Mathematics

1 answer:

natita [175]3 years ago

6 0

Answer:

C. 5

Step-by-step explanation:

Use the Distance Formula.

Substitute the values of x1 , y1 , x2 , and y2 .

|AB|² =|(1--2)²+(10-6)²|

|AB|² = |9+16|

|AB| = √ 25

|AB| =5

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will deposited $1550 in an account that pays 8% annually. How much would be in the account at the end of 24 months?

nikdorinn [45]

Answer:

$1798

Step-by-step explanation:

You take the money of 1550 and multiply the 16% for the 2 years, which is just 1.16 and you get 1798

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2 years ago

Read 2 more answers

Evaluate the difference quotient for the given function.

kherson [118]

Assuming you mean f(t) = g(t) × h(t), notice that

f(t) = g(t) × h(t) = cos(t) sin(t) = 1/2 sin(2t)

Then the difference quotient of f is

$\dfrac{\frac12 \sin(2(t+h)) - \frac12 \sin(2t)}h = \dfrac{\sin(2t+2h) - \sin(2t)}{2h}$

Recall the angle sum identity for sine:

sin(x + y) = sin(x) cos(y) + cos(x) sin(y)

Then we can write the difference quotient as

$\dfrac{\sin(2t)\cos(2h) + \cos(2t)\sin(2h) - \sin(2t)}{2h}$

$\boxed{\sin(2t)\dfrac{\cos(2h)-1}{2h} + \cos(2t)\dfrac{\sin(2h)}{2h}}$

(As a bonus, notice that as h approaches 0, we have (cos(2h) - 1)/(2h) → 0 and sin(2h)/(2h) → 1, so we recover the derivative of f(t) as cos(2t).)

7 0

2 years ago

Can somebody please help me?

9966 [12]

28+28+28= 84

28 x 28 x 28 = 21952

hope it helps!:)

4 0

2 years ago

Determine the value for x

Grace [21]

Answer:

5 im pretty sure

Step-by-step explanation:

if im right, can you mark me brainliest?

3 0

2 years ago

The graph of $y = ax^2 + bx + c$ is shown below. Find $a \cdot b \cdot c$. (The distance between the grid lines is one unit.)

Savatey [412]

Answer:

$a\cdot b\cdot c=\frac{15}{4}$

Step-by-step explanation:

Vertex is the minimum or maximum point of parabola

Vertex of parabola is (h,k)

Therefore, from given graph (-3,-2) is the lowest point.

Vertex of parabola is at (-3,-2).

Standard equation of parabola

$y-k=a(x-h)^2$

Substitute the values

$y-(-2)=a(x-(-3))^2=a(x+3)^2$

$y+2=a(x+3)^2$

(-1,0) lies on the parabola.

Therefore, it satisfied the equation of parabola.

$0+2=a(-1+3)^2=4a$

$a=2/4=1/2$

Now, using the value of a

$y+2=1/2(x+3)^2=1/2(x^2+6x+9)$

$y+2=\frac{1}{2}x^2+3x+\frac{9}{2}$

$y=\frac{1}{2}x^2+3x+\frac{9}{2}-2$

$y=\frac{1}{2}x^2+3x+\frac{9-4}{2}$

$y=\frac{1}{2}x^2+3x+\frac{5}{2}$

By comparing with

$y=ax^2+bx+c$

We get

$a=\frac{1}{2}, b=3, c=5/2$

$a\cdot b\cdot c=\frac{1}{2}\times 3\times \frac{5}{2}$

$a\cdot b\cdot c=\frac{15}{4}$

7 0

2 years ago