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4 years ago

8

If

1 answer:

Dovator [93]4 years ago

3 0

12^2=144

144 is divided by b which also is an odd interger, there only 2 numbers : 1 and 3

if b=3, a^2=48, then there will be no a available

if b=1 , a^2=144, then a is 12 and 12 can not be divided by 9

The answer is D

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Write a cosine function that has a midline of 3, an amplitude of 4 and a period<br> of pi.

sergejj [24]

Answer:

$f(x)=4\cos(2x)+3$

Step-by-step explanation:

Using the general equation $f(x)=a\cos(bx+c)+d$ , we already know that $a=4$ and $d=3$ , but we need to find $b$ from the period:

$\frac{2\pi}{|b|}=\pi\\ 2\pi=b\pi\\2=b$

Hence, the cosine function is $f(x)=4\cos(2x)+3$

8 0

2 years ago

The graphs below have the same shape. What is the equation of the blue graph?

postnew [5]

Answer:

<h2>C. G(x) = (x - 1)² - 3</h2>

Step-by-step explanation:

f(x) + n - shift the graph of f(x) n units up

f(x) - n - shift the graph of f(x) n units down

f(x - n) - shift the graph of f(x) n units to the right

f(x + n) - shift the graph of f(x) n units to the left

===================================

Look at the picture.

The graph of F(x) shifted 1 unit to the right and 3 units down.

Therefore the equation of the function G(x) is

$G(x)=(x-1)^2-3$

4 0

4 years ago

Find the distance between a point (– 2, 3 – 4) and its image on the plane x+y+z=3 measured across a line (x + 2)/3 = (2y + 3)/4

dimaraw [331]

Answer:

Distance of the point from its image = 8.56 units

Step-by-step explanation:

Given,

Co-ordinates of point is (-2, 3,-4)

Let's say

$x_1\ =\ -2$

$y_1\ =\ 3$

$z_1\ =\ -4$

Distance is measure across the line

$\dfrac{x+2}{3}\ =\ \dfrac{2y+3}{4}\ =\ \dfrac{3z+4}{5}$

So, we can write

$\dfrac{x-x_1+2}{3}\ =\ \dfrac{2(y-y_1)+3}{4}\ =\ \dfrac{3(z-z_1)+4}{5}\ =\ k$

$=>\ \dfrac{x-(-2)+2}{3}\ =\ \dfrac{2(y-3)+3}{4}\ =\ \dfrac{3(z-(-4))+4}{5}\ =\ k$

$=>\ \dfrac{x+4}{3}\ =\ \dfrac{2y-3}{4}\ =\ \dfrac{3z+16}{5}\ =\ k$

$=>\ x\ =\ 3k-4,\ y\ =\ \dfrac{4k+3}{2},\ z\ =\ \dfrac{5k-16}{3}$

Since, the equation of plane is given by

x+y+z=3

The point which intersect the point will satisfy the equation of plane.

So, we can write

$3k-4+\dfrac{4k+3}{2}+\dfrac{5k-16}{3}\ =\ 3$

$=>6(3k-4)+3(4k+3)+2(5k-16)\ =\ 18$

$=>18k-24+12k+9+10k-32\ =\ 18$

$=>\ k\ =\dfrac{13}{8}$

So,

$x\ =\ 3k-4$

$=\ 3\times \dfrac{13}{8}-4$

$=\ \dfrac{7}{4}$

$y\ =\ \dfrac{4k+3}{2}$

$=\ \dfrac{4\times \dfrac{13}{8}+3}{2}$

$=\ \dfrac{19}{4}$

$z\ =\ \dfrac{5k-16}{3}$

$=\ \dfrac{5\times \dfrac{13}{8}-16}{3}$

$=\ \dfrac{-21}{8}$

Now, the distance of point from the plane is given by,

$d\ =\ \sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2}$

$=\ \sqrt{(-2-\dfrac{7}{4})^2+(3-\dfrac{19}{4})^2+(-4+\dfrac{21}{8})^2}$

$=\ \sqrt{(\dfrac{-15}{4})^2+(\dfrac{-7}{4})^2+(\dfrac{9}{8})^2}$

$=\ \sqrt{\dfrac{225}{16}+\dfrac{49}{16}+\dfrac{81}{64}}$

$=\ \sqrt{\dfrac{1177}{64}}$

$=\ 4.28$

So, the distance of the point from its image can be given by,

D = 2d = 2 x 4.28

= 8.56 unit

So, the distance of a point from it's image is 8.56 units.

4 0

3 years ago

PLS ANSWER THIS SIMPLE MATH PROBLEM. (click on it to see the picture of it) and also pls dont troll I need a genuine answer. -CA

77julia77 [94]

Answer:

It’s either A or B

Step-by-step explanation:

Pretty sure it’s A though

4 0

3 years ago

Read 2 more answers

3 3/5 × 2 3/4 <br> Answer in mixed number

UkoKoshka [18]

Answer:

9 9/10

Step-by-step explanation:

I just used a calculator, sorry :(

anyways hope this helped! :)

6 0

3 years ago

Read 2 more answers