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Veseljchak [2.6K]
3 years ago
7

Figure these out cuz I can’t

Mathematics
1 answer:
ziro4ka [17]3 years ago
5 0

Answer:

Step-by-step explanation:

1) You have to create an equation by combining like terms. In the first one, it shows that 5*x + 2 is the same as or equal to 17. So, 5x+2=17. Then, all you have to do it simplify. Move the apples and oranges over (the different terms) so you get 5x=17-2 (whenever you move the numbers across the = sign, it turns into the opposite). 5x=15. X=15/5. X=3

also, for some reason, I can’t view the other images. Is there a way you can put them all into one or something if you want more answers?

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Which figure is the image produced by applying the composition T 0,3 o R0,90 to figure R?
Lyrx [107]

We are given original image R.

It is being translated by (0,3) first and then rotated by a positive angle 90 degrees.

Translation by (0,3) represents (x,y) --> (x, y+3) rule.

Positive 90 degree rotation represents, counterclockwise rotation.

The rule for counterclockwise rotation is (x,y) --> (-y,x).

Therefore, final rule for would be

(x,y) --> ( -y,x+3 )

Let us take a coordinate of R on y-axis as (0,-4).

Now if we apply rule (x,y) --> ( -y,x+3 ) we get

(0,-4)  --> (-(-4), 0+3) = (4,3).

Let us check the figure with coordinate (4,3).

We can clearly see that Figure H has transformed coordinate (4,3).

<h3>Therefore, correct option is first option A. figure H.</h3>
8 0
3 years ago
2. A cake in the shape of a circus tent is used as a centerpiece at a celebration. The cake consists of a cylinder and a cone. T
SashulF [63]
1808.64

Hope this helped!
8 0
2 years ago
Read 2 more answers
Simplify this expression 3-(7a+1)
Vinil7 [7]
3 - (7a + 1) =
3 - 7a - 1 =
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5 0
3 years ago
Show that the following functions are probability density functions for some value of k and determine k. Then, determine the mea
lord [1]

Answer:

a) 17.5

b) 15.6

c) 13.3

d) 21.51

Step-by-step explanation:

The given function is equal to:

f(x)=kx^2

where

\int\limits^y_0 {kx^{2} } \, =1

where y=23

Clearing k=0.00025

a) Ex=\int\limits^y_0 {xf(x)} \, dx =\int\limits^y_0 {x*0.00025x^{2} } \, dx =17.5

b)Vx=Ex^{2} -(Ex)^{2} =\int\limits^y_0 {x^{2}f(x) } \, dx-17.5^{2}  =\int\limits^y_0 {x^{2} *0.00025x^{2} } \, dx -17.5^{2} =321.82-306.25=15.6

c) The function is equal to:

f(x)=k(1+2x)

\int\limits^y_0 {k(1+2x)} \, =1

where y=20

k=0.0024

Ex=\int\limits^y_0 {xf(x)} \, dx =\int\limits^y_0 {x*0.0024(1+2x)} \, dx =13.3

d) Vx=Ex^{2} -(Ex)^{2} =\int\limits^y_0 {x^{2} f(x)} \, dx -13.3^{2}=\int\limits^y_0 {x^{2} *0.0024(1+2x)} \, dx-13.3^{2}   =198.4-176.89=21.51

8 0
3 years ago
Hello,
balu736 [363]

Answer:

yesss it makes sense!!!!!!!!!!!

6 0
2 years ago
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