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3 years ago

What is the value of y in the equation 4y-2(1-y)=-44?

Mathematics

1 answer:

LUCKY_DIMON [66]3 years ago

7 0

Hi there!

$\large\boxed{y = -7}$

4y - 2(1 - y) = -44

Begin by distributing -2:

4y - 2(1) - 2(-y) = -44

Simplify. A negative times a negative equals positive.

4y - 2 + 2y = -44

Combine like terms:

6y - 2 = -44

Add 2 to both sides:

6y = -42

Divide both sides by 6:

y = -7

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ABC and EDC are straight lines. EA is parallel to DB. EC = 8.1 cm. DC = 5.4 cm. DB = 2.6 cm. (a) Work out the length of AE. cm (

harkovskaia [24]

By applying the knowledge of similar triangles, the lengths of AE and AB are:

a. $\mathbf{AE = 3.9 $ cm}\\\\$

b. $\mathbf{AB = 2.05 $ cm} \\\\$

See the image in the attachment for the referred diagram.

The two triangles, triangle AEC and triangle BDC are similar triangles.
Therefore, the ratio of the corresponding sides of triangles AEC and BDC will be the same.

This implies that:

AC/BC = EC/DC = AE/DB

Given:

$EC = 8.1 $ cm\\\\DC = 5.4 $ cm\\\\DB = 2.6 cm\\\\AC = 6.15 $ cm$

a. Find the length of AE:

EC/DC = AE/DB

Plug in the values

$\frac{8.1}{5.4} = \frac{AE}{2.6}$

Cross multiply

$5.4 \times AE = 8.1 \times 2.6\\\\5.4 \times AE = 21.06$

Divide both sides by 5.4

$AE = \frac{21.06}{5.4} = 3.9 $ cm$

b. Find the length of AB:

$AB = AC - BC$

AC = 6.15 cm

To find BC, use AC/BC = EC/DC.

Plug in the values

$\frac{6.15}{BC} = \frac{8.1}{5.4}$

Cross multiply

$BC \times 8.1 = 6.15 \times 5.4\\\\BC = \frac{6.15 \times 5.4}{8.1} \\\\BC = 4.1$

Thus:

$AB = AC - BC$

Substitute

$AB = 6.15 - 4.1\\\\AB = 2.05 $ cm$

Therefore, by applying the knowledge of similar triangles, the lengths of AE and AB are:

a. $\mathbf{AE = 3.9 $ cm}\\\\$

b. $\mathbf{AB = 2.05 $ cm} \\\\$

Learn more here:

brainly.com/question/14327552

3 0

3 years ago

Help!! please if can i need answer

leva [86]

The answer is C

I think it is

8 0

3 years ago

What is the average rate of change for h(t) between t=0 and t=2

nordsb [41]

Part a: Option D 18 feet per second

Part b: increasing

Solution:

Height $h(t)=-16 t^{2}+50 t+3$

Part a: To find the average rate of change for h(t) between t = 0 and t = 2.

Substitute t = 0 in h(t).

$h(0)=-16 (0)^{2}+50 (0)+3$

h(0) = 3

Substitute t = 2 in h(t).

$h(2)=-16 (2)^{2}+50 (2)+3$

h(2) = 39

Average rate of change formula:

$$=\frac{h(b)-h(a)}{b-a}$

Here, a = 0 and b = 2.

$$=\frac{h(2)-h(0)}{2-0}$

$$=\frac{39-3}{2}$

$$=\frac{36}{2}$

= 18

Average rate of change = 18 feet per second

Option D is the correct answer.

Part b:

This means height of the ball is increasing for 0 < x < 2.

6 0

3 years ago

Manuel incorrectly claimed that the surface area of the cylinder shown is about 190.0 in.2. Use a formula to find the surface ar

Kitty [74]

Answer:

25.9%

Step-by-step explanation:

Note: you did not provide any attached image detailing the dimension of the cylinder however, let us arrange some dimension for explanation purpose

let the dimension of the cylinder be

radius= 3in

height= 5 in

Step two

The expression for the total surface area of a cylinder is

T. S. A. =2πrh+2πr2

substitute

T. S. A. =2*3.142*3*5+2*3.142*3^2

T. S. A. =94.26+56.556

T. S. A. =150.816 in^2

Now let us find the error

%error= actual -expected/expected*100

%error= 190 -150.816/150.816*100

%error= 39.184/150.816*100

%error= 0.259*100

%error= 25.9%

Now, you can use the dimension you have for the cylinder to calculate the total surface area, from there, you can solve for the error by following the steps above

3 0

3 years ago

Given a focus of (4, 5) and directrix of y= -3 , find the equation of the parabola.

andrey2020 [161]

Check the picture below.

notice, the focus point is at 4,5 whilst the directrix line is at y = -3, below the focus point, meaning the parabola is vertical and opening upwards.

keeping in mind that the vertex is "p" distance from either of these fellows, then the vertex is half-way between both of them, notice in the picture, the distance from y = 5 to y = -3 is 8 units, half that is 4 units, thus the vertex 4 units from the focus or 4 units from the directrix, that puts it at (4,1), whilst "p" is 4, since the parabola is opening upwards, is a positive 4 then.

$\bf \textit{parabola vertex form with focus point distance}\\\\
\begin{array}{llll}
(y-{{ k}})^2=4{{ p}}(x-{{ h}})
\\\\
\boxed{(x-{{ h}})^2=4{{ p}}(y-{{ k}})}
\end{array}
\qquad 
\begin{array}{llll}
vertex\ ({{ h}},{{ k}})\\\\
{{ p}}=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}\\\\
-------------------------------$

$\bf \begin{cases}
h=4\\
k=1\\
p=4
\end{cases}\implies (x-4)^2=4(4)(y-1)\implies (x-4)^2=16(y-1)
\\\\\\
\cfrac{1}{16}(x-4)^2=y-1\implies \cfrac{1}{16}(x-4)^2+1=y$

8 0

3 years ago