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almond37 [142]
4 years ago
10

35 points please help thank you❤️❤️

Mathematics
1 answer:
AveGali [126]4 years ago
8 0

Answer:

I'm not sure sorry


Step-by-step explanation:


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Please help. I need to know the strategy, the correct work, and to identify the error.
zysi [14]

Answer:

See below

Step-by-step explanation:

13x + 5 + 17x - 4.5 + x

NOTE: You can only add/subtract LIKE terms . Here we have  variable terms in x  (13x , 17x and x) and 2 constants ( 5 and 4.5).

Sarah's next line is

18x + 17x - 4.5 + x

This is incorrect . It looks like she  added 13x + 5 and got 18x!! Which is of course wrong because they are NOT LIKE terms. 5 is a constant and 13x is a variable term in x.

The third line is correct:-

35x - 4.5 + x.

She has added the 2 variables 17x and 18x.

Fourth line  is  30.5x + x. She's at it again - Trying to subtract 4.5 from 35x .

Last line 31.5x

She has added  the 2 terms in x correctly, but the final answer is wrong of course due to the earlier errors.


7 0
3 years ago
Find the area of a triangle whose vertices are (0,0), (4,2), (-1,2)​
Yakvenalex [24]

Answer:

Therefore area of a triangle whose vertices are (0,0), (4,2), (-1,2) is

5 units².

Step-by-step explanation:

Given:

Let the vertices be,

point A( x₁ , y₁) ≡ ( 0 , 0)

point B( x₂ , y₂) ≡ (4 , 2)

point C(x₃ , y₃ ) ≡ (-1 , 2)

To Find:

Area of Triangle = ?

Solution:

If the Vertices A( x₁ , y₁), B( x₂ , y₂) and C(x₃ , y₃ ) then the Area of Triangle is given by

\textrm{Area of Triangle}=\dfrac{1}{2}(x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2}))

Substituting the values we get

\textrm{Area of Triangle}=\dfrac{1}{2}(0(2-2)+4(2-0)+(-1)(0-2))

\textrm{Area of Triangle}=\dfrac{1}{2}(0+8+2))=\dfrac{10}{2}=5\ units^{2}

Therefore area of a triangle whose vertices are (0,0), (4,2), (-1,2) is

5 units².

7 0
3 years ago
Helppppppppp meeeeeeeee
zavuch27 [327]
Dividing two fractions is the same as multiplying the first fraction by the reciprocal of the second fraction. remember KFC: keep, change, flip.

a/b ÷ c/d *use KFC* = a*d/b*c
=ad/ bc

2/7 ÷ 4/5 = 2/7*5/4
=10/28
(simplify)
=5/14
5 0
3 years ago
All of these methods can be used to show that two triangles are similar EXCEPT
kvasek [131]
The answer is probably ss or SSA depending on your answer choices, if one of these isnt an answer choice let me know
7 0
3 years ago
Read 2 more answers
Find the volume of the solid obtained by rotating the region bounded by y=4x and y=2sqrt(x) about the line x=6.
Pie

Check the picture below.

so by graphing those two, we get that little section in gray as you see there, now, x = 6  is a vertical line, so we'll have to put the equations in y-terms and this is a washer, so we'll use the washer method.

y=4x\implies \cfrac{y}{4}=x\qquad \qquad y=2\sqrt{x}\implies \cfrac{y^2}{4}=x~\hfill \begin{cases} \cfrac{y}{4}=x\\\\ \cfrac{y^2}{4}=x \end{cases}

the way I get the radii is by using the "area under the curve" way, namely, I use it to get R² once and again to get r² and using each time the axis of rotation as one of my functions, in this case the axis of rotation will be f(x), and to get R² will use the "farthest from the axis of rotation" radius, and for r² the "closest to the axis of rotation".

\stackrel{R}{\stackrel{f(x)}{6}-\stackrel{g(x)}{\cfrac{y^2}{4}}}\qquad \qquad \stackrel{r}{\stackrel{f(x)}{6}-\stackrel{g(x)}{\cfrac{y}{4}}}~\hfill \stackrel{R^2}{\left( 6-\cfrac{y^2}{4} \right)^2}-\stackrel{r^2}{\left( 6-\cfrac{y}{4} \right)^2} \\\\\\ \stackrel{\textit{doing a binomial expansion and simplification}}{3y-3y^2-\cfrac{y^2}{16}+\cfrac{y^4}{16}}

now, both lines if do an equation on where they meet or where one equals the other, we'd get the values for y = 0 and y = 1, not surprisingly in the picture.

\displaystyle\pi \int_0^1\left( 3y-3y^2-\cfrac{y^2}{16}+\cfrac{y^4}{16} \right)dy\implies \pi \left( \left. \cfrac{3y^2}{2} \right]_0^1-\left. y^3\cfrac{}{} \right]_0^1-\left. \cfrac{y^3}{48}\right]_0^1+\left. \cfrac{y^5}{80} \right]_0^1 \right) \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \cfrac{59\pi }{120}~\hfill

7 0
2 years ago
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