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Nezavi [6.7K]
3 years ago
15

Factorise the following (x-y) (x²+4xy+10y²)​

Mathematics
1 answer:
Dmitrij [34]3 years ago
5 0

Answer:

=x3+3x2y+6xy2−10y3

Step-by-step explanation:

(x−y)(x2+4xy+10y2)

=(x+−y)(x2+4xy+10y2)

=(x)(x2)+(x)(4xy)+(x)(10y2)+(−y)(x2)+(−y)(4xy)+(−y)(10y2)

=x3+4x2y+10xy2−x2y−4xy2−10y3

A:=x3+3x2y+6xy2−10y3

Hope this helps!

:)

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Find the missing angle measure.
ycow [4]

Step-by-step explanation:

The sum of all inner angles in the shape should be 540°

(180° for triangles, 360° for squares and other simple 4-corner-shapes, the pattern is the number of corners minus 2 multiplied by 180°)

we can calculate

540-106-94-135=205

so we got 205 degrees for the two unclear corners and one of them has to be 5° greater.

x is 100°

x is 100°x+5 is 105°

(note that in the subtraction part we could have subtracted 5 more and would be left with 2x=200)

5 0
4 years ago
A plane can fly 1440 miles in the same time as it takes a car to go 400 miles. If the car travels 130 mph
patriot [66]

Answer:

The speed will be "144 mph".

Step-by-step explanation:

Let the speed of the plane be "x mph".

then,

The car's speed be "x-130 mph".

According to the question,

⇒ \frac{1440}{x} =\frac{400}{x-130}

By applying cross-multiplication, we get

⇒    1440(x-130)=400x

⇒ 1440x-187200=400x

⇒    1440x-140x=187200

⇒                1300x=187200

⇒                       x=\frac{187200}{1300}

⇒                          =144 \ mph

8 0
3 years ago
Find the equation of a line through the points (-1, 3) and (2,-5). show all work please!​
masha68 [24]
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3 0
3 years ago
How many zero pairs can be created in the model below?
Kryger [21]

Answer:

3

Step-by-step explanation:

Zero pairs are +/- pairs which add to 0. They are inverse of each other.

+x and -x

+1 and -1

+1 and -1

3 0
4 years ago
Read 2 more answers
write slope intercept form of equation for the line passing through the given points. and equation (-2,8) and (1,-3)
tresset_1 [31]

\bf (\stackrel{x_1}{-2}~,~\stackrel{y_1}{8})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{-3}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-3}-\stackrel{y1}{8}}}{\underset{run} {\underset{x_2}{1}-\underset{x_1}{(-2)}}}\implies \cfrac{-11}{1+2}\implies -\cfrac{11}{3}

\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{8}=\stackrel{m}{-\cfrac{11}{3}}[x-\stackrel{x_1}{(-2)}]\implies y-8=-\cfrac{11}{3}(x+2) \\\\\\ y-8=-\cfrac{11}{3}x-\cfrac{22}{3}\implies y=-\cfrac{11}{3}x-\cfrac{22}{3}+8\implies y=-\cfrac{11}{3}x+\cfrac{2}{3}

6 0
3 years ago
Read 2 more answers
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