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3 years ago

11

Factor the following by grouping. 12r^2+30ry-2xr-5xy

1 answer:

Darya [45]3 years ago

5 0

I'm sorry, I can't show much working for this, I just pictured it. Let me know if you need to see working:
(6r - x)(2r + 5y)

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Would it be written as <br><br><br><br> ± a=11<br> or a= ±11 <br><br> help

nataly862011 [7]

A= +11 It’s the second one

8 0

2 years ago

Rzqust [24]

Answer:

It will be 1:2 because of the step by step

Step-by-step explanation:

1

_

8

lcm is 2x2=4

And when dividing it will be 4x2=8

So it is 1:2

8 0

3 years ago

Put the linear inequality into slope-intercept form.

Ugo [173]

Answer:

y ≤ $\frac{7}{4}x - 5$

Step-by-step explanation:

We are to put the linear inequality;

7x - 4y ≥ 20

in slope intercept form.

We can rearrange the inequality as;

-4y ≥ - 7x + 20

Dividing throughout by -4 we get;

y ≤ $\frac{7}{4} x$ - 5

Note that the inequality sign '≥' changes to '≤' since we divided all through with a negative number.

8 0

3 years ago

I WILL GIVE BRAINLIEST!!!! - YEAR 9 MATHS!!

Serhud [2]

Answer: one point only.

Step-by-step explanation:

Since the length of the sides of the square is 9 cm, taking a point outside the square which is equidistant from point B and C will never give exact 6 cm from A.

The only way to achieve this is by taking a equidistant point inside.

You can achieve exactly 6 cm from point A from point between B and C

But can only achieve a single point if it has to be equidistant from point B and C.

3 0

3 years ago

solve the given matrix equation for X. Simplify your answers as much as possible. (In the words of Albert Einstein, "Everything

vova2212 [387]

a.

$XA^{-1}=A^3$

$(XA^{-1})A=A^3A$

$X(A^{-1}A)=A^4$

$X=A^4$

b.

$AXB=(BA)^2$

$A^{-1}(AXB)B^{-1}=A^{-1}(BA)^2B^{-1}$

$(A^{-1}A)X(BB^{-1})=A^{-1}(BA)^2B^{-1}$

$X=A^{-1}(BA)^2B^{-1}$

c.

$(A^{-1}X)^{-1}=(AB^{-1})^{-1}(AB^2)$

$X^{-1}A=(BA^{-1})(AB^2)$

$X^{-1}A=B(A^{-1}A)B^2$

$X^{-1}A=B^3$

$(XX^{-1})A=XB^3$

$XB^3=A$

$X(B^3(B^3)^{-1})=A(B^3)^{-1}$

$X=A(B^3)^{-1}$

d. Not totally sure what the equation is supposed to be, but I guess it's

$ABXA^{-1}B^{-1}=A$

$ABX(BA)^{-1}=A$

$((AB)^{-1}(AB))X((BA)^{-1}(BA))=(AB)^{-1}A(BA)$

$X=(AB)^{-1}A(BA)$

$X=(B^{-1}A^{-1})A(BA)$

$X=B^{-1}(A^{-1}A)(BA)$

$X=(B^{-1}B)A$

$X=A$

7 0

3 years ago