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2 years ago

What is the factor in this equation? 3+(4r + y)

Mathematics

1 answer:

zlopas [31]2 years ago

8 0

The factors are 3 and (4r+y)

Hope this helps :)

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What is 3,000 is 1/10 of

BartSMP [9]

Answer:

30000

Step-by-step explanation:

3000 / (1/10) = w

3000*10/1 = w

30000 = w

probe:

30000*1/10 = 3000

then:

30000

is 1/10

30000

4 0

2 years ago

a line is perpendicular to y=-4x-2 and intersects the point (0,9). what is the equation for this perpendicular line?

AysviL [449]

Answer:

y=(1/4)x + 9 OR y=x/4 +9

Step-by-step explanation:

Perpendicular lines have a slope/gradient of -1/n

Because the slope of this line is -4, the slope of a line perpendicular to it would be 1/4.

y=(1/4)x+b

(0,9) is the y-intercept, so b = 9

y=(1/4)x + 9 OR y=x/4 +9

3 0

3 years ago

Sur une photographie réalisée avec un

shtirl [24]

Answer:

0.3cm ?

Step-by-step explanation:

3 0

2 years ago

Divide the rational expressions and express in simplest form. When typing your answer for the numerator and denominator be sure

Veseljchak [2.6K]

Dividing by a fraction is equivalent to multiply by its reciprocal, then:

$\begin{gathered} \frac{3y^2-7y-6}{2y^2-3y-9}\div\frac{y^2+y-2}{2y^2+y-3^{}}= \\ =\frac{3y^2-7y-6}{2y^2-3y-9}\cdot\frac{2y^2+y-3}{y^2+y-2}= \\ =\frac{(3y^2-7y-6)(2y^2+y-3)}{(2y^2-3y-9)(y^2+y-2)} \end{gathered}$

Now, we need to express the quadratic polynomials using their roots, as follows:

$ay^2+by+c=a(y-y_1)(y-y_2)$

where y1 and y2 are the roots.

Applying the quadratic formula to the first polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{7\pm\sqrt[]{(-7)^2-4\cdot3\cdot(-6)}}{2\cdot3} \\ y_{1,2}=\frac{7\pm\sqrt[]{121}}{6} \\ y_1=\frac{7+11}{6}=3 \\ y_2=\frac{7-11}{6}=-\frac{2}{3} \end{gathered}$

Applying the quadratic formula to the second polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot2\cdot(-3)}}{2\cdot2} \\ y_{1,2}=\frac{-1\pm\sqrt[]{25}}{4} \\ y_1=\frac{-1+5}{4}=1 \\ y_2=\frac{-1-5}{4}=-\frac{3}{2} \end{gathered}$

Applying the quadratic formula to the third polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{3\pm\sqrt[]{(-3)^2-4\cdot2\cdot(-9)}}{2\cdot2} \\ y_{1,2}=\frac{3\pm\sqrt[]{81}}{4} \\ y_1=\frac{3+9}{4}=3 \\ y_2=\frac{3-9}{4}=-\frac{3}{2} \end{gathered}$

Applying the quadratic formula to the fourth polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot1\cdot(-2)}}{2\cdot1} \\ y_{1,2}=\frac{-1\pm\sqrt[]{9}}{2} \\ y_1=\frac{-1+3}{2}=1 \\ y_2=\frac{-1-3}{2}=-2 \end{gathered}$

Substituting into the rational expression and simplifying:

$\begin{gathered} \frac{3(y-3)(y+\frac{2}{3})2(y-1)(y+\frac{3}{2})}{2(y-3)(y+\frac{3}{2})(y-1)(y+2)}= \\ =\frac{3(y+\frac{2}{3})}{2(y+2)}= \\ =\frac{3y+2}{2y+4} \end{gathered}$

8 0

8 months ago

What is the value of x?

bekas [8.4K]

Answer:

I can not see the picture.

Step-by-step explanation:

5 0

2 years ago