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3 years ago

The equation of a straight line is 2y+6x=-8.find the gradient of the line

Mathematics

1 answer:

Illusion [34]3 years ago

7 0

Step-by-step explanation:

Y= mx + c

2y = 3x - 8

2y = -3× - 8 ( divide all yerms by two )

y = -3 × 4

y = -3

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For an independent-measures research study (i.e., independent measures t Test), the data show an 8-point difference between the

zloy xaker [14]

Answer:

C. 2

Step-by-step explanation:

Cohen's d is a parameter used to express the standardised difference between two means. It is defined as the difference between the means divided by the pooled standard deviation.

In this case, the difference between both means (M2-M1) is 8. As for the pooled standard deviation, simply take the square root of the given pooled variance:

$S=\sqrt{S^2}\\S=\sqrt{16} \\S= 4$

Therefore, the value of Cohen's d (d) is:

$d=\frac{M_{2} -M_{1}}{S} \\d=\frac{8}{4}\\d=2$

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3 years ago

David measures how many inches his tomato plant grows every week. What is the independent variable?

Norma-Jean [14]

Answer:

It is tomato plant

Step-by-step explanation:

- The independent variable is the variable the experimenter changes or controls and is assumed to have a direct effect on the dependent variable. - - - The dependent variable is the variable being tested and measured in an experiment, and is 'dependent' on the independent variable.

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3 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:

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

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1 year ago

Two pipes are connected to a tank. when working together, the two pipes can fill the tank in 4 hours. the larger pipe, working a

ziro4ka [17]

Ten hours maybe?????

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3 years ago

What do we do to both sides of the equation? X-8=3

Alik [6]

Answer:

$\huge\fbox\red{a}\huge\fbox\orange{n} \huge\fbox\pink{s}\huge\fbox\green{w} \huge\fbox\blue{e}\huge\fbox\purple{r}\$