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3 years ago

What is the product ? (4y-3)(2y^2+3y-5)

Mathematics

2 answers:

DIA [1.3K]3 years ago

5 0

Answer:

=8y^3+6y^2−29y+15

Step-by-step explanation:

(4y−3)(2y^2+3y−5)

=(4y+−3)(2y^2+3y+−5)

=(4y)(2y^2)+(4y)(3y)+(4y)(−5)+(−3)(2y^2)+(−3)(3y)+(−3)(−5)

=8y^3+12y^2−20^y−6y^2−9y+15

=8y^3+6y^2−29y+15

yuradex [85]3 years ago

4 0

Answer:

8y^3 + 6y^2 - 29y + 15

Step-by-step explanation:

8y^3 + 12y^2 - 20y - 6y^2 - 9y + 15

then

8y^3 + 6y^2 - 29y + 15 is the product

Best regards

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In spring 2014, faculty from the City University of New York (CUNY) reported on a randomized controlled trial they conducted. Th

Tems11 [23]

Answer:

Explanatory variable: Extra support by faculties

Response variable: whether community college students who have assess into elementary algebra could be successful if they were placed directly into college-level statistics.

Confounding Variable: Ability of students to get passing grades.

It is an experiment to check if the hypothesis is true.

Step-by-step explanation:

An explanatory variable is a type of independent variable. An explanatory variable is that variable that explains changes in another variable. It can be anything that might affect the response variable. The explanatory variable is used to predict or explain differences in the response variable. In an experimental or research study, the explanatory variable is the variable that is manipulated by the researcher.

From the information given,

Explanatory variable: Extra support by faculties.

The response variable is also known as the dependent/outcome variable. Here, its value can be predicted or its variation can be explained by the explanatory variable. In an experimental study, this is the outcome that is measured after manipulation of the explanatory variable

From the question above,

Response variable: whether community college students who have assess into elementary algebra could be successful if they were placed directly into college-level statistics.

A confounding variable is that outside influence that changes the effect of a dependent and independent variable. This extraneous influence is often used to influence the outcome of an experimental design. The Confounding Variable in the above is:

Confounding Variable: Ability of students to get passing grades.

It is an experiment to check if the hypothesis is true.

6 0

2 years ago

Quadrilateral ABCD is inscribed in circle 0 What is m<A

valentinak56 [21]

We know that
A quadrilateral is inscribed in a circle if and only if the opposite angles are supplementary. ( Inscribed Quadrilateral Theorem)
so
m∠B+m∠D=180°
2x+(3x-5)=180
5x=180+5
5x=185
x=185/5
x=37°

m∠A=x+5-----> 37+5------> 42°

the answer is
m∠A is 42°

6 0

3 years ago

A box with a rectangular base and open top must have a volume of 128 f t 3 . The length of the base is twice the width of base.

noname [10]

Answer:

$Width = 4ft$

$Height = 4ft$

$Length = 8ft$

Step-by-step explanation:

Given

$Volume = 128ft^3$

$L = 2W$

$Base\ Cost = \$9/ft^2$

$Sides\ Cost = \$6/ft^2$

Required

The dimension that minimizes the cost

The volume is:

$Volume = LWH$

This gives:

$128 = LWH$

Substitute $L = 2W$

$128 = 2W * WH$

$128 = 2W^2H$

Make H the subject

$H = \frac{128}{2W^2}$

$H = \frac{64}{W^2}$

The surface area is:

Area = Area of Bottom + Area of Sides

So, we have:

$A = LW + 2(WH + LH)$

The cost is:

$Cost = 9 * LW + 6 * 2(WH + LH)$

$Cost = 9 * LW + 12(WH + LH)$

$Cost = 9 * LW + 12H(W + L)$

Substitute: $H = \frac{64}{W^2}$ and $L = 2W$

$Cost =9*2W*W + 12 * \frac{64}{W^2}(W + 2W)$

$Cost =18W^2 + \frac{768}{W^2}*3W$

$Cost =18W^2 + \frac{2304}{W}$

To minimize the cost, we differentiate

$C' =2*18W + -1 * 2304W^{-2}$

Then set to 0

$2*18W + -1 * 2304W^{-2} =0$

$36W - 2304W^{-2} =0$

Rewrite as:

$36W = 2304W^{-2}$

Divide both sides by W

$36 = 2304W^{-3}$

Rewrite as:

$36 = \frac{2304}{W^3}$

Solve for $W^3$

$W^3 = \frac{2304}{36}$

$W^3 = 64$

Take cube roots

$W = 4$

Recall that:

$L = 2W$

$L = 2 * 4$

$L = 8$

$H = \frac{64}{W^2}$

$H = \frac{64}{4^2}$

$H = \frac{64}{16}$

$H = 4$

Hence, the dimension that minimizes the cost is:

$Width = 4ft$

$Height = 4ft$

$Length = 8ft$

8 0

2 years ago

Which set of parametric equations over the interval 0 ≤ t ≤ 1 defines a line segment with initial point (–5, 3) and terminal poi

Ksivusya [100]

Given:

A line segment with initial point (–5, 3) and terminal point (1, –6).

To find:

The set of parametric equations over the interval 0 ≤ t ≤ 1 which defines the given line segment.

Solution:

Initial point is (–5, 3). So,

$x(0)=-5,y(0)=3$

Terminal point is (1, –6).

$x(1)=1,y(1)=-6$

Check which of the given set of parametric equations satisfy $x(0)=-5,y(0)=3,x(1)=1,y(1)=-6$ .

Put t=1 in each set of parametric equations.

In option A,

$y(1)=3-6(1)=3-6=-3\neq -6$

So, option A is incorrect.

In option B,

$y(1)=1-6(1)=1-6=-5\neq -6$

So, option B is incorrect.

In option C,

$y(1)=3-9(1)=3-9=-6$

$x(1)=-5+6(1)=-5+6=1$

Put t=0, in this set of parametric equations.

$x(0)=-5+6(0)=-5$

$y(0)=3-9(0)=3$

So, option C is correct.

In option D,

$y(1)=1-7(1)=1-7=-3\neq -6$

$x(1)=-5+8(1)=-5+8=3\neq 1$

So, option D is incorrect.

8 0

3 years ago

If a positive integer that is not a multiple of 5 is divided by 5, what is the least possible remainder

solniwko [45]

I would say 1 would be the least possible remainder...

11/5 = 2 remainder 1
21/5 = 4 remainder 1
31/5 = 3 remainder 1

and since ur dealing with positive integers (whole numbers), 1 is gonna be the smallest

8 0

3 years ago