Help me please ill give points and stuff

Goryan [66]

Hi friend,

This is a perfect square trinomial.

It can be recognised because it is of the form:

a^2−2ab+b^2=(a−b)^2

with a=3x and b=4

9x2−24x+16=(3x)2−(2⋅(3x)⋅4)+42

=(3x−4)2

6 0

3 years ago

Thank you guys so much

jeka57 [31]

Answer:

B. City P is above sea level and City R is below sea level

Step-by-step explanation:

A. City R is negative meaning it is below sea level however City Q is 0 meaning it is at sea level, so this statement is false.

C. Once again, City P is positive meaning it is above sea level but City Q is 0 meaning it is at sea level, so this statement is false.

D. Like before, City P is above sea level and City Q is at sea level not below, so this is, once again, false.

6 0

3 years ago

You have several options to add to your brand new car: leather seats, heated seats, sunroof, tinted windows, and back up camera.

scZoUnD [109]

Answer:

The number of different ways the car can be built is 31 different ways

Step-by-step explanation:

The given options are'

1), Leather seats, 2) Heated seats, 3) Sunroof, 4) Tinted windows, 5), back up camera

Therefore, the number of different ways the car can be built, n, is given as follows;

n = ₅C₁ + ₅C₂ + ₅C₃ + ₅C₄ + ₅C₅

n = 5 + 10 + 10 + 5 + 1 = 31 different ways

The number of different ways the car can be built = n = 31 different ways

3 0

2 years ago

The polynomial P(x) = 2x^3 + mx^2-5 leaves the same remainder when divided by (x-1) or (2x + 3). Find the value of m and the rem

Zigmanuir [339]

Answer:

m=7

Remainder =4

If q=1 then r=3 or r=-1.

If q=2 then r=3.

They are probably looking for q=1 and r=3 because the other combinations were used earlier in the problem.

Step-by-step explanation:

Let's assume the remainders left when doing P divided by (x-1) and P divided by (2x+3) is R.

By remainder theorem we have that:

P(1)=R

P(-3/2)=R

$P(1)=2(1)^3+m(1)^2-5$

$=2+m-5=m-3$

$P(\frac{-3}{2})=2(\frac{-3}{2})^3+m(\frac{-3}{2})^2-5$

$=2(\frac{-27}{8})+m(\frac{9}{4})-5$

$=-\frac{27}{4}+\frac{9m}{4}-5$

$=\frac{-27+9m-20}{4}$

$=\frac{9m-47}{4}$

Both of these are equal to R.

$m-3=R$

$\frac{9m-47}{4}=R$

I'm going to substitute second R which is (9m-47)/4 in place of first R.

$m-3=\frac{9m-47}{4}$

Multiply both sides by 4:

$4(m-3)=9m-47$

Distribute:

$4m-12=9m-47$

Subtract 4m on both sides:

$-12=5m-47$

Add 47 on both sides:

$-12+47=5m$

Simplify left hand side:

$35=5m$

Divide both sides by 5:

$\frac{35}{5}=m$

$7=m$

So the value for m is 7.

$P(x)=2x^3+7x^2-5$

What is the remainder when dividing P by (x-1) or (2x+3)?

Well recall that we said m-3=R which means r=m-3=7-3=4.

So the remainder is 4 when dividing P by (x-1) or (2x+3).

Now P divided by (qx+r) will also give the same remainder R=4.

So by remainder theorem we have that P(-r/q)=4.

Let's plug this in:

$P(\frac{-r}{q})=2(\frac{-r}{q})^3+m(\frac{-r}{q})^2-5$

Let x=-r/q

This is equal to 4 so we have this equation:

$2u^3+7u^2-5=4$

Subtract 4 on both sides:

$2u^3+7u^2-9=0$

I see one obvious solution of 1.

I seen this because I see 2+7-9 is 0.

u=1 would do that.

Let's see if we can find any other real solutions.

Dividing:

1 | 2 7 0 -9

| 2 9 9

-----------------------

2 9 9 0

This gives us the quadratic equation to solve:

$2x^2+9x+9=0$

Compare this to $ax^2+bx+c=0$

$a=2$

$b=9$

$c=9$

Since the coefficient of $x^2$ is not 1, we have to find two numbers that multiply to be $ac$ and add up to be $b$ .

Those numbers are 6 and 3 because $6(3)=18=ac$ while $6+3=9=b$ .

So we are going to replace $bx$ or $9x$ with $6x+3x$ then factor by grouping:

$2x^2+6x+3x+9=0$

$(2x^2+6x)+(3x+9)=0$

$2x(x+3)+3(x+3)=0$

$(x+3)(2x+3)=0$

This means x+3=0 or 2x+3=0.

We need to solve both of these:

x+3=0

Subtract 3 on both sides:

x=-3

----

2x+3=0

Subtract 3 on both sides:

2x=-3

Divide both sides by 2:

x=-3/2

So the solutions to P(x)=4:

$x \in \{-3,\frac{-3}{2},1\}$

If x=-3 is a solution then (x+3) is a factor that you can divide P by to get remainder 4.

If x=-3/2 is a solution then (2x+3) is a factor that you can divide P by to get remainder 4.

If x=1 is a solution then (x-1) is a factor that you can divide P by to get remainder 4.

Compare (qx+r) to (x+3); we see one possibility for (q,r)=(1,3).

Compare (qx+r) to (2x+3); we see another possibility is (q,r)=(2,3).

Compare (qx+r) to (x-1); we see another possibility is (q,r)=(1,-1).

6 0

3 years ago

Find the highest common factor using prime factorization for 27y and 54y³<br>

amid [387]

Given:

27y and 54y³

To find:

The highest common factor (HCF) using prime factorization.

Solution:

We have,

27y and 54y³

Using the prime factorization, we get

$27y=3\cdot 3\cdot 3\cdot y$

$54y^3=2\cdot 3\cdot 3\cdot 3\cdot y\cdot y\cdot y$

Now, the common prime factors are 3, 3, 3 and y. So,

$HCF=3\cdot 3\cdot 3\cdot y$

$HCF=27y$

Therefore, the highest common factor of the given terms is 27y.

5 0

3 years ago