What is the total cost of a $720 ipad that is on sale at 15% off if the local sales tax is 6%

I NEED EXPERT..ILL GIVE BRAINLEST

sashaice [31]

I’m here to help what’s the problem?

3 0

2 years ago

Can you solve this? I need help. I don't understand.

ZanzabumX [31]

Answer:

2 sqrt(5) OR 4.5

Step-by-step explanation:

You have to know Pythagorean theorem to solve this question.

a^2 + b^2 = c^2

To use this theorem you have to have a right triangle. There are two right triangles in your image. The lower (larger) one has two sides labeled, so you can use Pythagorean thm to find the third side. There's a short cut, bc some right triangles have easy-to-memorize lengths of the sides. 3-4-5 is one of these number sets. A multiple of this is 6-8-10. We could've solved:

b^2 + 8^2 = 10^2

But it would've come out the same. The unlabeled side is 6.

We can use the 6 and the 4 on the smaller right triangle and use the Pythagorean thm again to solve for x.

4^2 + x^2 = 6^2

16 + x^2 = 36 subtract 16 from both sides.

x^2 = 20

Take the square root of both sides.

sqrt (x^2) = sqrt 20

x = 2 sqrt(5) which is approximately 4.472.

2 sqrt(5) is an exact answer if that is what they are asking for. 4.472 is an approximation to the nearest thousandth. It would be 4.47 to the nearest hundredth or 4.5 to the nearest tenth.

8 0

2 years ago

What does supplementary mean and how do you know if angles are supplementary

IRISSAK [1]

Answer:

Supplementary angles are those angles that sum up to 180 degrees. For example, angle 130° and angle 50° are supplementary angles because sum of 130° and 50° is equal to 180°.

Step-by-step explanation:

6 0

2 years ago

Solve 3 [-x + (2 x + 1)] = x - 1.<br> X=

Anna35 [415]

Answer:

-2

Step-by-step explanation:

1. Distribute: -3x + 6x + 3 = x-1

2. Combine like terms: 3x + 3 = x-1

3. Subtract 3 on both sides: 3x = x-4

4. Subtract x on both sides: 2x = -4

5. Divide each side by 2: x = -2

6. Check by plugging in -2 to the equation:

4 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