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2 years ago

What is the simplified form? (4/7)^3 a. 343/64 b. 21,952 c. 407 d. 64/343

Mathematics

2 answers:

charle [14.2K]2 years ago

8 0

=(4/7)^3

=(4/7)*(4/7)*(4/7)

=4*4*4 / 7*7*7

=4^3 / 7^3

=64/343

Answer is D.

olga55 [171]2 years ago

8 0

$( \frac{4}{7} )^{3} = \frac{4^{3} }{ 7^{3}} = \frac{4*4*4}{7*7*7} = \frac{64}{343}$

The answer is D.

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Jack reads his book for 8 minutes before school and 10 minutes after school. He attends school Monday through Friday. Which equa

sweet-ann [11.9K]

Answer:

Step-by-step explanation:

Even though the options to choose from are not given in this question we can try and lay our hand on the most likely equation for the number of minutes Jack reads his book.

firstly on a daily Jack reads a total of = 8+10 = 18 mins

He attends school from Mon- fri = 5 days

Now on a weekly basis jack reads = 5*18

in other words, the equation is simply the number of days times the time spent to read his book per day

hence this is = 90 min or 1hr 30 mins

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3 years ago

Which expression is equivalent to x + x + x + 5 + 5 no mater what value substituted in for x

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Answer:

3x+10

Step-by-step explanation:

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2 years ago

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Solve for M in the formula S= C+M. <br> A. M= S+ C<br> B. M= S - C<br> C. M= SC<br> D. M= s/c

Effectus [21]

Answer:

im pretty sure its a

Step-by-step explanation:

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2 years ago

(-10x^2+2)+(-2x^2+9x+7)

Harlamova29_29 [7]

Answer:

- 12x² + 9x + 9

Step-by-step explanation:

(- 10x² + 2) + (- 2x² + 9x + 7 ) ← remove parenthesis

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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

2 years ago