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Svetllana [295]

3 years ago

10

IF you want your questions answered faster yall should offer more then 5 points for a difficult question anyway.. solve 2.5x-10.

5=64(0.5*) for x

1 answer:

svetoff [14.1K]3 years ago

6 0

This means that the answer is 2

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Antonius the chef wants to cook several batches of lasagna and spaghetti for his restaurant. He plans to use at least 4.54.5 kil

yaroslaw [1]

Answer: No he does not meet both of his expectation by cooking 10 batches of spaghetti and 4 batches of lasagna.

Step-by-step explanation:

Since here S represents the number of batches of spaghetti and L represents the total number of lasagna.

And, the chef planed to use at least 4.5 kilograms of pasta and more than 6.3 liters of sauce to cook spaghetti and lasagna.

Which is shown by the below inequality,

$0.3S+0.65L \geq 4.5$ ----------(1)

And, $0.25S+0.8L >6.3$ --------(2)

By putting S = 10 and L = 4 in the inequality (1),

$0.3\times 10+0.65\times 4 \geq 4.5$

⇒ $5.6\geq 4.5$ (true)

Thus, for the values S = 10 and L = 4 the inequality (1) is followed.

Again By putting S = 10 and L = 4 in the inequality (2),

$0.25\times 10+0.8\times 4 >6.3$

⇒ $5.7>6.3$ ( false)

But, for the values S = 10 and L = 4 the inequality (2) is not followed.

Therefore, Antonius does not meet both of his expectations by cooking 10 batches of spaghetti and 4 batches of lasagna.

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M(x) = 5x + 4 n(x) = 6x - 9

Part 1 (m + n)(x) = 5x + 4 + 6x - 9
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Part 2 (m * n)(x) = (5x + 4)(6x - 9) = 30x^2 - 21x - 36

Part 3 m[n(x)] = 5(6x - 9) + 4
= 30x - 45 + 4
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Step-by-step explanation:

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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).

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3 years ago